A common occurrence is the **collision** of a moving object with another object, after which both objects stick together. Because there is no bounce, such a **collision** is referred to as** inelastic**.

**Inelastic collisions** occur when** kinetic energy** is lost. An **elastic collision**, on the other hand, is one in which the **kinetic energy **after is equal to the** kinetic energy** before.** Inelastic collisions** occur when there is some "bounce" but the** final kinetic energy **is less than the** initial kinetic energy**. A **collision **with no bounce is sometimes referred to as completely** inelastic**.** Collisions **that are completely** inelastic** result in the objects sticking together and lose the most** kinetic energy**.

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a 2000 kg elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of the shaft. the spring is intended to stop the elevator, compressing 2.00 m as it does so. during the motion a safety clamp applies a constant 17,000 n friction force to the elevator.

A. The **speed** of the **elevator** after it has moved **downward** 1.00 m from the point where it first contacts a spring is 3.65m/s

B. The **acceleration** when the **elevator** is 1.00 {\rm m} below point where it first contacts a **spring** is 4m/s²

In calculating the speed of the elevator and acceleration, first we have to find the force of **gravity** F on the elevator, which is the force pulling the elevator in downward direction. Using the equation for force of gravity which is:

F = mg

Where:

Mass of the elevator; m= 2000kg

Acceleration due to gravity; g = 9.8m/s

2000kg × 9.8m/s²= 19600N

F = 19600

Force of opposing friction clamp of gravity = 17000N

Net force on the elevator = force of gravity - Force of opposing friction clamp

Net **force** on the elevator = 19600 - 17000

Net force on the elevator = 2600 N

We will also find the kinetic energy K.E; of the elevator at the point of contact with the spring using:

K.E = 1/2 mv²

Where

Mass of the elevator; m = 2000kg

Velocity of the elevator = 4.00m/s

K.E = (1/2)*2000kg*(4m/s)²

K.E = 16000J

The kinetic energy and energy gained will be absorbed by the spring across the next 2m

Therefore,

Energy; E = K.E + P.E

Where:

Kinetic energy K.E = 16000J

Potential Energy P.E = ?

P.E of spring = net force absorbed × distance at compression

Where:

Net force absorbed = 2600N

Distance at compression = 2.0m

P.E = 2600*2

P.E = 5200J

E = 16000J + 5200J

E = 21200J

Spring constant = k

To find k

Using:

E = (1/2)*k*(x)²

Where:

E = 21200J

k = ?

x = 2m

21200J = (1/2)*k*(2m)²

21200J*2 = (4m)k

K = 42400J/4m

K = 10600N/m

Therefore,

Acceleration at 1m compression = ?

Using:

F = K*X

Where

F is force provided by the spring = 10600N/m,

K = 10600 N/m

X = 1m

F = 10600N/m * 1m

F = 10600N (upward)

A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?

Using:

Original Kinetic energy + net force on the elevator = final kinetic energy + spring energy

16000N + 2600N = (1/2)mv² + (1/2)k x²

18600 = (1/2)(2000)(v²) + (1/2)(10600N)(1²)

18600 = 1000(v²) + 5300

18600 - 5300 = 1000(v²)

13300 = 1000(v²)

V² = 13.300

V =3.65m/s

B. The acceleration of the elevator is 1.00m below point where it first contacts a spring

Spring constant = net force on the elevator + resultant force

Where:

Spring constant = 10600N

Net force on the elevator = 2600N

Resultant force = ?

10600N = 2600N + resultant force

Resultant force = 10600N - 2600N

Resultant force = 8000N

Using the equation for Newton's 2nd law where F = ma,

a = F/m

Where:

Resultant force; F =8000N

Mass of the elevator; m =2000kg)

a = 8000 / 2000

a = 4m/s²

Here's the **complete** question:

In a "worst-case" design scenario, a 2000kg elevator with broken cables is falling at 4.00m/s when it first contacts a cushioning spring at thebottom of the shaft. The spring is supposed to stop the elevator,compressing 2.00m as it does so. During the motion a safety clampapplies a constant 17000N frictional force to the elevator.

