The slope of the graph current vs voltage can be explained with the next

[tex]\begin{gathered} V=IR \\ I=\frac{V}{R} \end{gathered}[/tex]In this case, current vs voltage means that current is y axis, and voltage is x axis. Based n that, the equation is the second one

[tex]I=\frac{V}{R}[/tex]And the slope is the **inverse of the resistance.** Dont forget that current - voltage is different to voltage - current

Diffraction is:A.the difference in density of the compression and rarefaction parts of a sound wave.B.the change of frequency heard by an observer when sound waves come from a moving source.C.when waves suddenly appear in a medium without a source.D.the apparent bending of sound waves around obstacles.

Diffraction is a phenomenon that occurs when a wave hits an object or it passes through a small gap.

The propagation of the wave will have a circular pattern after the diffraction effect:

Therefore **the correct option is D.**

Students were experimenting with objects in a collision. Ball A moves with a constant acceleration by sliding down a frictionless incline plane before colliding with Ball B. Students used motion detectors to measure the velocity of Ball A at various points. Ball A (mass of 1.00 kg) began from rest, at position 1, on a ramp at a height of 1.25 m. At position 2, ball A was moving at 5.00 m/s. Ball A continues to roll at a constant 5.00 m/s when it collides with Ball B (mass of 1.00 kg) at position 3. Ball A comes to a complete stop. Ball B moves at a constant velocity after the collision.FORMULAS: PE = m•g•h (g=9.8m/s²) KE = 1/2•m•v² momentum = m•v

**Given data:**

* The mass of ball A is m_1 = 1 kg.

* The mass of ball B is m_2 = 1 kg.

* The initial velocity of ball A is u_1 = 5 m/s.

* The final velocity of ball A is v_1 = 0 m/s.

* The initial velocity of the ball B is u_2 = 0 m/s.

**Solution:**

According to the law of conservation of momentum, the net momentum of the system before the collision is equal to the net momentum of the system after the collision.

Thus,

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Substituting the known values,

[tex]\begin{gathered} 1\times5+1\times0=1\times0+p_2 \\ 5+0=0+p_2 \\ p_2=5\text{ kgm/s} \end{gathered}[/tex]where p_2 is the momentum of the ball B,

**Thus, the momentum of ball B after the collision is 5 kgm/s.**

**Hence, the third option is the correct answer.**

n Fig. P9.82, the cylinder Figure P9.82 and pulley turn without friction about stationary horizontal axles that pass through their centers. A light rope is Pulley wrapped around the cylinder, passes over the pulley, and has a 3.00 kg box Cylinder Box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 kg and radius 40.0 cm. The pulley is a uniform disk with mass 2.00 kg and radius 20.0 cm. The box is released from rest and descends as the rope unwraps from the cylinder. Find the speed of the box when it has fallen 2.50 m.

The **speed** of the box when it has fallen 2.50 m is **4.22 m/s.**

The **speed** of the box when it has fallen through the given height is calculated as follows;

Apply the principle of** conservation of energy** to determine the speed of the box.

ΔK.E = ΔP.E

K.Ef - K.Ei = mg(hf - hi)

K.Ef - 0 = mg(hf - 0)

K.Ef = mghf

where;

K.E is the final kinetic energy = rotational + translational kinetic energyhf is the final height of the box¹/₂mv² + ¹/₂I_pω² + ¹/₂I_cω²= mghf

¹/₂mv² + ¹/₂(I_p + I_c)ω² = mghf

where;

I_p is moment of inertia of the pulleyI_c is the moment of inertia of the cylinderω is the angular speed of the boxm is mass of the boxI_p = ¹/₂MR²

where;

M is mass of the pulleyR is the radius of the pulleyI_p = ¹/₂(2)(0.2)² = 0.04 kgm²

I_c = MR²

I_c = (5)(0.4)²

I_c = 0.8 kgm²

¹/₂mv² + ¹/₂(I_p + I_c)(v/r)² = mghf

¹/₂mv² + ¹/₂r²(I_p + I_c)v² = mghf

¹/₂v²[m + 1/r²(I_p + I_c)] = mghf

v²[m + 1/r²(I_p + I_c)] = 2mghf

v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)

v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)

8.25v² = 147

v² = 147/8.25

v² = 17.8

v = √17.8

v = 4.22 m/s

Learn more about** speed of box around pulley** here: https://brainly.com/question/13896956

#SPJ1

A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°. How far is the ball from the football player when it lands? How much farther would the ball go if he kicked it with the same speed, but at a 45° angle? Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?

