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FinnZ [79.3K]
3 years ago
7

A 5 kg ball is sitting on the floor without moving. What is the momentum of the ball?

Physics
1 answer:
natita [175]3 years ago
5 0

Answer:

The ball has no momentum

Explanation:

The given parameters are;

The mass of the ball = 5 kg

The velocity of the ball = 0 (The ball is sitting on the floor without moving)

The momentum of the ball = The mass of the ball × Velocity of the ball

Therefore, the momentum of the ball = 5 kg × 0 m/s = 0

The momentum of the ball is zero, the ball has no momentum.

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Suppose that two objects attract each other with a gravitational force of 50N. If the mass
vagabundo [1.1K]

Answer:

112.5 N

Explanation:

50 = GMm/r^2

Let F be the new force of attraction

F/50 = ( G(3M)(3m)/(2r)^2 ) / (GMm/r^2)

[Elimiating G,M,m,r]

F = 112.5 N

7 0
3 years ago
3. Calculate the wavelength of wave that has a frequency of 4.75 x 1012Hz.
Klio2033 [76]
  • Frequency=v=4.75×10^12Hz
  • Wavelength=?

We know

\boxed{\sf \lambda=\dfrac{C}{V}}

\\ \sf\longmapsto \lambda=\dfrac{3\times 10^8ms^{-1}}{4.75\times 10^{12}s^{-1}}

\\ \sf\longmapsto \lambda=0.631\times 10^{-4}m

\\ \sf\longmapsto \lambda=6.31\times 10^{-5}m

4 0
3 years ago
Both the moon and the sun influence the tides on Earth. The moon has a much greater influence though. Why is that
3241004551 [841]
Because even though the moon is smaller, therefore a weaker gravitational pull, the moon is much closer to the earth than the sun, thus having a greater gravitational pull
3 0
3 years ago
What is the wavelength of a wave traveling at 300 m/s if the frequency is 10 Hz?
lions [1.4K]

Answer:

30 m

Explanation:

The wavelength of a wave is found by the velocity divided by the frequency. Therefore, the wavelength is (300 m/s)/(10 Hz) = 30 m

I hope this helps! :)

8 0
3 years ago
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
3 years ago
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