1. What is the speed of the elevator after it has moved downward 1.00m from the point where it first contacts aspring?

2. When the elevator is 1.00m below point where it first contacts a spring, what is its acceleration?

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According to Newton's first law of motion, what will an object in motion do when no external force acts on it?

come to a stop

move at the same velocity

speed up

change direction

Mark this and return

**Answer:**

Move at the same velocity

**Explanation:**

According to newton's 1st law of motion, an object at rest remains at rest or if in motion, it will remain in motion at constant velocity unless acted upon by a net external force

a block is given an initial velocity of 5.00 m/s up a frictionless incline of angle 20 degrees. the magnitude of the block’s acceleration is 2.5 m/????2 while it is on the ramp. how far up the incline does the block get before sliding down again?

The block will incline up with **projectile motion** until** 0.085 m **before sliding down.

We need to know about the **projectile motion** to solve this problem. The **projectile motion** is known as parabolic motion and the velocity is divided by 2 axes.

vox = vo cosA

voy = vo sinA

where vox is initial velocity of x axis, voy is initial velocity of y axis, vo is initial velocity and A is the angle.

From the question above, the given parameters are

vo = 5 m/s

A = 20⁰

By using the uniform motion, where vty = 0 m/s. Hence,

vty² = voy² - 2gh

0 = 5.sinA - 2 . 10 . h

20h = 5 . sin20⁰

h = 1/4 . 0.34

h = 0.085 m

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if the parachutist comes to rest with constant acceleration over a distance of 0.660 mm , what force does the ground exert on her?

The **force** exerted by the parachutist on the **ground **is 471.24 N when she comes to rest at a distance of 0.660 m

Let t = Time taken

u = Initial **velocity** , v = Final velocity

s = Displacement, a = **Acceleration**

As given in the problem initial velocity u = 3.85 m/s

Final velocity v = 0 (rest)

s = 0.660 m

From the equation of **motion** v² = u² + 2as

we have a = v²-u² / 2s

⇒a = 0-3.85² / (2* 0.660)

⇒a = -11.22 m/s² (negative sign shows deceleration)

we know force F = mass m * acceleration a

Given **mass** of **parachutist** m = 42 kg

∴**Force exerted **by parachutist on **ground **

F = 42 * 11.22 = 471.24 N

If the distance is shorter, acceleration increases in magnitude and so does the force exerted by the parachutist.

*The question is incomplete and the full question is probably " A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant acceleration over a distance of 0.660 m, what force does the ground exert on her? "*

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Which of the following scenarios would most likely be considered pseudoscience?

A team of scientists replicating another experiment.

Generating a claim based on empirical evidence.

The theory of relativity.

The belief that the stars determine our fate.

The belief that the **stars** determine our fate is considered to be **pseudoscience .**

Pseudoscience, which is defined as an activity that purports to be scientific but is actually not, is defined as seems science. Science does not engage in extraterrestrial affairs it simply studies the natural world. Science cannot provide an answer to morality, immorality, or questions of meaning or purpose, but it does require ethical behavior. In other hand astronomy is regarded as pseudoscience since no link between the locations of the stars and human behavior can be demonstrated using evidence from science.

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**Answer: The answer is The belief that the stars determine our fate or C**

**Explanation:**

While on the moon, Buzz Aldrin carried on his back a support system that would

weigh over 1,775 N on Earth. What did the backpack weigh on the moon? (gravity

on the moon is 1.62 m/s^2)

**Answer:**

W' = 287.55 N

**Explanation:**

We know that weight force is defined as follows:

**W = m × g**

So:

On Earth (W = 1775 N):

1775 = m × 9.8

m = 1775 / 9.8

m = 181.12 kg

On the Moon (g' = 1.62 m/s²):

W' = m × g'

W' = 181.12 × 1.62

W' = 293.42 N

A rifle is aimed horizontally at a target 53 m away. The bullet hits the target 2.3 cm below the aim point.

The** time of flight** of bullet is 0.0685 seconds and the** speed** when it emerges out of the gun is 775.

As the bullet is fired horizontally from the gun at a distance of 53m away. But the bullet hits the target 2.3cm below the target.