A football player kicks a football off a tee with a speed of **16 m/s** at an angle of **63°**

The horizontal and vertical speed of the ball is given by

[tex]\begin{gathered} v_x=v\cos (\theta) \\ v_y=v\sin (\theta) \end{gathered}[/tex]Where v = 16 m/s and θ = 63°

[tex]\begin{gathered} v_x=16\cos (63\degree)=7.26\; \frac{m}{s} \\ v_y=16\sin (63\degree)=14.26\; \frac{m}{s} \end{gathered}[/tex]How far is the ball from the football player when it lands?

The range of the ball is given by

[tex]x=v_x\times t[/tex]Where **t** is the time the ball remains in the air.

The time (t) can be found as

[tex]y=v_yt+\frac{1}{2}at^2[/tex]y = 0 when the ball is in the air.

The acceleration is due to gravity (-9.8 m/s²)

[tex]\begin{gathered} 0=14.26t+\frac{1}{2}(-9.8)t^2 \\ 0=14.26t-4.9t^2 \\ 0=t(14.26-4.9t) \\ 0=14.26-4.9t \\ 4.9t=14.26 \\ t=\frac{14.26}{4.9} \\ t=2.91\; s \end{gathered}[/tex]Finally, the range is

[tex]x=v_x\times t=7.26\times2.91=21.13\; m[/tex]Therefore, the ball will land **21.13 meters** far from the football player.

How much farther would the ball go if he kicked it with the same speed, but at a 45° angle?

We need to repeat the above calculations

The horizontal and vertical speed of the ball is given by

[tex]\begin{gathered} v_x=v\cos (\theta)=16\cos (45\degree)=11.31\; \frac{m}{s} \\ v_y=v\sin (\theta)=16\sin (45\degree)=11.31\; \frac{m}{s} \end{gathered}[/tex]The time (t) is given by

[tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=11.31_{}t+\frac{1}{2}(-9.8)t^2 \\ 0=11.31_{}t-4.9t^2 \\ 0=11.31_{}-4.9t \\ 4.9t=11.31_{} \\ t=\frac{11.31_{}}{4.9} \\ t=2.31\; s \end{gathered}[/tex]Finally, the range is

[tex]x=v_x\times t=11.31\times2.31=26.13\; m[/tex]Therefore, the ball will land **26.13 meters** far from the football player.

Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?

The ball kicked at 16 m/s and at a 63° angle takes **2.91 s** to land.

The ball kicked at 9 m/s and at a 45° angle will take

[tex]v_y=9\sin (45\degree)=6.36\; \frac{m}{s}[/tex][tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=6.36t+\frac{1}{2}(-9.8)t^2 \\ 0=6.36t-4.9t^2 \\ 0=6.36-4.9t \\ 4.9t=6.36 \\ t=\frac{6.36}{4.9} \\ t=1.30\; s \end{gathered}[/tex]So, the ball kicked at 9 m/s and at a 45° angle takes **1.30 s** to land.

Therefore, the ball kicked at 9 m/s and at a 45° angle will land first.

mick took his friend out to dinner the bill was $40 but he applied a coupon then the total price was $33 what was the % off?

**ANSWER**

**17.5 %**

**EXPLANATION**

We have to find the percent change, given that the initial price was $40 and the final price was $33.

[tex]\text{ \% }change=\frac{final.price-initial.price}{initial.price}\times100[/tex][tex]\text{\% }change=\frac{33-40}{40}\times100=\frac{-7}{40}\times100=-0.175\times100=-17.5\text{\%}[/tex]The coupon was for **17.5% off**

I just started a new lesson and have this study guide, but the material is unfamiliar to me and

Answer:

2.79 eV

Explanation:**Given**

The wavelength, λ = 445 nm

The planck's constant, h = 6.626 x 10^-34 m² kg/s

The speed of light, c = 3 x 10^8 m/s

**The energy of the photon is calculated as:**

But,** 1 eV = 1.6 x 10^-19J**

What is the relationship of the force on the spring and stretch of the spring.(This is the spring we are looking at weights can be attached to it)

We will have the following:

We know that for a simple spring the following is true:

[tex]\begin{cases}F=-kx \\ \\ F=ma\end{cases}[/tex]So:

[tex]-kx=ma[/tex]So, the stretch of the spring is directly proportional to the force.

