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o-na [289]
3 years ago
11

A baseball has a mass of 0.45 kg and is thrown with a speed of 25 m/s. what is the momentum of the baseball?

Physics
2 answers:
amm18123 years ago
7 0
The momentum<span> of a particle is defined as the product of its mass times its velocity. It is a vector quantity. Therefore, the momentum of the baseball would be as follows:

Momentum = mass x velocity
Momentum = 0.45 kg (25 m/s)
Momentum = 11.25 kg m/s</span>
Ray Of Light [21]3 years ago
4 0
The answer given is correct, but it would be rounded to 11 on the site.  
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How does mass affect the gravitational pull of the Sun, Earth, and the Moon?
madreJ [45]
Larger mass creates a stronger pull
3 0
3 years ago
Read 2 more answers
The charges on two metallic balls are 5.0 and 7.0 coulombs respectively. They are kept 1.2 meters apart. What is the force of in
Marysya12 [62]
We know, F = k * q₁ * q₂ / r²

Substitute the known values, 
F = 9 * 10⁹ * 5 * 7 / (1.2)²

F = 315 * 10⁹ / 1.44

F = 218.75 * 10⁹ N

F = 2.1875 * 10¹¹ N   [ Final Answer ]

Hope this helps!
5 0
3 years ago
(a) Calculate the acceleration due to gravity on the surface of the Sun.
Ira Lisetskai [31]
<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

                      g=\frac{GM}{r^2}

 Here we need to find acceleration due to gravity of Sun,

                G = 6.67259 x 10⁻¹¹ N m²/kg²

    Mass of sun, M = 1.989 × 10³⁰ kg

    Radius of sun, r = 6.957 x 10⁸ m

Substituting,

                g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

   Ratio of gravity = 274.21/9.81 = 27.95

   Weight = mg

  Factor of increase in weight = 27.95

8 0
3 years ago
A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

3 0
3 years ago
An ambulance driver traveling at 31.0 m/s (69.3 mph) honks his horn as he sees a motorist ahead on the highway traveling in the
IrinaVladis [17]

Answer:

fo = 378.52Hz

Explanation:

Using Doppler effect formula:

f'=\frac{C-Vb}{C-Va}*fo

where

f' = 392 Hz

C = 340m/s

Vb = 20m/s

Va = 31m/s

Replacing these values and solving for fo:

fo = 378.52Hz

4 0
3 years ago
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