So, the** vertical displacement** of the bullet is 2.3cm. The accelerations in the vertical direction will be g (acceleration due to gravity).

(a) As we know, in case of gravity, we can use** equations of motion,**

So, using** equation,**

S = ut + 1/2gt²

Where, u is the initial velocity of the bullet which is 0m/s in this case,

S is the vertical displacement of the bullet,

g is **acceleration due to gravity** and,

t is **time of flight** of the bullet.

Putting all the values,

S = 1/2gt²

2.3/100 = 1/2(9.8)(t²)

0.023 = 4.9t²

t² = 0.00469387

t = √0.0046987

t = 0.0685 seconds.

(b) The velocity as it emerges from the gun is equal to the muzzle velocity of the gun,

So, we can write,

**Muzzle velocity V** =** Horizontal distance** covered/**time taken.**

V = 53/0.0685

V = 773.72 m/s.

One thing that should be noted here is that the initial velocity that we took as zero in part (a) is taken with respect to the vertical motion and nit the Horizontal motion. So, one should not confuse the muzzle velocity in part (b) with the initial velocity mentioned in the part(a)

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Complete question - A rifle is aimed horizontally at a target 53 m away. The bullet hits the target 2.3 cm below the aim point. Find

(a) the bullets time of flight and (b) its speed as it emerges from the rifle?

the trees in a wooded area near a factory look like burnt toothpicks. what might be causing the problem?

Acid rain, smoke, and exhaust from the factory.

Hope this helps pls mark brainliest if it did :)

Hope this helps pls mark brainliest if it did :)

two small, identical steel balls collide completely elastically. initially, ball 1 is moving with velocity ????1 directly toward ball 2, and ball 2 is stationary. what are the final velocities of ball 1 and ball 2, respectively?

When two small ball **collide elastically** in which first ball is moving with **velocity, v** with second ball which is in** rest **than after **collision** the second ball will** move** with** velocity, v** and **first** ball will come to **rest**.

An **elastic collision** is one in which the system does not experience a net **loss** of **kinetic energy** as a result of the **collision**. In e**lastic collisions**, **momentum** and **kinetic energy** are both **conserved**.

An** elastic collision** is one in which the **momentum**, **p** and the** kinetic energy, KE** are both conserved.

[tex]p_{i} =p_{f}[/tex]

[tex]KE_{i} = KE_{f}[/tex]

Given in the question two identical **steel ball** that means have same mass and one is moving with **velocity, v** another is in the rest initially, after **collision** the **momentum** of first ball will transfer to second ball so second ball will move with **velocity, v** and first ball will come to** rest**.

So final **velocity** of first ball is **zero** and second ball is v.

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Can anyone give me the step by step solution for this question?

**Answer:**

ans a

**Explanation:**

The answer would be a as it would have to be the two wave heights added together which equals 15

A cliff jumper runs off a 15\,\text m15m15, start text, m, end text high cliff. Rocks extend 3.0\,\text m3.0m3, point, 0, start text, m, end text past the cliff bottom. We can ignore air resistance.

1.7 m/s is the **horizontal velocity** that is required to clear the rocks.

As the jumper runs off a 15 m high cliff.

And according to the question, starts with 0 velocities in the start direction thus he covers 15 m.

Given,

x = 15 m

a = g = 9.8 m, which is the **acceleration** due to **gravity**.

iv is the initial velocity which is 0, vo = 0

vf is the final velocity,

Using 2nd equations of motion,

x = iv × t + (1 ÷ 2) a × t²

15 = 0 + ((1 ÷ 2) × ( 9.8) × t²)

t = 1.75

Thus, the time is 1.75 seconds.

Using the same equation, **horizontal velocity** will be:

x = iv × t + (1 ÷ 2) a × t²

3 = vf × (1.75) + 1 ÷ 2 × (0) × (1.75)

vf = 1.7

Therefore, The horizontal velocity required to cover the rocks is 1.7 m/s.

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What do we call the illusion of movement that results from two or more stationary, adjacent lights blinking on and off in quick succession?.

In **1912 Wertheimer** discovered the **phi** **phenomenon**, an **optical illusion** in which stationary objects shown in rapid **succession**, transcending the **threshold** at which they can be perceived separately, appear to move.

