The cannon ball will travel the **highest distance **when the angle of projection is **35 degrees**.

The** maximum height** reached by a projectile is calculated using the following formula.

H = u²sin²θ/2g

where;

u is the initial velocity of the projectile θ is the angle of pojectiong is acceleration due to gravitywhen the angle of projection is 25 degrees;

H = (15² (sin 25)²) / (2 x 9.8)

H = 2.05 m

when the angle of projection is 35 degrees;

H = (15² (sin 35)²) / (2 x 9.8)

H = 3.78 m

Thus, the cannon ball will travel the **highest distance **when the angle of projection is **35 degrees**.

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what is velocity ratio

**Answer:**

Definition of velocity ratio

: the ratio of a distance through which any part of a machine moves to that which the driving part moves during the same time.

pls brainlies me me new pls

**Explanation:**

Answer:

Le rapport de vitesse, parfois appelé rapport de distance, est une comparaison de la quantité de force qu’un objet, comme une voiture, crée par rapport aux autres forces autour de lui qui agissent contre lui.

Explanation:

Lorsque le premier engrenage (le conducteur ou l'engrenage d'entrée) tourne, le deuxième engrenage (l'engrenage entraîné ou en sortie) se transforme en réponse. La différence entre les vitesses des deux vitesses est appelée rapport de vitesse ou rapport de réduction.

Le rapport de vitesse, parfois appelé rapport de distance, est une comparaison de la quantité de force qu’un objet, comme une voiture, crée par rapport aux autres forces autour de lui qui agissent contre lui.

Explanation:

Lorsque le premier engrenage (le conducteur ou l'engrenage d'entrée) tourne, le deuxième engrenage (l'engrenage entraîné ou en sortie) se transforme en réponse. La différence entre les vitesses des deux vitesses est appelée rapport de vitesse ou rapport de réduction.

A ball rolls off a table and falls 0.75m to the floor, landing with a speed of 4 m/s. (A) What is the acceleration of the ball just before it strikes the ground? (B) What was the initial speed of the ball? (C) What initial speed must the ball have if it is to land with a speed of 5 m/s?

The **acceleration** of the ball is 21.33m/s²,

We are given that,

The ball fall's from height = d = 0.75m

The final **speed **of the ball = Vf = 4m/s

So that to know the acceleration of the ball we can calculate by the equation of motion in term of velocity and acceleration i.e. given as,

V = u + at

Where, V is the final** velocity **, u is the initial velocity , t is the time and a is the acceleration of the object.

t = d/v

t = (0.75m)/(4m/s)

t = 0.1875s

Thus, the value of t, V, and initial velocity is zero putting in equation of motion to get acceleration,

a = (4m/s)/(0.1875s)

a = 21.33m/s²

The acceleration of the ball would be 21.33m/s²

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In a parallel circuit with a 12 V battery and three 6 Ohm resistors, what is the total current in the entire circuit? Select one:a.36 Ampsb.6 Ampsc.4 Ampsd.2 Amps

First, let's find the equivalent resistance. Since they are in parallel, we can find it as follows:

[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3} \\ so: \\ \frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2} \\ Hence: \\ R_{eq}=2\Omega \end{gathered}[/tex]Now, we can use ohm's law to calculate the current:

[tex]\begin{gathered} V=IR \\ so: \\ I=\frac{V}{R}=\frac{12}{2}=6 \end{gathered}[/tex]**Answer:**

**b.**

**6 Amps**

a 30.0 kg child starting from rest slides down a water slide with the vertical height of 10.0 m what is the child speed (a )halfway down the slides vertical distance and (b) 3/4 of the way down

We know that

• The mass is m = 30.0 kg.

,• The vertical height is h = 10.0 m.

(a)We have to use the conservation of energy theorem, which states that mechanical energy is constant all the time. Also, halfway down means a height of 5.0 m. It's important to know that at the top the total energy is potential, while halfway is distributed as kinetic and potential, the expression below shows this

[tex]E_{p1}=E_{k1}+E_{p2}[/tex]Then, using the definition of each energy, we have

[tex]mgh_1=\frac{1}{2}mv^2+mgh_2[/tex]Now, we use the given values to find the speed.