Number 1. Part b: what are the final kinetic energy of the system

Given that there is a cart of mass, m = 0.12 kg moving with initial speed of, u1 = 0.45 m/s and it collides with another cart of mass, m = 0.12 kg with initial speed, u2 = 0 m/s

We have to find the initial and final kinetic energy.

(a) Initial kinetic energy,

[tex]\begin{gathered} K\mathrm{}E.1=\frac{1}{2}mv^2 \\ =\frac{1}{2}\times0.12\times(0.45)^2 \\ =0.012\text{ J} \end{gathered}[/tex]According to the conservation of linear momentum,

[tex]mu1+mu2=2mv[/tex]Here, v is the final speed.

[tex]\begin{gathered} 0.12\times0.45=2\times0.12\times v \\ v=\frac{0.45}{2} \\ =0.225\text{ m/s} \end{gathered}[/tex]Here, the final speed is 0.225 m/s.

(b) The formula to find kinetic energy is

[tex]K\mathrm{}E\mathrm{}=\frac{1}{2}(2m)v^2[/tex]Substituting the values, we get

[tex]\begin{gathered} K\mathrm{}E\mathrm{}=0.12\times(0.225)^2 \\ =6.075\times10^{-3}\text{ J} \end{gathered}[/tex]Hence the kinetic energy is 6.075 x 10^(-3) J.

You throw an object up with an initial velocity of Voy = 11 m/s from a height of y = 25 m. How long, in seconds, does it take for the object to reach the ground? What is the objects final velocity, in meters per second, as it impacts the ground? Find the time, in seconds, if you instead threw the object DOWN with the same velocity Voy

Calculate the ball's greatest height using the **vertical motion **model, h = -16t2 + vt + s, where v is the beginning **velocity** in feet/second and s is the height in feet.

A **ball** is thrown into the air, where it gradually loses **speed** until it abruptly comes to a rest at the peak of the **motion**. The body is travelling upward against gravity at the top, hence the **acceleration** there is 9.8 ms2. For example, g=9.8 ms2 is the formula for the **acceleration** caused by **gravity**.

Only at the greatest point of a body being hurled **vertically** upwards would velocity be **zero** because of the constant downward acceleration brought on by the gravitational force. As a result, velocity is zero throughout the rest of the **motion**.

A ball is originally moving **upward** when it is tossed into the air, for instance.

Learn more about **vertical motion **refer

**https://brainly.com/question/27526920**

**#SPJ13**

A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. What isthe acceleration of the skier?(Unit = m/s?)Enter

A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. The **acceleration** of the skier is approximately 2.97 m/s².

To find the **acceleration** of the skier, we need to use Newton's second law of motion and consider the forces acting on the skier.

Identify the forces acting on the skier:

The forces acting on the skier are the force of gravity (mg) and the force exerted by the rope (T), where m is the mass of the skier (72.5 kg) and g is the acceleration due to gravity (9.8 m/s²). The force exerted by the rope is parallel to the slope and can be calculated using the given value of 383 N.

Resolve the** forces**:

Since the rope is parallel to the slope, we need to resolve the force of gravity into components parallel and perpendicular to the slope. The component parallel to the slope is m * g * sin(21.7°).

Apply Newton's second law of motion:

**Newton's second law** states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, Fnet = ma.

Determine the net force:

The net force acting on the skier is the difference between the force exerted by the rope and the component of the force of gravity parallel to the slope. Fnet = T - m * g * sin(21.7°).

Calculate the acceleration:

Using Newton's second law, we can rearrange the equation Fnet = ma to solve for the acceleration (a). a = Fnet / m.

Substitute the values and solve:

Substitute the known values into the equation to find the acceleration.

Therefore, the acceleration of the skier is approximately 2.97 m/s².

For more such questions on **acceleration**, click on:

https://brainly.com/question/30595126

#SPJ8

**Answer:**

about 0.567 m/s²

**Explanation:**

You want the **acceleration** of a **72.5 kg skier** up a **21.7° slope** with µk = 0.120, towed by a rope exerting a **force of 383 N**.

