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katrin2010 [14]
2 years ago
7

A volleyball experiences 494 Ns of impulse over a time period of 7 seconds. What was the magnitude of the force that acted on th

e volleyball during this time period?
Physics
1 answer:
expeople1 [14]2 years ago
6 0

Answer:

70.6N

Explanation:

Given parameters:

Impulse  = 494Ns

Time  = 7s

Unknown:

Force applied = ?

Solution:

To solve this problem, we use the formula of impulse;

      Impulse = Force x time

Now insert the parameters and solve;

       494  = Force x 7

            Force  = \frac{494}{7}  

          Force  = 70.6N

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If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of t
emmasim [6.3K]

Answer:

Angular acceleration = 6.37rad/sec²

Approximately, Angular acceleration =

6.4 rad/sec²

Explanation:

Length of the rod = 2.0m long

Inclination of the rod (horizontal) = 30°

Mass of the rod is not given so we would refer to it as = M

Rotational Inertia of the Rod(I) = 1/3ML²

Angular Acceleration = ?

There is an equation that shows us the relationship between Torque and Angular acceleration.

The equation is :

Torque(T) = Inertia × Angular Acceleration

Angular acceleration = Torque ÷ Inertia

Where:

Torque = L/2(MgCosθ)

Where M = Mass

L = Length = 2.0m

θ = Inclination of the rod (horizontal) = 30°

g = Acceleration due to gravity = 9.81m/s²

Inertia = 1/3ML²

Angular Acceleration =  (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²

Angular Acceleration =

(3 × g × cos 30°) ÷ 2× L

Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L

Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m

Angular Acceleration = 6.37rad/sec²

Approximately Angular Acceleration =

6.4rad/sec²

5 0
3 years ago
Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N
vovangra [49]
Refer to the diagram shown below.

g = 9.8 m/s², and air resistance is ignored.

For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.

For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.

Let a =  the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²

Answer:  0.665 m/s²  (nearest thousandth)

7 0
3 years ago
Somebody please help I need exuse to not go to pe because I injured my self anybody know what to say
algol13

Answer:

Say you feel like you twisted ur ankle or can move your foot or something in that category

3 0
2 years ago
In order to be _______________ forces, their effects must cancel each other out and not cause a change in the object's motion
Tju [1.3M]
Balancing.

When the same force is applied from both sides,the forces cancel out each other.
4 0
3 years ago
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