[tex]\begin{gathered} \text{mgh}_1=m(\frac{1}{2}v^2+gh_2) \\ gh_1=\frac{1}{2}v^2+gh_2 \\ 9.81m/s^2\cdot10m=\frac{1}{2}v^2+9.81m/s^2\cdot5m \\ 98.1m^2/s^2=\frac{1}{2}v^2+49.05m^2/s^2 \\ 98.1m^2/s^2-49.05m^2/s^2=\frac{1}{2}v^2 \\ 2\cdot49.05m^2/s^2=v^2 \\ v=\sqrt[]{98.1m^2/s^2} \\ v\approx9.9m/s \end{gathered}[/tex]Therefore, the speed of the child halfway down is 9.9 meters per second.(b)In this case, we just have to use as the second height of the equation the magnitude 2.5 meters because that's 3/4 of the way down. So, let's use the same process and expression

[tex]\begin{gathered} gh_1=\frac{1}{2}v^2+gh_2 \\ 9.81m/s^2\cdot10m=\frac{1}{2}v^2+9.81m/s^2\cdot2.5m \\ v=\sqrt[]{2(98.1m^2/s^2-24.53m^2/s^2)} \\ v\approx12.1m/s \end{gathered}[/tex]Therefore, the speed of the child 3/4 of the way down is 12.1 meters per second.Two forces F1 = -6.00i + 7.90j and F2 = 6.80i + 5.30j are acting on an object with a mass of m = 4.10 kg. The forces are measured in newtons, i and j are the unit vectors. What is the magnitude of the object's acceleration?

The **magnitude **of object's **acceleration **is 3.26m/s².

The **mass **of the body is 4.10 lg.

The two **forces **that are acting on the object are F₁ = -6i + 7.9j newton and F₂ = 6.8i + 5.3j **Newton**.

We know that the force acting on an object is,

F = Ma

Where,

F is the **force** acting,

M is the mass of the object and,

a is the **acceleration **of the object.

As we can see, two forces are acting on the body,

We can simplify the forces in** x direction** and** y direction,**

The forces are F₁ = -6i + 7.9j N and F₂ = 6.8i + 5.3j N.

So, the total force in x-direction,

Fₓ = (-6+6.8)i

Fₓ = 0.8i

Fᵧ = (7.9+5.3)j

Fᵧ = 13.2j

So, the net force Fₙ on the object is Fₙ = (0.8i + 13.2j) N

Now, putting value of force and mass in the **formula**,

F = Ma

0.8i + 13.2j = 4.1a

a = 0.19i + 3.21j m/s².

The magnitude of acceleration is,

|a| = √[(0.19)²+(3.21)²]

|a| = 0.361 +10.3

|a| = 3.26m/s².

So, the magnitude of **acceleration **is 3.26m/s².

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Emily and Gemma did a Reaction time lab. Emily dropped the ruler while Gemma tried to catch it. She caught the ruler 5 times and her average catching distance is 0.12 m. What is Gemma's reaction time?

The average** reaction time** of Gemma is 0.1564 seconds.

As we know, Gemma is catching the scale and Emily is dropping the scale.

The whole experiment is taking place under gravity, so the **acceleration **is constant.

As we know, the scale is dropped, it means that the initial velocity of the scale is zero.

We can use the **equation of motion,**

The equation is,

S = Ut + 1/2at²

Where,

S is the **displacement**, which is 0.12 m in our case,

U is **initial velocity** which is 0m/s because the stone is dropped,

t is the **time taken**, this is equal tot he reaction time here,

a is the **acceleration due to gravity** whose value is 9.8m/s.

Now, putting all the values,

0.12 = 1/2(9.8)(t)²

t² = 0.24/9.8

t = 0.1564 seconds.

Gemma reacts in 0.1564 seconds to catch the scale.

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In Young's double-slit experiment, two slits are separated by 5.0 mm and illuminated by light with a wavelength of 480 nm. The screen is 3.0 m from the plane of the slits. Calculate the separation between the eighth bright fringe and the third dark fringe observed with respect to the central bright fringe.

We are given the following information.

