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zavuch27 [327]
2 years ago
9

A rock is rolling down a hill. At position 1, it’s velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, it’s velo

city is 44.0 m/s. What is the acceleration of the rock?
42.0 m/s^2
3.7 m/s^2
3.8 m/s^2
3.5 m/s^2
Physics
1 answer:
mr Goodwill [35]2 years ago
5 0

Answer

Hi,

correct answer is {D} 3.5 m/s²

Explanation

Acceleration is the rate of change of velocity with time. Acceleration can occur when a moving body is speeding up, slowing down or changing direction.

Acceleration is calculated by the equation =change in velocity/change in time

a= {velocity final-velocity initial}/(change in time)

a=v-u/Δt

The units for acceleration is meters per second square m/s²

In this example, initial velocity =2.0m/s⇒u

Final velocity=44.0m/s⇒v

Time taken for change in velocity=12 s⇒Δt

a= (44-2)/12  = 42/12

3.5 m/s²

Best Wishes!

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A person driving a car suddenly applies the brakes. The car takes 4 s to come to rest while traveling 20 m at constant accelerat
Zinaida [17]

Answer:

Yes we can find the initial velocity of car without finding acceleration.

u = 10 m/s.

Explanation:

Given that

s=20 m

Car takes 4 s to come in rest.

We know that when acceleration is constant then we can apply motion equation

v=u+at        ----------1

s=ut+\dfrac{1}{2}at^2       ------2

From equation 1 and 2

s=ut+\dfrac{1}{2}t^2\left (\dfrac{v-u}{t} \right )

So we can say that

s= \left(\dfrac{v+u}{2}\right)t

Given that the velocity of car at final condition will be zero (v=0)

s= \left(\dfrac{0+u}{2}\right)t

s= \left(\dfrac{u}{2}\right)t

From the above equation we can find the initial velocity of car without finding the acceleration

20= \dfrac{u}{2}\times 4

u = 10 m/s

7 0
2 years ago
A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it
Lisa [10]

Answer:

Vertical distance=  3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time=  0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

3 0
2 years ago
A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block
solniwko [45]
We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
4 0
2 years ago
An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

where in the second equation, the term m\frac{v^2}{R} represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the  following expression:

tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

In this second situation, the cars moves around the track at a speed

w>v

This means that the centripetal force term

m\frac{v^2}{R}

is now larger than before, and therefore, the horizontal component of the normal reaction, N sin \theta, is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

N sin \theta + F = m\frac{w^2}{R}

And re-arranging for F,

F=m\frac{w^2}{R}-N sin \theta (3)

But from eq.(2) in the previous part we know that

N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
2 years ago
10 kg cart and a 5 kg cart are placed on identical surfaces. The 10 kg cart experiences a net force of 12 N to the left, while t
SashulF [63]
F=ma

For the first (10kg) cart,
12=10a
a=6/5 m/s^2 to the left

For the second (5kg) cart,
8=5a
a=8/5 m/s^2 to the left

Therefore, the lighter (5kg) cart experiences a greater acceleration.
7 0
2 years ago
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