When two neighbouring optical **stimuli** are presented in alternation with a relatively high **frequency**, an apparent **motion** is seen that is known as the **phi** **phenomenon**. At higher frequencies, beta movement is visible, but the stimuli themselves don't seem to move.

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a race car driver is driving his car at a constant speed of 51.0 m/s on a circular track with a radius of 235 m. (a) what is the angular speed (in rad/s) of the car? rad/s (b) what are the magnitude (in m/s2) and direction of the car's acceleration? magnitude m/s2 direction

0.19362 rad/s is the **angular speed** and 8.8096 m/s2 is car's **acceleration.**

[tex]angular speed=\omega=v/r=45.5m/s/235m=0.19362rad/sec[/tex]

[tex]car acceleration =a=v^2/r=45.5^2m/s^2/235m=8.8096m/s^2[/tex]

The definition of angular speed is the rate of **change **of **angular displacement**, which is the angle a body travels along a circular route. The ratio of the number of rotations or revolutions performed by a body to the time taken is used to compute angular speed. The** Greek letter ** **Omega**, stands for angular speed.** Rad/s** is the angular speed unit in the SI.

As is common knowledge, **angular speed** is defined as the rate of change of displacement divided by the time. Therefore, the equation for angular speed is = /t. Despite the **aforementioned **method, there is a different and more popular **formula **for calculating angular speed when it comes to competitive tests. As** ω = θ/t**

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Bode made a rocket using an empty plastic bottle and fins made out of cardboard. After partially filling the bottle with water, he pumps air into the bottle using a pump attached to the rocket with plastic tubes. When the pressure in the bottle becomes high enough, the water gushes out and launches the rocket into the air, as shown in the diagram below.

How does the force that causes the water to leave the rocket compare to the force that causes the rocket to launch?

A.

They are unequal and act in opposite directions.

B.

They are equal and act in the same direction.

C.

They are equal and act in opposite directions.

D.

They are unequal and act in the same direction.

The force that causes the water to leave the **rocket **compared to the **force **that causes the rocket to launch is that they are unequal and act in opposite directions.

According to Newton's third law of **motion**, every action has a corresponding and opposing response. In the instance of Bode's rocket, the pressured air in the bottle causes the water to rush downward out of the bottle, causing the rocket to fly upward into the air. The upward force on the **rocket **is equivalent to the force of the water streaming down. Thus, the two forces are equivalent and exert opposing **forces**.

When you pump, more **air **is pumped into the bottle while maintaining the same **volume**. By doing this, you raise the pressure inside the bottle, increasing the force acting on the interior of the bottle. Eventually, when the bottle can no longer withstand the pressure, the cork will be violently ejected from the bottle's bottom by the force of the **pressure**.

The air will be forced out of the **bottle **as rapidly as possible by the abrupt pressure drop caused by the new opening. When the air leaves the bottle, the pressure that was previously holding it under pressure is used to drive all of the **water **out with it.

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**Answer:**

They are equal and act in opposite directions.

**Explanation:**

**I just answered the question on study island and got it right :P**

Newton's Law with Friction (Force on Angle) please help

The **Acceleration **of the system as depicted in the diagram is **1.38 m/s².**

**Acceleration **is the rate of change of velocity. The S.I unit of **acceleration **is m/s²

To calculate the **acceleration **of the system, we use the formula below.

Formula:

a = (Fcos∅-mgμ)/m.......... Equation 1Where:

a =From the question,

Given:

m = 1.1 kgF = 5.4 N∅ = 54.6°μ = 0.15g = 9.8 m/s²Substitute these values into equation 1

a = {(5.4cos54.6°)-(1.1×9.8×0.15)}/1.1a = (3.13-1.617)/1.1a =Hence, the **acceleration **of the system is** 1.38 m/s².**

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A merry-go-round spins freely when janice moves quickly to the center along a radius of the merry-go-round. It is true to say that.

No, its fake the instant of **inertia** of the **machine** decreases and the **angular** pace will increase.