Seperation between slits: d = 5.0 mm

Wavelength of light: λ = 480 nm

Distance from the plane of slits: D = 3.0 m

We are asked to calculate the separation between the 8th bright fringe and the 3rd dark fringe observed with respect to the central bright fringe.

The position of the **8th bright fringe** is given by

The position of the **3rd dark fringe** is given by

Finally, the separation between the 8th bright fringe and the 3rd dark fringe is

[tex]\begin{gathered} x_8-x_3=2.304\times10^{-3}-7.2\times10^{-4} \\ x_8-x_3=1.584\times10^{-3}\;m \end{gathered}[/tex]Therefore, the separation between the eighth bright fringe and the third dark fringe observed with respect to the central bright fringe is **1.584×10⁻³ m**.

Exercise 1 :On a circuit, a pilot covers 600 m in 7.2 s.1. Calculate its speed in m / s.2. Convert this speed to km / h in two different ways.

**Given data**

*The distance covers by the pilot is **d** = 600 m

*The given time is** t** = 7.2 s

**(1)**

The formula for the speed is given as

[tex]s=\frac{d}{t}[/tex]Substitute the known values in the above expression as

[tex]\begin{gathered} s=\frac{600}{7.2} \\ =83.3\text{ m/s} \end{gathered}[/tex]Hence, the speed is **s** = 83.3 m/s

**(2)**

The speed converted into km/h as

[tex]\begin{gathered} s=83.3\times(\frac{5}{18}) \\ =299.88\text{ km/h} \end{gathered}[/tex]The second way to convert the speed into kilometer per hour as,

[tex]undefined[/tex]When light goes from a slower medium to a faster medium; what way does the light bend relative to the normal?

[tex]The\text{ light bends with a greater angle than the incidence angle.}[/tex]

**Answer: A.**

It moves away from the normal.

**Explanation: ed mentum or plato**

if an astronaut weighs 981 N on Earth and only 160 N on the moon, then what is the mass on the moon

If an astronaut weighs 981 N on Earth and only 160 N on the Moon, then his mass on the Moon will be** 98.1 kg.**

Let's calculate the mass of Earth as per the Earth's acceleration due to gravity.

Now, considering the acceleration due to gravity as **10m/s².**

Mass = Weight/Acceleration due to gravity

Mass = 981/10

Mass = 98.1 kg

Now, as per the established fact, the mass is independent of acceleration due to the gravity of the planet i.e. The mass of the person on earth and on the moon is same.

Hence, his mass on the Moon will be **98.1 kg.**

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1) Prove that for the adiabatic process of gas pVr =const ; r=Cp/Cv *

2) Write down the expression for the number of molecules of N2 gas having speeds in the interval (100m/s ; 102m/s) for one mole of gas at T=300K

3) Demonstrate how to obtain the Boltzmann Distribution of Energies from the Maxwell-Boltzmann distribution of speeds.*

For an **adiabatic **process of gas PV(^r)= **Constant**, where r = Cp/Cv , Cp, and Cv are constants for a particular gas.

It is an imaginary **gas **for which the volume occupies by it is negligible, this gas does not exist in a practical situation and the concept of an **ideal **gas is only the theoretical one.

Universal Gas **Equation **for the ideal gas,

PV=nRT

By differentiating on both sides: P.dV + V.dP = n.R.dT…. (1)

For the adiabatic process, the expansion or compression of gas occurs at a very fast rate which makes no **heat transfer**. Therefore, dQ= 0

A/c to First Law of TD: dQ=dU+ dW, implies dU= -dW=-P.dV ….. (2)

But dU= n.Cv. dT …. (3) Combining 2 and 3, we get: n.dT= -P.dV/Cv …..(3)

Using R= Cp-Cv, the gas law gives: from 1 and 3.

n.dT= (P.dV + V.dP)/(Cp-Cv) = -PdV/Cv

On rearranging terms we get, (dP/P)+((Cp/Cv)*(dV/V))=0

Cp and Cv are **constants **for a particular gas, hence integrating,

Ln(P)+ r.Ln(V)=constant (replacing Cp/Cv = r or Gamma)

Hence, P.V(^r)= constant is termed an **adiabatic **equation.

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A spring with spring constant 175 N/m has 20 J of EPE stored in it. How much is it compressed?