The force up the slope is 383 N.

The forces down the slope will be the sum of the force due to gravity and the friction force.

GravityThe force down the slope due to gravity is ...

F = m·g·sin(θ) = (72.5 kg)(9.8 m/s²)(sin(21.7°) ≈ 262.7 N

FrictionThe force due to friction will be the product of µk and the force normal to the slope:

F = m·g·cos(θ)·µk = (72.5 kg)(9.8 m/s²)cos(21.7°)·0.120 ≈ 79.22 N

Net ForceThe net force up the slope is ...

383 N -262.7 N -79.22 N ≈ 41.08 N

This will accelerate a mass of 72.5 kg in the amount of ...

A = F/m = 41.08 N/(72.5 kg) ≈ 0.567 m/s²

**The acceleration of the skier is about 0.567 m/s² up the slope**.

__

*Additional comment*

We don't have to figure the forces. Rather we can figure the gross acceleration due to the tow rope, then subtract the accelerations due to gravity and friction. This saves a few math operations as we don't have to multiply, then divide, by 72.5 kg.

<95141404393>

A resistor uses energy at a rate of 2.50W when there is a current of 4.00A passing through it. What must be the potential difference across the resistor?1.25V1.50V0.625V10.0V

As we know

[tex]P=\text{ V}\times I;[/tex]Where,

P= electric power= 2.50 W

V= potential difference= ?

I= current = 4.00A

Using above formula we get,

[tex]\begin{gathered} P=V\times I; \\ \therefore2.50=\text{ V}\times4; \\ V=\text{ }\frac{2.50}{4}=\text{ 0.625V} \end{gathered}[/tex]**Final answer is :- 0.625 V**

When is a secondary source more helpful than a primary source?

A. When you want to confirm the conclusion with your own experiment

B. When you are not an expert in the field being studied in the experiment

C. When you do not need to know the conclusion of the experiment

D. When you want to know the exact data values

**Option ****A**** **is** **the correct answer: A **secondary** source is more helpful than a primary source when you want to confirm the conclusion with your own **experiment**.

Primary and secondary sources work well together to support the **argument** you are trying to make. Although secondary sources demonstrate how your work connects to earlier research, **primary** sources are more reliable as evidence. When you want to confirm the **conclusion** of your experiment, secondary sources prove to be more useful.

Original **research** is built on primary sources. You are able to discover novel information, offer solid justification for your claims, and provide reliable **facts** on your subject.

Secondary sources are useful for getting a comprehensive **perspective** of your subject and learning other researchers' methods. They frequently combine numerous primary **materials** that would take a lot of time and effort to obtain independently.

To know more about **primary** sources, refer to the following link:

https://brainly.com/question/19649709

#SPJ9

A net force of 43.1 N causes a mass to accelerate at a rate of 0.2 m/s2. Determine the mass in kilograms.

The **mass **of the an **object **caused to accelerate at 0.2m/s² by a 43.1 Newton force is 215.5 kilograms.

A **force **is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.

Force, according to **Newton's** Second Law is expressed as;

F = m × a

Where a is acceleration and m is the mass.

Given the data in the question;

Force applied F = 43.1N = 43.1kgm/s²Acceleration a = 0.2m/s²Mass m = ?Plug the given values into the formula above and solve for m.

F = m × a

43.1kgm/s² = m × 0.2m/s²

m = 43.1kgm/s² / 0.2m/s²

Mass m = 215.5kg

Therefore, the **mass **of the **object **is 215.5 kilograms.

Learn more about **force** here: brainly.com/question/27196358

#SPJ1

Suppose you walk 16 m straight east and then 22 m straight south. At what angle, in degrees South of East, is a line connecting your starting point to your final position?

**53.9 angle**, in degrees South of East and **27.20 m** is a line connecting your **starting** **point** to your **final** **position**.

The distance in decimeters between the starting point and the **ending** **position** is in a straight line. The distance between an object's original position and its ultimate position is known as **displacement**.

