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AVprozaik [17]
3 years ago
14

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.
Physics
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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Answer:

4.32\cdot 10^5 J

Explanation:

Power is related to energy by the following relationship:

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t is the time elapsed

In this problem, we know that

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t = 1 h \cdot (60 min/h)(60 s/min)=3600 s

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E=Pt=(120 W)(3600 s)=4.32\cdot 10^5 J

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A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o
Cerrena [4.2K]

Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j  D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of  the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the  projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

D) Finally, in order to get the speed of the projectile when it hit the wall, we need just to find the magnitude of the velocity, as we get the magnitude of any vector given its vertical and horizontal components:

v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

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