Given:

• Spring constant = 175 N/m

,• Energy = 20 J

Let's find by how much it is compressed.

Apply the formula:

[tex]E=\frac{1}{2}kx^2[/tex]Where:

E is the energy = 20 J

k is the sring constant = 175 N/m

x is the compression in meters

Rewrite the formula for x:

[tex]x=\sqrt{\frac{2E}{k}}[/tex]Input values and solve for x:

[tex]\begin{gathered} x=\sqrt[]{\frac{2\times20}{175}} \\ \\ x=\sqrt[]{\frac{40}{175}} \\ \\ x=\sqrt[]{0.2285} \\ \\ x=0.48\text{ m} \end{gathered}[/tex]**ANSWER:**

**0.48 m**

Remember the experiment done by Arthur Holly Compton that demonstrated the particle nature of light (X-rays) definitively. The reaction was:γ+e→γ+e (1) where the outgoing gamma was an X-ray of aa. higher?b. lower?frequency than the initial gamma. Circle your choice.

The correct option is **(b)**

The outgoing gamma rays are of lower frequency than that of the initial gamma-ray. While investigating the scattering of X-rays, Compton observed that the outgoing rays lose some of their energy in the scattering process and emerge with slightly decreased frequency.

Right answer b

The outgoing rays lose some of their energy.

The outgoing rays lose some of their energy.

Which of the following is not a unit of a force?

Answer: **B**

Explanation:

*One of the common units of force is Newtons, N*

Recall,

Force = ma

where

m is the mass of the body

a is the acceleration of the body

If acceleration, a = ms^-2 and mass = kg, then

*Force = kgms^-2*

Also, joule is the unit of work

recall,

work = force x distance = J

If we divide work by distance(meters), it becomes joule/meter

Thus,

*jm^-1 is a unit of force*

Thus, **the option that is not a unit of force is**

**B. Js^-1**

Four charges are arranged in a square formation. Take q to be 1 C of charge and a to be 2 cm in length. Four charges are arranged in a square formation. Find the net electric field at the center of the square.

We have to calculate the electric field in the center, so we need the distance to the center R and the interaction of each charge with that point

[tex]E=\sum_{n\mathop{=}0}^{\infty}E_i[/tex]To calculate the distance R we have a triangle

[tex]R=\sqrt{\frac{2a^2}{4}}=\frac{a}{\sqrt{2}}=0.014m=1.41cm[/tex][tex]\begin{gathered} \sum_{n\mathop{=}0}^{\infty}Ex=\frac{9\cdot10^9}{2\cdot10^{-4}}\cdot(cos45)\cdot(6C)=1.91\cdot10^{14}N/C \\ \sum_{n\mathop{=}0}^{\infty}Ey=4.5\cdot10^{13}\cdot(cos45\degree)(2C)=0.636\cdot10^{14}N/C \\ Etot=\sqrt{Ex^2+Ey^2}=2.01\cdot10^{14}N/C \end{gathered}[/tex]Is important to remember that E has direction so you have to calculate each axis, x, and y

How many kilometers does the space shuttle have to travel to complete one orbit? In terms of a circle, what is this distance called? Explain.

**40,840.7 ** kilometers the space shuttle have to travel to complete one **orbit**. In terms of a **circle**, this distance is termed as the circumference of the circle.

The **circumference **of a circle is the length measured around its edge. The diameter of a circle is the **distance **from the center to the outside.

Here we need to find the distance of the space shuttle that completed one circle, ie, the circumference of the **orbit**. The Circumference or distance covered by the **space shuttle** can be denoted by [tex]C_{SS}[/tex] and can be calculated by application of the below formula,

[tex]C_{SS}[/tex] = 2πr

where r is the **radius **of earth ie, 6500Km

Therefore, the equation becomes:

[tex]C_{SS}[/tex] = 2×π×r

= 2×π×6500

=40,840.7

So, the **kilometer **required to travel is **40,847.7Km**

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Gravitation always does negative work. Is this true or false?

The given statement 'Gravitation always does negative work' is false.

The work done by the gravitation depends upon the reference level of the system. Therefore, according to reference level it is taken as positive or negative.

Which of the following best explains why Venus has a higher temperature than Mercury? And Why?