You walk 16 m straight east and then 22 m straight south. This forms a right angled triangle with a horizontal distance of 16 m, a vertical distance of 22 m and the hypotenuse is the distance between the **ending** and starting point. Let x represent the distance between the ending and starting point. Using **Pythagoras** **theorem**:

x² = 16² + 22²

x² = 256 + 484

x² = 740

Taking square root of both sides:

√x² = √740

x = √740

**x = 27.20 m** = distance between the **ending** and **starting** **point**.

Now use **trigonometry**:

sinθ=B/R

sinθ=22/27.20

sinθ= 0.808

**θ = 53.9 degree**. This is your **angle**.

To know more about** Final position** visit:

https://brainly.com/question/11942199

**#SPJ13**

In a physics lab a student discovers that the magnitude of the magnetic field in a specific location near a long wire is 21.919 microTesla. If the wire carries a current of 35.483 A, what is the distance from the wire to that location ?

We will have the following:

First, we have that the permeability of free space is:

[tex]\mu_0=4\pi\ast10^{-7}Tm/A[/tex]Then:

[tex]\begin{gathered} B=\frac{\mu_0I}{2r\pi}\Rightarrow r=\frac{\mu_0I}{2B\pi} \\ \\ \Rightarrow r=\frac{(4\pi\ast10^{-7}Tm/A)(35.483A)}{2\pi(2.1919\ast10^{-5}T)}\Rightarrow r=0.3237647703...m \\ \\ \Rightarrow r\approx0.32m \end{gathered}[/tex]So, **the distance is approximately 0.32 m**.

5. Was you hypothesis supported by the data collected for question 2? Why or why not ?

**Answer:**

**Your hypothesis was not supported by the data collected**

**Explanation:**

On question 2 you said that the point with the greatest kinetic energy would be point 5. However, when you analyze the data, the point with the greatest velocity was point 2 which means that this is the point with the greatest kinetic energy.

Therefore, your hypothesis was not supported by the data collected because based on the picture, it is hard to say where is the lowest point of the roller coaster.

Given two 2.00μC charges on the horizontal axis are positioned at x=0.8m, and the

other at x=-0.8m, and a test charge q = 1.28x10-18C at the origin.

(a) What is the net force exerted on q by the two 2.00μC charges? [5]

(b) What is the electric field at the origin due to 2.00μC charges? [5]

(c) What is the electric potential at the origin due to the two 2.00μC charges

**Answer:Question 1 **

** Given q1=2µC **

** q2=2µC**

** q= 1.2×10^-18C at origin **

**Net force exerted by two charges on q**

**F_1= force on q due to q1**

**F_2= force on q due to q2**

**F_net= F_(1-) F_2 **

** = (Kqq_1)/r^2 - (kqq_2)/r^2 Then q_1=q_2=〖2×10〗^(-6) **

**F_net=0N**

**b) The electric field at charge q **

**E_net= E_1- E_2**

** = (kq_1)/r^2 - (kq_2)/r^2 **

**Then q_1=q_2**

** E)_net= 0 N/C **

**c) The electric potential at origin due to two charge **

**V_net= V_(1 )- V_2**

** = (kq_1)/r - (kq_2)/r **

**Then q1= q2**

**V_net= 0 V**

**Explanation:**

Set the cannon to have an initial speed of 20 m/s. For which situation do you think the cannon ball will go father: if it is set at a 60-degree angle, or if it is set at a 70-degree angle?

Question 2 options:

60 degrees

70 degrees

The **cannon ball** will go **farther** if the the angle of projection is set at 60 degrees

How to determine which **angle** will result in** farther** distance

Case 1:

Initial velocity (u) = 20 m/sR = u²Sine(2θ) / g

R = 20² × Sine (2×60) / 9.8

R = 346.41 / 9.8

R = 35.35 m

Case 2:

Initial velocity (u) = 20 m/sR = u²Sine(2θ) / g

R = 20² × Sine (2×70) / 9.8

R = 257.12 / 9.8

R = 26.24 m

From the above calculations, we can conclude that the ball will go **farther**, if the **angle** is 60 °

Learn more about **projectile motion:**

https://brainly.com/question/19128146

#SPJ1

It takes 225 kJ of work to accelerate a car from 20.1 m/s to 28.1 m/s. What is the car's mass?

