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Anika [276]
3 years ago
8

a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landi

ng?
Physics
1 answer:
vovangra [49]3 years ago
4 0

5.2m/s

Explanation:

Given parameters:

Mass of baseball = 0.15kg

Momentum of baseball = 0.78kgm/s

Unknown:

Speed of baseball = ?

Solution:

The momentum of the baseball is a function of the product of the mass and velocity. It is a vector quantity:

        Momentum = mass x velocity

 Since the speed of the ball is unknown:

           Velocity  =  \frac{momentum }{mass}

                           = \frac{0.78}{0.15}

                           = 5.2m/s

The speed of the baseball before it lands is 5.2m/s

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

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Inertia is responsible for :
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B. the reason we must wear seat belts
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Which of the following is a field (action-at-a-distance) force?
Kobotan [32]

Answer:

F_{grav} is a field force, as gravity does not physically apply force and does not require proximity.

Explanation:

3 0
3 years ago
Margaret wants to go for a swim, and decides to jump in using the diving board that measures 3-m long.
otez555 [7]

Answer:

The horizontal component of her velocity is approximately 1.389 m/s

The vertical component of her velocity is approximately 7.878 m/s

Explanation:

The given question parameters are;

The initial velocity with which Margaret leaps, v = 8.0 m/s

The angle to the horizontal with which she jumps, θ = 80° to the horizontal

The horizontal component of her velocity, vₓ = v × cos(θ)

∴ vₓ = 8.0 × cos(80°) ≈ 1.389

The horizontal component of her velocity, vₓ ≈ 1.389 m/s

The vertical component of her velocity, v_y = v × sin(θ)

∴ v_y = 8.0 × sin(80°) ≈ 7.878

The vertical component of her velocity, v_y ≈ 7.878 m/s.

6 0
3 years ago
Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its
antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem$\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

\mathrm{x}=\frac{\sqrt{2}}{4} L$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$$I_x=I_8+M x^2$$

$$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$, Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

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5 0
1 year ago
Answer the following question​
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Answer:

A) OA, AB, BC

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C) see explanation

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Explanation:

From the Velocity time graph shown:

The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.

Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.

-ve slope = BC

B) Acceleration of body in path OA.

Acceleration = change in Velocity / time

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C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).

D) Length of BC

BC corresponds to the distance moved, that velocity / time

Velocity = 150 ; time = 6

Therefore Distance (BC) = 150/6 = 25

E.) Velocity =0 ; Hence body is at rest

5 0
2 years ago
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