A **merry**-go-spherical spins freely while **janice** actions fast to the middle alongside a radius of the merry-go-spherical. Moment of inertia is a amount withinside the rotational movement which performs a position analogous to position performed with the aid of using mass withinside the linear **movement**. Hence, it's also referred to as angular **mass** or the rotational momentum.

Janice decreases her contribution to the instant of inertia of the machine with the aid of using transferring towards the **rotation** axis. This decreases the whole second of inertia of the machine. The angular **momentum** L of the machine does now no longer extrade seeing that no outside **torque** is applied. The **angular** pace ω will increase sinceω = L/I.

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An astronaut jumped on Mars. His initial velocity

was 1.88 m/s. If the gravitational acc. is -3.5-m/s²

on the moon,

a) What is the TOTAL time the astronaut is in the

atmosphere?

b) How high did the astronaut jump?

Due to the **law of gravit**y, or** gravitation,** which draws all objects with mass in its vicinity toward it, **Mars**' gravitation is a natural occurrence.

.53 s the astronaut is in the atmosphere.

astronaut jump form .49 m.

What is Mars famous for?The rusty red color offrom motion Equation : v= u + at

then we can find t = [tex]\frac{v-u}{a}[/tex] = -1.88/ 3.5 = .53 sec

s = ut + ½at²

put u = 1.88 a= -3.5 and t= .53

s = .49 m

.53 s the astronaut is in the atmosphere.

astronaut jump form .49 m.

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7) You push on box G that is next to box H, causing both boxes to slide along the floor, as shown in the figure.

The reaction force to your push is

Push

GH

A) the push of box G against you. B) the acceleration of box G.

C) the push of box H on box G. D) the push of box G on box H.

**Answer:**

c

**Explanation:**

because whenever you push g box into h box and then h displaces according to third law of motion h box will apply the force on g box so that's why

a 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. it accelerates upward at 30 m/s2 for 34 s, then runs out of fuel. ignore any air resistance effects.

A rocket with 200kgs is loaded extra 100kgs then the Rocket maximum altitude is **93005 m** and total **Time-of-Flight** is **294 s**

The **Time-of-Flight** concept (ToF) is a technique for determining distances between **sensors **and objects based on the amount of time that passes between a signal being sent and returning to the **sensor **after being reflected by an object. The Time-of-Flight principle can be employed with a variety of signals (also known as carriers), with sound and light being the most popular ones. The time it takes for photons to travel between two points—from the **sensor's **emitter to a target and back—in order to calculate distance for all Time-of-Flight (ToF) sensors.

Direct and indirect ToF both have unique advantages in particular situations. Each pixel in a scene can be simultaneously measured for distance and intensity.

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What acceleration will result when a 44.7-N net force applied to a 7.7-kg object?

The **acceleration **when a 44.7 N net force is applied to 7.7 kg object is 5.8 m / s²

According to Newton's second law of motion,

**F = m a **

F = Force

m = Mass

a = Acceleration

m = 7.7 kg

F = 44.7 N

a = F / m

a = 44.7 / 7.7

a = 5.8 m / s²

**Newton's second law of motion** states that the force is equal to the rate of change of momentum. It can also be said as acceleration of an object is directly proportional to the net force acting on the object.

Therefore, the acceleration of the object is **5.8 m / s²**

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arunner jack 12km north then turns and runs 16km in 3hours calculate his average velocity

A runner jack 12km north then turns and runs 16km in 3hours Therefore his **average velocity **is 1.3 km/h.

**What is average velocity?**

The change in position or displacement (x) divided by the time intervals (t) during which the displacement happens is the definition of **average velocity**. Based on the direction of the displacement, the **average velocity **might either be positive or negative. Meters per second (m/s or ms-1) is the SI measure for** average velocity**.

Given:

A runner jack 12km north then turns and runs 16km in 3hours .

Therefore **average velocity**= (16 - 12)/ 3 km/h

=1.3 km/h

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the moderate temperatures of islands throughout the world has much to do with water's vast supply of internal energy. high evaporation rate. absorption of solar energy. high specific heat. poor conductivity.