**Answer:**

THE REMAINIG WILL BE 75

**Explanation:**

HOPE IT HELPS YOU

A superelastic collision is one in which1) kinetic energy before the collision equals kinetic energy after the collision.2) kinetic energy after the collision is zero.3) kinetic energy before the collision is less than kinetic energy after the collision.4) kinetic energy before the collision is greater than kinetic energy after the collision

A superelastic collision is the the collision in which the potential energy of the system is converted into the kinetic energy such the kinetic energy of the system after the collision is more than the kinetic energy of the system before the collision.

**Hence, 3rd option is the correct answer.**

A 12 N force acts at a 25 degrees and an 8 N force acts at 65 degrees. Determine the magnitude and direction (include angle) of the resultant . Scale is 1cm = 1N

NEEDDD HELLPP ASAPPPPP

A 12 N** force** acts at a 25-degree angle, while an 8 N force acts at a 65-degree angle. The** magnitude** and** direction** (including angle) of the **resultant** force are 21.43.

Consider the **formula** for a** force**, F=M×Cos∝

As M is a **Mass** of an object and, ∝ Is the** angle** at which **force** is acting on an object.

[tex]F_{1}[/tex]**=M × Cos ∝**

M=12N, and Cos∝= Cos25 = 0.906

[tex]F_{1}[/tex] = 12× 0.906

∴ [tex]F_{1}[/tex]= 10.87N

For [tex]F_{2}[/tex] = 25×Cos65

= 10.56N

According to** Superposition** resulting force is the** Sum** of **total forces**.

[tex]F_{RESULTANT} = F_{1} + F_{2}[/tex]

= 10.87+ 10.56

=21.43N

∴The **resultant force** is 21.43.N

To know more about **Resultant force**, use the link.

https://brainly.com/question/22260425

#SPJ1

009 (part 1 of 2) 10.0 points When a water gun is fired while being held horizontally at a height of 1.19 m above ground level, the water travels a horizontal distance of 1.98 m. Find the initial velocity of the water. The acceleration of gravity is 9.81 m/s^2

Answer in units of m/s.

010 (part 2 of 2) 10.0 points A child, holding the same gun in a horizontal

position, slides down a 33.0◦incline at a constant speed of 1.40 m/s. The child fires the gun when it is 4.36 m above the ground and the water takes 0.868 s to reach the ground. How far will the water travel horizontally?

Answer in units of m.

**Answer:**

**Speed = 4 m/s**

**Explanation:**

Given:

**009 (part 1 of 2)**

H =1 .19 m

L = 1.98 m

g = 9.81 m/c²

____________

V₀ - ?

**Equation of motion horizontally:**

L = V₀*t (1)

**Equation of vertical motion:**

H = g*t² / 2 (2)

From equation (2):

**Time:**

t = √ (2*H / g) = √ ( 2*1.19 / 9.81) ≈ 0,49 s

**From equation (1):**

**Horizontal speed:**

V₀ = L / t = 1.98 / 0.49 ≈ 4 m/s

3. A boulder rolls with speed of 3.5 m/s off a cliff. It hits the ground 2.25 m from the base ofthe ledge. A) How high is the ledge? B) How long did it take the boulder to fall to the bottomof the cliff?DrawingVerticalHorizontal

**Given data**

*The given distance from the base of the ledge is** R** = 2.25 m

*The given speed is **v** = 3.5 m/s

The diagram is given below

**(a)**

Let **(h)** be the height of the edge

The formula for the distance from the base of the ledge is given as

[tex]\begin{gathered} R=v\times t \\ R=v\times\sqrt[]{\frac{2h}{g}} \\ h=\frac{R^2\times g}{2v^2} \end{gathered}[/tex]Substitute the known values in the above expression as

[tex]\begin{gathered} h=\frac{(2.25)^2\times9.8}{2\times(3.5)^2} \\ =2.025\text{ m} \end{gathered}[/tex]Hence, the height of the ledge is** h** = 2.025 m

**(b)**

The formula for the time taken by the boulder to fall to the bottom of the cliff is given as

[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]Substitute the known values in the above expression as

[tex]\begin{gathered} t=\sqrt[]{\frac{2\times2.025}{9.8}} \\ =0.642\text{ s} \end{gathered}[/tex]Hence, the time taken by the boulder to fall to the bottom of the cliff is **t** = 0.642 s

make G the subject of the formula

F = GMM²/1²

HENCE WRITE THE DIMENSION FOR G

The value of G is FR²/M₁M₂. and the **dimension **of G is [M⁻¹L³T⁻²]..

