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Sladkaya [172]
3 years ago
11

Shelly experiences a backward jolt when the driver starts the school bus. Which of the following explains this phenomenon? A. th

e inertia of Shelly B. the inertia of the bus C. the normal force of the seats on Shelly D. the net force acting on Shelly in a backward direction E. a zero net force acting on Shelly
Physics
2 answers:
JulijaS [17]3 years ago
6 0

A.

Newton's first law. When the bus starts, Shelly is still at rest so she resists motion. (When the bus stops, she'll jolt forward)

Mnenie [13.5K]3 years ago
5 0

Answer: A. The inertia of Shelly

Explanation:

We know from Newton's first law that an object does not changes its state of rest or motion until an unbalanced external force exerts on it. This is also known as law of inertia. Initially Shelly and school bus are in rest state. when the driver starts the school bus and moves forward, Shelly experiences a backward jolt. This is because initially Shelly was in rest state but the bus started moving so she would feel a backward pseudo force.

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Ossicles

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7 0
3 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

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The force is repulsive because the points charged have the same sign.

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