A. The atmosphere on Venus is thicker than the atmosphere on Mercury.

B. The atmosphere on Venus is thinner than the atmosphere on Mercury.

C. The length of time it takes Venus to revolve around the Sun is shorter than Mercury.

D. The length of time it takes Venus to revolve around the Sun is longer than that of Mercury.

**Answer: The reason why Venus is hotter than Mercury is because it has a thick atmosphere primarily made up of carbon dioxide, which is a greenhouse gas that helps retain the heat from the Sun. In comparison to this, Mercury has almost no atmosphere, so any heat that beats down on the planet isn’t retained.**

**Explanation: Searched**

What is the electric field strength at a distance of 0.9 m from a charge of 5.71 x 10^-6 C?

Given:

[tex]\begin{gathered} Q=5.71\times10^{-6}\text{ C} \\ r=0.9\text{ m} \end{gathered}[/tex]The electric field strength is given as,

[tex]E=\frac{KQ}{r}[/tex]Here, *K* is the electrostatic constant.

Putting the values,

[tex]\begin{gathered} E=\frac{9\times10^9\times5.71\times10^{-6}}{(0.9)^2} \\ =63444.44\text{ N/C} \end{gathered}[/tex]Therefore, **the electric field strength is 63444.44 N/C.**

Fill in the blanks on machines efficiency

Efficiency = ____________ energy _________ / ____________ energy ______________

**Machine Efficiency** = output energy/ input energy

The percent of input work that becomes work done by the machine is called **efficiency**. The output work is always less as compared to the input work because some of the input work gets used in overcoming friction, therefore the efficiency is always less than 100 percent.

**Efficiency** do not have units. It is written as a decimal or as a percentage. Energy efficiency is the use of less amount of energy to perform the same task. **Energy-efficien**t homes and buildings use lesser energy to heat, cool, and run any appliances and energy-efficient manufacturing facilities use lesser energy to produce goods.

Efficiency measures work or energy that can conserved in a process. We can also say that efficiency is about comparing the **output of the energy **to the input of the **energy**.

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When doing a chin-up, a student lifts her body with a force of 400N and a distance of 0.25 meters in 2 seconds. What is the power delivered by the students biceps?

**Given:**

The student lifts her body with a force F = 400 N

The distance is d = 0.25 N

The time is t = 2 s

**To find: **The power delivered by student's biceps.

**Explanation:**

The formula to calculate power is

[tex]P=\frac{F\times d}{t}[/tex]Substituting the values, the power will be

[tex]\begin{gathered} P=\frac{400\times0.25}{2} \\ =\text{ 50 W} \end{gathered}[/tex]**Final Answer: The power delivered by student's biceps is 50 W.**

The potential of a 5.0 cm radius conducting sphere is -100 V. What is the charge density on its surface?Group of answer choices-1.8x10-8 C/m23.5x10-8 C/m22.2x10-8 C/m2-2.2x10-8 C/m2-3.5x10-8 C/m2

First lets calculate the surface area

[tex]S=4\cdot\pi\cdot r^2=4\pi\cdot0.05^2m=0.0314m^2[/tex]Now to know the surface charge density we need the next formula:

[tex]CD=\frac{q}{A}[/tex]But we are missing the amount of charge, we only have the potential

So in this case, we going to apply a different formula

[tex]V=\frac{q}{4\cdot\pi\cdot\xi\cdot r}[/tex]q=5.56*10^-10

[tex]CD=-5.56\cdot\frac{10^{-10}}{0.0314}=-1.77\cdot\frac{10^{-8}C}{m^2}[/tex]The anwer might be -1.8x10-8

The x-component of a force on a 46-g golf ball by a 7-iron versus time is plotted in the following figure: a. Find the x-component of the impulse during the intervals i. [0, 50 ms], and ii. [50 ms, 100 ms] b. Find the change in the x-component of the momentum during the intervals iii. [0, 50 ms], and iv. [50 ms, 100 ms]

The x-component of the **impulse **during the intervals [0, 50 ms] is **750 Nms.**

The x-component of the **impulse **during the intervals [50, 100 ms] is **1,500 Nms.**

The change in the **x-component** of the **momentum **during the intervals [0, 50 ms] is **0.75 kgm/s.**

The change in the **x-component** of the **momentum **during the intervals [50, 100 ms] is **1.5 kgm/s.**

What is the impulse experienced by the ball?