The provided formula is of** gravitational force F** between two objects,

F = GM₁M₂/R²

Where** M₁ **is the mass of first object and **M₂** is the mas of the other object while R is the distance between there centers and G is the **universal gravitation constant.**

To find the **dimension of G**, making G the subject of formula,

G = FR²/M₁M₂.

As we know, unit of mass is Kilogram (Kg), unit of force is Newton (N) and unit of distance is Meter (M).

Putting all the values, **Units **in the place of quantities,

G = N.R²/Kg.Kg

Now, using** Dimensional analysis**, and writing the **dimensions **of all the other units,

G = [MLT⁻²][L²]/[M][M]

G = [ML³T⁻²]/[M²]

G = [M⁻¹L³T⁻²]

The dimensions of G are [M⁻¹L³T⁻²].

To know more about **Dimensional Analysis,** visit,

https://brainly.com/question/28571069

#SPJ9

Information givenknown: mass of Christine=60 kgmass of cart= 22 kg mass of hailey=69The two girls on the cart to the left pushing off of Conner take .3833 s to reach a distance of 0.3m. Conner reaches the same distance in 0.2333s. What is the mass of Conner?

We will have the following:

We use conservation of momentum to solve, that is:

[tex]\begin{gathered} (60kg+22kg+69kg)(0.3m/0.3833s)=(m+22kg)(0.3m/0.233s) \\ \\ \Rightarrow118.1841899kg\ast m/s=(m+22kg)(\frac{300}{233}m/s) \\ \\ \Rightarrow m+22kg=91.78968976kg\Rightarrow m=69.78968976... \\ \\ \Rightarrow m\approx69.8 \end{gathered}[/tex]So, **Conner's mass is approximately 69.8 kg**.

A tensile load of 190 kN is applied to a round metal bar with a diameter of 16mm and a gage length Of 50mm. Under this load the bar elastically deforms so that the gage length increases to 50.1349 mm and the diameter decreases to 15.99 mm. Determine the modulus of elasticity and Poisson s ratio for this metal.

The elasticity and Poisson's **ratio** for this metal is 0.232.

What is **ratio?**

The realation between two numbers which shows how much bigger one quantity is than another.

Sol-

As per the given question

P=190KN

d=16 mm

Lo=50mm

X=50.1349-50=0.1349mm

Y=15.99-16=-0.01mm

The formula-

E=ó/€

Ó=P/A

A=r/4 d^2 =π/4(16)^2=201.062 mm

ó={190(1000)}201.062=944.982 Mpa

E=944.982/0.002698=350.253 GPa

€y=-0.000625

v=0.232(answer)

To know more about **ratio **click

https://brainly.com/question/2914376

#SPJ9

Calculate the depth in the ocean at which the pressure is three times the atmospheric pressure

**ANSWER:**

20.17 meters

**STEP-BY-STEP EXPLANATION:**

Given:

Pressure = Po = 1.013x10^5 Pa

Pressure at depth = P = 3Po

Density of sea water = 1025 kg/m^3

We can calculate the depth as follows:

[tex]\begin{gathered} P=P_o+d\cdot g\cdot h \\ \text{ we solve for h} \\ dgh=P-P_o \\ h=\frac{P-P_o}{d\cdot g} \end{gathered}[/tex]We replacing and calculate the depth:

[tex]\begin{gathered} h=\frac{3\cdot P_o-P_o}{1025\cdot9.8}=\frac{2P_o}{10045}=\frac{2\cdot1.013\cdot10^5}{10045} \\ h=20.17\text{ m} \end{gathered}[/tex]Therefore, the depth is equal to 20.17 meters

The tortoise and the hare A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s.

The **time **taken for the hare to catch up with the tortoise is **10 seconds.**

The **time **taken for the hare to catch up with the tortoise is calculated by applying the principle of **relative velocity** as shown below.

Distance travelled by hare + Distance travelled by tortoise = Total distance

V₁t + V₂t = d

where;

V₁ is the velocity of hareV₂ is the velocity of tortoiset is the time taken for them to meetd is the distance between themSince they are moving in the same direction, the relative velocity becomes

V₁t - V₂t = d

(V₁ - V₂)t = d

(4 - 1)t = 30

3t = 30

t = 30/3

t = 10 seconds

Thus, the **time **taken for the hare to catch up with the tortoise is **10 seconds.**

Learn more about **relative velocity** here: https://brainly.com/question/17228388

#SPJ1

The complete question is below:

A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s. At what time does the hare catch up with the tortoise?

