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3 years ago

B. Jerome plays middle linebacker for South's varsity football team. In a game against

Physics

1 answer:

weqwewe [10]3 years ago

4 0

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

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You are driving at 75 km/h. Your sister follows in the car behind at 75 km/h. When you honk your horn, your sister hears a frequ

OLga [1]

The ideal concept for solving this question is based on the Doppler effect, for which it is indicated that the source's listening frequency changes as the distance and the relative speed between the receiver and the transmitter are also changed. However, if the relative velocity between the two objects is zero as in the particular case presented (since both travel at 75km / h) we have that there will be no change in frequency.

Therefore the frequency that I hear and that my sister would listen would be the same.

5 0

3 years ago

As the arm lifts the mass, does elbow angle increase or decrease?

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In increases depends how you move the muscle

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3 years ago

A 25.0-kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0°

SCORPION-xisa [38]

(a) 139.7 J

The potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

$U=mg\Delta h$

where

m = 25.0 kg is the mass of the child

g = 9.8 m/s^2

$\Delta h$ is the difference in height between the initial position and the bottom position

We are told that the rope is L = 2.20 m long and inclined at 42.0° from the vertical: therefore, $\Delta h$ is given by

$\Delta h = L - L cos \theta =L(1-cos \theta)=(2.20 m)(1-cos 42.0^{\circ})=0.57 m$

So, her potential energy is

$U=(25.0 kg)(9.8 m/s^2)(0.57 m)=139.7 J$

(b) 3.3 m/s

At the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

$U=K=\frac{1}{2}mv^2$

where

m = 25.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(139.7 J)}{25.0 kg}}=3.3 m/s$

(c) 0

The work done by the tension in the rope is given by:

$W=Td cos \theta$

where

T is the tension

d is the displacement of the child

$\theta$ is the angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. Therefore, $\theta=90^{\circ}$ and $cos \theta=0$ , so the work done is zero.