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3 years ago

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a 45 kg ice skater initially skating at a velocity of 3 m/s speeds up to a velocity of 5 m/s. calculate the difference in the ma

gnitude of momentum of the skater.

2 answers:

ad-work [718]3 years ago

4 0

Answer: 90 kgm/s

Explanation:

The momentum (linear momentum) $p$ is given by the following equation:

$p=m.V$

Where:

$m=45 kg$ is the mass of the skater

$V$ is the velocity

In this situation the skater has two values of momentum:

Initial momentum: $p_{1}=m.V_{1}$

Final momentum: $p_{2}=m.V_{2}$

Where:

$V_{1}=3 m/s$

$V_{1}=5 m/s$

So, if we want to calculate the difference in the magnitude of the skater's momentum, we have to write the following equation(assuming the mass of the skater remains constant):

$p=p_{2}-p_{1}=m.V_{2}-m.V_{1}$

$p=m(V_{2}-V_{1})$

$p=45 kg(5 m/s - 3 m/s)$

Finally:

$p=90 kgm/s$

ExtremeBDS [4]3 years ago

4 0

Answer: 90m

Explanation:I hope this helps

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starting from rest , a formula one car accelerates uniformly at 25m\s2 for 30secs. what distance does it cover in the last one s

Anestetic [448]

The distance covered in the last second of motion is 737.5 m

Explanation:

The motion of the car is a uniformly accelerated motion, so we can use the suvat equations.

First of all, we have to find the velocity of the car when the last second of motion starts, that is the velocity of the car after t = 29 s. We can use the equation:

v = u + at

where

u = 0 is the initial velocity

$a=25 m/s^2$ is the acceleration

Substituting t = 29 s,

$v=0+(25)(29)=725 m/s$

Now we can find the distance covered in the last second of motion by using

$s=ut+\frac{1}{2}at^2$

where

u = 725 m/s is the velocity at the beginning of the last second

t = 1 s is the time interval considered

$a=25 m/s^2$ is the acceleration

Substituting,

$s=(725)(1)+\frac{1}{2}(25)(1)^2=737.5 m$

Note that the acceleration of $25 m/s^2$ is not realistic for a car, but I still have used the data of the problem.

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3 years ago

The atmosphere of earth does not escape into the space but remains attached to earth's surface why??

Alexeev081 [22]

Just like any other gas or mixture of gases, the gas molecules are
zipping around in all different directions and with a whole range of
different speeds.

Those that happen to be moving at a speed greater than the Earth's
"escape velocity", AND are pointed away from Earth, AND don't hit
any other molecules before they escape, are lost.

With the combination of Earth's escape velocity, and the temperatures,
thickness, and density of the atmosphere, that process happens slowly
enough to have maintained an atmosphere around this planet until now.

Personally, I hope it hangs around for a while longer. But with the constant
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3 years ago

A 75.0-kg man standing on a bathroom scale in an elevator. Calculate the scale in N, reading if the elevator moves upward at a c

Salsk061 [2.6K]

The scale in N, reading if the elevator moves upward at a constant speed of 1.5 m/s^2 is 862.5 N.

weight of man = 75kg

speed of elevator, a = 1.5 $m/ s^{2}$

$F - w = ma \\$

$F = w + ma$

$F = m ( a +g )$

$F = 75 ( 1.5 + 10 ) \\$

$F = 75 ( 11.5 )$

$F = 862.5 N$

So, the scale reading in the elevator is greater than his 862.5 N weight. This indicates that the person is being propelled upward by the scale, which it must do in order to do so, with a force larger than his weight. According to what you experience in quickly accelerating or slowly moving elevators, it is obvious that the faster the elevator acceleration, the greater the scale reading.

Speed can be defines as the pace at which the position of an object changes in any direction. Since speed simply has a direction and no magnitude, it is a scalar quantity.

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2 years ago

When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the

topjm [15]

Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\sqrt{\dfrac{k}{m}}$

Where, m = mass

k = spring constant

Put the value into the formula

$\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}$

$\omega=4.74\ rad/s$

We need to calculate the period of oscillation,

Using formula of time period

$T=\dfrac{2\pi}{\omega}$

Put the value into the formula

$T=\dfrac{2\pi}{4.74}$

$T=1.33\ sec$

Hence, The period of oscillation is 1.33 sec.

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3 years ago

A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fir

bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.

So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

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3 years ago