The **impulse **experienced by the ball is calculated from the product of **force **and **time **of motion of the ball.

J = Ft

where;

F is the applied forcet is the time of motionThe x-component of the **impulse **during the intervals i. [0, 50 ms] is calculated as follows;

From the diagram, the impulse between (0, 50 ms) is the area of the triangle.

Jₓ = ¹/₂(b)(h)

where;

b is the base of the triangle = 50 ms h is the height of the triangle = 30 NJₓ = ¹/₂(50 ms)(30 N) = 750 N.ms

The **impulse **during [50 ms, 100 ms] is the area of the rectangle,

Jₓ = Lb

where;

L is the length = 100 ms - 50 ms = 50 msb is the breadth = 30 NJₓ = 50 ms x 30 N

Jₓ = 1,500 Nms

**Impulse **is the change in **momentum **of an object.

The change in the **x-component** of the **momentum **during the intervals [0, 50 ms] is calculated as follows;

ΔP = Jₓ = ¹/₂(50 ms)(30 N) = 750 N.ms = 0.75 Ns = 0.75 kgm/s

For interval of [50 ms, 100 ms];

ΔP = 1,500 Nms = 1.5 Ns = 1.5 kgm/s

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**Answer:**

We had this question yesterday, let me check my book quickly

(A) The **acceleration** of the ball is 9.81 m/s² just before it strikes the ground.

(B) The** initial speed** of the ball is equal to 1.14 m/s.

(C) The** initial speed **must be 3.21 m/s if it is to land with a **speed** of 5 m/s.

The** equation of motion** is the way to represent the relation between the time, **acceleration**, initial and final **velocity**, and **distance** covered by a moving object.

The** three equations of motion** are:

[tex]v= u+ at\\v^2 = u^2 +2aS\\S = ut +(1/2) at^2[/tex]

The **acceleration** of the ball just before it strikes the ground is equal to **gravitational acceleration**, g = 9.81 m/s².

Given, the **final velocity** of the ball, v = 4m/s

The **height** of the table, h = 0.75 m

The initial **velocity** of the ball, [tex]u = \sqrt{v^2-2gh}[/tex]

[tex]u = \sqrt{(4)^2-2\times 9.8\times 0.75}[/tex]

u = 1.14 m/s

When the **final velocity **of the ball, v = 5m/s

The **initial velocity** will be :[tex]u = \sqrt{(5)^2-2\times 9.8\times 0.75}[/tex]

u = 3.21 m/s.

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Which of these does sound travel the fastest Through?A. steel SolidB. water LiquidC. Air gasD. Sound travels the same through all mediums

Speed of sound is maximum on solid.

Correct option.

A) steel Solid

A flat coil of wire has an area of 0.020 m2 and contains 50 turns. Initially the coil is oriented so that the normal to its surface is parallel to and in the same direction as a constant magnetic field of 0.18 T. The coil is then rotated through an angle of 30o in a time of 0.10 s. What is the average induced emf? -0.44 V +0.44 V +0.24 V

**ANSWER:**

3rd option: +0.24 V

**STEP-BY-STEP EXPLANATION:**

Given:

N = 50

Area = 0.020 m^2

B = 0.18 T

θf = 30°

time = 0.10 s

We can calculate the average induced emf by the following formula

[tex]\epsilon=N\cdot B\cdot A\cdot\left(\frac{\cos\theta_i-\cos\theta_f}{t}\right)[/tex]We replacing:

[tex]\begin{gathered} \epsilon=\left(50\right)\left(0.18\right)\left(0.02\right)\left(\frac{\cos\:0\degree\:-\cos\:30\degree}{\:0.1}\right) \\ \epsilon=0.241\cong0.24\text{ V} \end{gathered}[/tex]The correct answer is 0.24V

A roller coaster car is traveling through a loop at 13 m/s. If the loop has a 23m radius, what centripetal force will the 53kg rider feel?