The figure shows a standing wave oscillating at 100 Hz on a string. What is the wave speed ?

**Answer:**

**The wavelength is 60 cm**.

**Explanation:**

**Speed = frequency x wavelength = 100 x 60 = 6000 cm/s**

**or if you wanted the answer in m/s**

**Speed = frequency x wavelength= 100 x 0.60 = 60 m/s**

A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is.......

A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new **speed** is....... 400 J.

What is **Speed**?

**Speed** is the time rate at which an object is travelling along a path, whereas velocity is the pace and direction of an object's movement. In other words, velocity is a vector, whereas speed is a scalar valu

Work = Change in Kinetic energy

= ½m(v² - u²)

= ½ × 20 kg × [(7 m/s)² - (3 m/s)²]

= 10 kg × 40 m²/s²

= 400 J

To know more about** Speed**

brainly.com/question/13943409

#SPJ9

At an animal shelter, 3/8 of the animals are dogs. Of thedogs, 2/5 have spots. What portion of the animals at theshelter are dogs that have spots?
Kelsey ran the 40 meter track at her school in 91.4 seconds running at the same speed how long did it take for the run 100 meter
Guys help with this chart please
The table below shows the densities of the elements in Group 0 of the periodic table. Noble Gas helium neon argon krypton xenon radon Density (kg/m) 0.2 0.9 1.8 3.7 5.9 9.7 Use the data from the table to draw a graph to show the trend in density as you move down Group 0.
Please help Which position shows the sun setting? 3 1 4 2
hi need help with this
Premises:If I'm a student, then I go to school. If I go to school, then I learn.Conclusion:If I'm a student, then I learn.This is an example of the Law of_________
A current of 17.0A is maintained in a single circular loop of 2.00m circumference. A magnetic field of 0.800T is directed parallel to the plane of the loop. (a) Calculate the magnetic moment of the loop. [5] (b) What is the magnitude of the torque exerted on the loop by the magnetic field?
Describe three styles of menu boards that a counter-service restaurant might use.
Nosotros les a cenar en casa. Mariana nos que iba a venir pronto. Usted dijo que no poda venir porque las llaves de su casa. Antonio ya otros planes con sus amigos. Roberto y diana s venan porque ya de sus vacaciones. Rosa y t cenado. Ustedes al cine. A las nueve todava no nadie. A las diez la cena se fra. Cuando llegaron todos, nosotros.
Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 29.0 below the horizontal. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
DUE TODAY ASAP!! Megan has $50 and saves $5.50 each week. Connor has $18.50 and saves $7.75 each week. After how many weeks will Megan and Connor have saved the same amount? Include step by step explanation for brainliest.
What is one issue with the use of living wills?
in the tftp protocol: if the client changes its port number on a subsequent connection, but the server does not, what prevents an old-duplicate data packet sent by the server from being accepted by the new client? if the server changes its port number on a subsequent connection, but the client does not, what prevents an old-duplicate data packet sent by the server from being accepted by the new client?
HELP!!!! i dont understand
Based on the below experiment What is the purpose of the liquid detergent?
You are a presidential candidate developing your campaign.Come up with a slogan and tell me 3 propaganda techniques that you will use on the campaign trail, and describe how you will use them to get into office.
Which inequality represents the values of x that ensure triangle abc exists? a triangle with side lengths of 6.5 units, 3.5 units, and 2.5 + x units
how were the native people of Washington State impacted by the arrival of early settlers, such as fur trappers, missiona pioneers? O They welcomed the new arrivals and were able to successfully integrate them into their society O They were presented with agricultural knowledge, leading to a change in their lifestyles. O They were forced to move east in search of land and new economic opportunities for their people O They suffered from new diseases that caused many deaths and a population decline
. Lisa, Bret, and Gary harvested apples.Lisa filled 3 carts with apples. Bret alsofilled 3 carts with apples. Gary filledanother 3 carts with apples. Write amultiplication equation and a divisionequation for this story.