**High specific heat** of the water. Option (c)

**What is Specific heat?**

The amount of heat required to increase the temperature of one gram of a substance by one degree Celsius is referred to as the substance's **specific heat**. Typically, calories or joules are used per gram and degree Celsius when referring to the units of **specific heat. **

The moderate temperature of islands has much to do with the water's high **specific heat**. The typical off-water is more significant than this clear land or soil. Due to this fact, water absorbs and releases eat more slowly. In comparison to the land.

Hence, the water has high** specific heat.**

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a plane accelerates from 4 m/s/s for 28.7 s until it finally lifts on the ground. Assuming the plane started from the rest, what was the planes displacement before takeoff?

There were 1647 before liftoff. 38 m** moving **planes.

An object is considered to be in** motion **if its location throughout time changes in relation to its surroundings. It is the gradual change in an object's location. The only kind of **motion** that exists is **linear motion**.

The three equations are as follows: S = ut + 12at2, v = u + at, v2 = u + 2as.

The airplane's **initial speed** is U=0 m/s.

The plane accelerates at a 4 m/s rate.

with a time stamp of 28.7 seconds.

Distance traveled using the formula S=ut+a

⇒ S=0(28.7)+0.5(4) (4)

=1647.38 ~ 1647m

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The largest and the smallest balls used in the experiment are with diameter 9. 52 mm, and 2. 38 mm respectively. For a glycerin with viscosity 1. 0 pa. S, what is the time necessary for each ball to reach a velocity 95% of the terminal velocity? density of the ball material is given in the text. Round the result to three decimal places.

The **time **necessary for the larger ball is 66.904 x 10⁻³ s and for the smaller ball is 4.166 x 10⁻³ s.

We have ,

density of balls(ρ) = 1.42 g/cm³

**density **of glycerin(σ) = 1.30 g/cm³

**acceleration **due to gravity = 9.8 m/s²

Diameter of larger ball = 9.52 mm

diameter of smaller ball = 2.38 mm

we know that the formula for terminal velocity is given by

[tex]V_{t}[/tex] =( [tex]\frac{2r^{2}g}{9n}[/tex] ) (ρ - σ)

now for larger ball **terminal velocity **is given by

[tex]V_{1}[/tex] = (2 [9.52/2 x 10⁻³]² x 9.8 x(1.42 x 10⁻³ - 1.30 x 10⁻³ )) **/ **9 x 1.0

∴ [tex]V_{1}[/tex] = 5.92 x 10⁻³ m/s

now for smaller ball terminal velocity is given by

[tex]V_{s}[/tex] = (2 [2.38/2 x 10⁻³]² x 9.8 x(1.42 x 10⁻³ - 1.30 x 10⁻³ )) **/ **9 x 1.0

[tex]V_{s}[/tex] = 0.37 x 10⁻³ m/s

Now 95 % of the terminal velocity of larger ball will be = 5.62 x 10⁻³ m/s

95 % of the terminal velocity of smaller ball will be = 0.35 x 10⁻³ m/s

Also the acceleration = 0.084 m/s⁻²

using the equation of motion

v= u + at ( v= 95% velocity and u = initial velocity (0) )

as u = 0

∴ v = at

t = v/a

so we can calculate **time **as follows

t (larger ball) = 5.62 x 10⁻³/0.084 = 66.904 x 10⁻³ seconds

t (smaller ball)= 0.35 x 10⁻³/ 0.084 = 4.166 x 10⁻³ seconds

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A 75 kg student drives a 650 kg car. What is the applied force on the car if the car

starts from rest and accelerates to 60 m/s in 8 seconds?

**Answer:**

5437.5

**Explanation:**

I took the quiz.

help

What causes polar jet streams to form? (1 point)

A.rapid changes in wind direction

B.air masses moving away from each other

C.warm and cold air masses

D.high-pressure systems

**Polar jet streams** are formed because of the warm and cold air masses.

**Polar front jet **stream, also called **polar front jet** or mid-latitude jet stream stream, a belt of powerful upper level winds that sits at top the polar front.

The winds are strongest in the** troposphere**, which is the upper boundary of the **troposphere**, and move in a generally westerly direction in mid-latitudes.

