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1 year ago

Determine the magnitude of the momentum of a ... 107-kg halfback moving eastward at 8 m/s.The halfback's momentum in kgm/s is:

Physics

1 answer:

Veronika [31]1 year ago

3 0

Given:

The mass of the halfback is m = 107 kg

The speed of the halfback is v = 8 m/s

To find the momentum.

Explanation:

The momentum of the halfback is

$\begin{gathered} p=\text{ mv} \\ =\text{ 107}\times8 \\ =\text{ 856 kg m/s} \end{gathered}$

Thus, the momentum of the halfback is 856 kg m/s

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A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou

koban [17]

Answer:

The value is $f = 0.05 \ Hz$

Explanation:

From the question we are told that

The mass of the car is $m = 615 \ kg$

The period of the circular motion is $T = 20 \ s$

The radius is $r = 80 \ m/s$

Generally the frequency of the circular motion is

$f = \frac{1}{T }$

=> $f = \frac{1}{ 20 }$

=> $f = 0.05 \ Hz$

3 0

3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

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4 years ago

Anaerobic metabolism:

goblinko [34]

Explanation:

B. leads to muscle strain.

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3 years ago

This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters, what is the correspon

adelina 88 [10]

So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.

So this appears to be a right triangle so we can find the area of a triangle as:

0.5bh = A

Since our area is 10 meters lets alter our formula a bit to fit the situation:

Our base here is time and our height is velocity so:

0.5tv = Δx

So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s

So we have v, and Δx so lets isolate for time by dividing by v and 0.5

t = Δx / 0.5v

Now lets plug all that in:

t = 10 / 0.5(2)

t = 10 seconds

Hope this helped!

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3 years ago

A student jogs for a mile when their heartbeat starts racing and the student feels too fatigued to keep jogging. The student sto

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Answer:

Explanation

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8 0

3 years ago

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