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3 years ago

10

The Voyager 1 spacecraft, shown below, completed its space exploration mission in 1980 and has since been traveling away from Ea

rth at a speed of about 39,000 miles per hour.
If no unbalanced forces act on Voyager 1, then

A. the direction of its motion will reverse.
B. it will continue traveling at the same speed.
C. its speed will slowly decrease.
D. its speed will slowly increase.

2 answers:

Andreas93 [3]3 years ago

8 0

its b hope it helps

have a good day

aleksklad [387]3 years ago

4 0

I think b hope this helps

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A 65 N boy sits on a sled weighing 52 N on a horizontal surface. The coefficient of friction between the sled and the snow is 0.

pogonyaev

Answer:

1.40 N

Explanation:

The magnitude of the frictional force is given by:

$F=\mu N$

where

$\mu$ is the coefficient of friction

N is the magnitude of the normal reaction

The coefficient of friction for this problem is $\mu=0.012$ . The magnitude of the normal reaction is equal to the combined weight of the boy and the sled, because the surface is horizontal, so

$N=65 N+52 N=117 N$

Therefore, the frictional force is

$F=\mu N=(0.012)(117 N)=1.40 N$

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BaLLatris [955]

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Kobotan [32]

When one body(sun) exerts a force on a second body(planet), the second body simultaneously exerts a force equal in magnitude and opposite in direction of the first body. Which makes the planet orbit in path C.

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3 years ago

A 2000 kg car moves along a horizontal road at speed vo

cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

$\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a$ (Eq. 1)

$\Sigma F_{y} = N-m\cdot g = 0$ (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

$a = -\mu_{k}\cdot g$ (Eq. 3)

Where:

$a$ - Deceleration of the car, measured in meters per square second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $\mu_{k} = 0.0735$ and $g = 9.807\,\frac{m}{s^{2}}$ , then deceleration of the car is:

$a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = -0.721\,\frac{m}{s^{2}}$

The stopping distance of the car ( $\Delta s$ ), measured in meters, is determined from the following kinematic expression:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (Eq. 4)

Where:

$v_{o}$ - Initial speed of the car, measured in meters per second.

$v$ - Final speed of the car, measured in meters per second.

If we know that $v_{o} = 15.9\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $a = -0.721\,\frac{m}{s^{2}}$ , stopping distance of the car is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}$

$\Delta s = 175.319\,m$

The shortest possible stopping distance of the car is 175.319 meters.

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4 years ago

Can someone help with question.

Katarina [22]

Answer:

IDK

Explanation:

IDK

5 0

3 years ago