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DiKsa [7]
1 year ago
10

Whipple is confused about the connection between the velocity and acceleration of the tennis ball. he decides to compare the vel

ocity of the ball and the acceleration of the ball at the apex of the ball's motion while varying the speed of the elevator. he runs the experiment for a wide range of positive speeds of the elevator and records the speed and acceleration of the ball at the top of its motion. what can whipple conclude from these data? the speed of the ball is always positive and the acceleration is always equal to −g at the instant the ball reaches the top of its motion. the speed of the ball and the acceleration are both zero when the ball reaches the top of its motion. the speed of the ball is always zero and the acceleration is always −g at the instant the ball reaches the top of its motion. the speed of the ball is always negative and the acceleration is always equal to −g at the instant the ball reaches the top of its motion.
Physics
1 answer:
tamaranim1 [39]1 year ago
7 0

The speed of the ball is always zero and the acceleration is always -g when it reaches the top of its motion. This is because when the ball is free, only gravity acts on it which is always downwards, hence g is the net acceleration and it is always negative. However the velocity does not direction change instantly, negative acceleration first slows down the ball with a positive velocity, until that point the ball keeps moving up, then the ball velocity becomes zero just before changing direction and becoming negative after which the ball will now go down along gravity. Hence the ball velocity is zero at the top (neither going up nor down). Mathematically this can be seen as velocity is the integration of acceleration.

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A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
2 years ago
1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour
grigory [225]

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 kg \cdot cm^2

Explanation:

<u>Given</u>:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

<u>To Find:</u>

The moment of inertia of Alex's rolling hoop = ?

<u>Solution</u><u>:</u>

The moment of inertia  = mr^2

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia  = (0.350)(0.75)^2

=> moment of inertia  = (0.350)(0.5625)

=> moment of inertia  = (0.197)

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 kg m^2 s^{-1}

If she pulls her arms in to 0.12 m, her new linear speed  is  18.33 m/s^2

Explanation:

Given:

Mass  = 2.0 kg

Radius = 0.8 m

Velocity =  1.2 m/s

a.The combined angular momentum of the masses:

L = r \cdot m \cdot v_1

Substituting the values,

L = 0.8 \cdot 2.0 \cdot 1.1

L= 1.76 kg m^2 s^{-1}

b. If she pulls her arms in to 0.12 m, what is her new linear speed

0.12 \cdot 0.8 \cdot v_2 = 1.76

0.096 cdot v_2 = 1.76

v_2 = \frac{1.76}{0.096}

v_2 = 18.33 m/s^2

6 0
2 years ago
An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to
ankoles [38]

Answer:

The magnitude of the centripetal acceleration during the turn is a=12.04\ m/s^2.

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , T=t\times 4=60\ s.

Now , Velocity ,

v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.

Also , we know :

Centripetal acceleration ,

a=\dfrac{v^2}{r}

Putting all values we get :

a=12.04\ m/s^2.

Hence , this is the required solution .

5 0
1 year ago
A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80
Sidana [21]

Explanation:

It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

7 0
2 years ago
Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.
Rufina [12.5K]

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

4 0
2 years ago
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