Given,

The velocity of the car, v=13 m/s

The radius of the loop, r=23 m

The mass of the rider, m=53 kg

The centripetal force is the force that keeps an object in its circular path.

The centripetal force is given by,

[tex]F=\frac{mv^2}{r}[/tex]On substituting the known values,

[tex]\begin{gathered} F=\frac{53\times13^2}{23} \\ =389.43\text{ N} \end{gathered}[/tex]**Therefore, the centripetal force on the rider is 389.43 N**

Compare(how they are the same) and contrast(how they are different) psychoanalysis and behaviorism as two of the early schools of psychology. Offer the major names associated with each and explain how each explained behavior.

**Abstract **of American Intercontinental University

This paper will compare and contrast three of the **10 main **early psychology views. This** assignment's** three approaches are behavioral, humanistic, and cognitive. Three Early **Psychology** Perspectives are **compared **and **contrasted**. Psychology, like anything else, offers a plethora of **theories** and methods. One idea may be **beneficial **to one patient while being ineffective to another. The trick is to discover the **optimal** method for each patient. The concept that behaviors arise as a result of conditioning is known as **behaviorism**. This theory does not acknowledge the presence of interior mental factors such as thoughts, feelings, and moods, nor does it take free will into** account.**

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QUESTION 2 (NOVEMBER 2014) Two blocks of masses 20 kg and 5 kg respectively are connected by a light inextensible string, P. A second light inextensible string, Q, attached to the 5 kg block, runs over a light frictionless pulley. A constant horizontal force of 250 N pulls the second string as shown in the diagram below. The magnitudes of the tensions in P and Q are T, and T, respectively. Ignore the effects of air friction.

2.3 Calculate the magnitude of the tension T, in string P. (6)

The magnitude of **tension** T in string P is 250 N when a constant horizontal force of 250 N pulls the second string.

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draw each of the following vectors, label an angle that specifies the vectors direction, then find its magnitude and direction.a. B= -4.0 I+ 4.0jb. r= (-2.0i-1.0j) cmc. v= (-10-100j) m/s d. a= (20i+10j) m/s^2the I's and j's have the hat and its the vector simble on the letters

[tex]\begin{gathered} A) \\ B=-4.0i+4.0j \\ |B|=\sqrt{(-4.0)^2+(4.0)^2} \\ |B|=\sqrt[]{16^{}+16^{}} \\ |B|=\sqrt[]{32^{}} \\ |B|=4\sqrt{2} \\ \text{The magnitude of B is }4\sqrt[]{2} \\ \theta=\tan ^{-1}(\frac{4.0}{-4.0})=135\text{ \degree} \\ The\text{ angle of B is }135\text{ \degree} \\ B) \\ R=(-2.0i-1.0j)cm \\ |R|=\sqrt[]{(-2.0)^2+(-1.0)^2} \\ |R|=\sqrt{4.0+1.0} \\ |R|=\sqrt[]{5.0} \\ \text{The magnitude of R is }\sqrt[]{5.0}cm \\ \theta=\tan ^{-1}(\frac{-1.0}{-2.0})=26.57\text{ \degree, but it is below of negative x-axis, hence} \\ \theta=180+26.57=206.57\text{ \degree} \\ \text{The angle of R is }206.57\text{ \degree} \\ C) \\ V=(-10i-100j)\text{ m/s} \\ |V|=\sqrt{(-10)^2+(-100)^2} \\ |V|=\sqrt[]{100+10000} \\ |V|=\sqrt[]{10100}=10\sqrt{101}\approx100.5\text{ m/s} \\ \text{The magnitude of V is }100.5\text{ m/s} \\ \theta=\tan ^{-1}(\frac{-100}{-10})\approx264.29 \\ \text{The angle of V is }264.29\text{ \degree} \\ \\ D) \\ A=(20i+10j)m/s^2 \\ |A|=\sqrt{20^2+10^2} \\ |A|=\sqrt{400+100} \\ |A|=\sqrt{500}=10\sqrt{5}\approx22.36m/s^2 \\ \text{The magnitud of A is }22.36m/s^2 \\ \theta=\tan ^{-1}(\frac{10}{20})=26.57\text{ \degree} \\ \text{The angle of A is }26.57\text{ \degree} \end{gathered}[/tex]

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