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3 years ago

The change in momentum experienced by a object is equivalent to the...

Physics

1 answer:

Papessa [141]3 years ago

8 0

Answer: C (impulse acting on the object)

The momentum is defined as it is the impulse acting on the force . Change in momentum is known as Impulse. Impulse is used to increase or decrease the momentum of object.

From Newtons II law

F = m. a

= m. v/t <em>since a = rate of change of velocity.</em>

<em> </em>F . t = m . v

F . t is known as impulse momentum

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3 years ago

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A toy car is moving along with 0.40 joules of kinetic energy if it’s speed is double then it’s new kinetic energy will be ?

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3 years ago

Transverse waves are being generated on a rope under constant tension. By what factor is the required power increased or decreas

ICE Princess25 [194]

Answer:

a) when the length of the rope is doubled and the angular frequency remains constant: The power increases by factor of two

b) when the amplitude is doubled and the angular frequency is halved: The power is the same

c) when both the wavelength and the amplitude are doubled: The power increases by a factor of 8

d) when both the length of the rope and the wavelength are halved: The power increases by factor of two

Explanation:

For a sinusoidal mechanical wave (Transverse wave), the time-averaged power is the energy associated with a wavelength divided by the period of the wave.

$P_{avg} =\frac{E \lambda}{T}=\frac{1}{2} \mu A^2 \omega^2 \frac{\lambda}{T}$

where;

A is the Amplitude

ω is the angular frequency

λ is the wavelength

a) when the length of the rope is doubled and the angular frequency remains constant

L = λ/2, λ = 2L

The power increases by factor of two

b) when the amplitude is doubled and the angular frequency is halved

$P_{avg} ={\frac{1}{2} \mu (2A)^2 (\frac{\omega}{2}) ^2 \frac{\lambda}{T} = \frac{1}{2} \mu A^2 \omega^2 \frac{\lambda}{T}$

The power is the same

c) when both the wavelength and the amplitude are doubled

$P_{avg} =\frac{1}{2} \mu (2A)^2 \omega^2 \frac{(2\lambda)}{T} = \frac{1}{2} \mu (4A^2) \omega^2 \frac{(2\lambda)}{T} = 8 (\frac{1}{2} \mu A^2 \omega^2 \frac{\lambda}{T})$

The power increases by a factor of 8

d) when both the length of the rope and the wavelength are halved

L = λ/2

when both are halved

L/2 = λ/4, λ = 2L

The power increases by factor of two

8 0

3 years ago

A person throws a ball upward into the air with an initial velocity of 15.0 m/s. calculate (a) how high it goes, and (b) how lon

eduard

In the vertical direction, we know that acceleration (a) is g = 9.8m/s^2 (this will be negative since it is going in the opposite direction of the initial velocity), we know our initial velocity (vi) = 15m/s and our final velocity (vf) = 0m/s if the ball stops at its max height. Using this kinematics equation: vf^2 = vi^2 + 2a*h, where h is the max height of the ball, we can find h by solving: h = (vf^2 - vi^2)/2a, now we plug in numbers h = (0^2 - 15^2)/2(-9.8) = 11.48 m.

Now we can find the time, but let's use a trick, we know the acceleration will be the same going up and down and that it will cover the same distance. Using this logic, we can know that the ball will be going the exact same speed we threw it up at when it comes back down to our hand (provided our hand is at the same height). We can then say it will take the same amount of time to reach its peak after leaving our hand as it will take to go from its peak back down to our hand. Using this, we can just get the time it takes to get to the top of its arc and then multiple it by 2. Using d = (vf + vi)*t / 2, we can solve for t, so t = 2d/(vf+vi) = 2(11.48)/(15) = 1.53 s for the trip up, doubling it for the trip down, we get a total of 3.06 s to go up and back down to our hand.

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3 years ago