The vertical wind shear which extends below the core of this** jet** **stream** is associated with horizontal temperature gradients that extend to the surface. As a consequence, this** jet **manifests itself as a front that marks the division between colder air over a deep layer and warmer air over a deep layer.

**Polar jet streams** are formed due to warm and cold air masses.

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Answer:

C

Explanation:

Isaac is practicing his volleyball skills by volleying a ball straight up and down, over and over again. His teammate

Marie notices that after one volley, the ball rises 3.6 m above Isaac's hands. What is the speed with which the ball left

Isaac's hand? (8.4 m/s) ANSWER FAST PLEASE!

The action of volleying a ball straight up and down comes under the free-fall motion of the ball. **Free-fall **is defined as the motion of a body which falls freely due to the gravitational pull of the earth.

The** velocity **(v) of the ball undergoing free-fall can be calculated using the free-fall formula,

[tex]v^{2}[/tex] = 2gh

where, g is the acceleration due to **gravity** and h is the height of the freely falling body.

Given that, the **height **to which ball rises, h = 3.6m. (g = 9.8 m/[tex]s^{2}[/tex])

Substituting h and g in the **velocity **formula,

[tex]v^{2}[/tex] = 2 × 9.8 m/[tex]s^{2}[/tex] × 3.6m

[tex]v^{2}[/tex] = 70.56 [tex]m^{2}/s^{2}[/tex]

The** velocity** of volley ball is,

v = [tex]\sqrt{70.56m^{2}/s^{2} }[/tex]

∴ v = 8.4 m/s

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Clare de Iles is shopping. She walks 16m North to the end of an aisle. She then makes a right hand turn and walks 21m Eastward to the end of the aisle. Determine th emagnitude of Claire's resultant displacement. And the angle of her resultant.

The magnitude and direction of Clare's **resultant displacement** is **26.4 m** and **37.3⁰** respectively.

The magnitude of Clare's **resultant displacement** is calculated as follows;

her initial position, a = 16 m north

Final position, b = 21 m east

Resultant displacement, d = √a² + b²

d = √(16² + 21²)

d = 26.4 m

The **direction **of the resultant displacement is calculated as follows;

tan θ = 16 / 21

where;

θ is the direction of the displacement obtained by formingtan θ = 0.762

θ = arc tan(0.762)

θ = 37.3⁰

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a rotor rotating about a fixed axis through its mass center, the rotor has a mass of 60 kg, radius of gyration of 70 mm, and angular acceleration of 9 rad/s2. determine the moment the fixed axis.

Given

60 kg of mass, 70 mm or 0.07 meters of **gyroscopic** radius, and 9 rad/sec2 of **angular acceleration**

**Inertia**, It is an object's resistance to any change in motion. The amount of inertia is directly related to an object's mass; in fact, as we frequently observe, the heavier an object is, the harder it is to change its direction of motion.

The moment of inertia must be identified. EM = I > I > MK2

I stands for inertia mass moment.

I = 60 kg/m2 M = I = 0.216 kg/m2 = 11 kg/m2 M = 2.376 nm

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**Answer:**

**Explanation:**

Given:

m = 60 kg

R = 70 mm = 0.070 m

ε = 9 rad/s²

__________________

M - ?

**The moment:**

M = J·ε = m·R²·ε

M = 60·0.070²·9 ≈ 2.65 N·m

it takes 4.0 μj of work to move a 15 nc charge from point a to b. it takes -5.0 μj of work to move the charge from c to b. part a what is the potential difference vc−va ? express your answer in volts.

W is the work done by charge from A to B and from C to B then the **potential difference** (vc−va)=**552.62V.**

The difference between the energy levels that charge carriers have at two places in a circuit is known as the **potential difference**.

**Voltage**, commonly known as potential difference (p.d. ), is a unit of electrical potential. The charge carriers in a circuit pass through the electrical components, transferring energy to them. We gauge the potential difference using a voltmeter (or **voltage**).

V = I x, the potential difference formula

The quantity of current times the resistance equals the potential difference (also known as **voltage**), which is equal to that. When a charge travels between two locations in a circuit, a potential difference of one Volt is equivalent to one Joule of energy being used by that charge.

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