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MariettaO [177]

3 years ago

6

A space probe on the surface of Mars sends a radio signal back to the Earth, a distance of 8.70 ✕ 107 km. Radio waves travel at

the speed of light (3.00 ✕ 108 m/s). How many seconds does it take for the signal to reach the Earth?

1 answer:

Ahat [919]3 years ago

3 0

Answer: 2.9 × 10 E 3 sec

Explanation: The explanation is attached

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What must occur for work to be done on an object?

stiv31 [10]

The answer is D since work done must have displacement..if theres force but no displacement there will be no work done based on W=Fs

4 0

2 years ago

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During the flight of the ball, what is the direction of its Acceleration due to gravity

vekshin1

Regardless of what direction an object is moving, the acceleration
due to gravity is always directed toward the center of the Earth.
That's the direction commonly known as "down".

4 0

2 years ago

A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of

GREYUIT [131]

Answer:

$a=190\ m/s^2$

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

$a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2$

So, the acceleration of the hockey puck is $190\ m/s^2$ .

5 0

3 years ago

If you walk 6mph how far would you get in 10 seconds?

abruzzese [7]

Answer:

0.017

Explanation:

6÷3600×10=

6÷360=0.00166666666

=0.017(rounded)

3 0

2 years ago

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A solid uniform cylinder of mass 4.1 kg and radius 0.057 m rolls without slipping at a speed of 0.79 m/s. What is the cylinder’s

Amiraneli [1.4K]

Answer:

The cylinder’s total kinetic energy is 1.918 J.

Explanation:

Given that,

Mass = 4.1 kg

Radius = 0.057 m

Speed = 0.79 m/s

We need to calculate the linear kinetic energy

Using formula of linear kinetic energy

$K.E_{l}=\dfrac{1}{2}mv^2$

$K.E_{l}=\dfrac{1}{2}\times4.1\times(0.79)^2$

$K.E_{l}=1.279\ J$

We need to calculate the rotational kinetic energy

$K.E_{r}=\dfrac{1}{2}\times I\omega^2$

$K.E_{r}=\dfrac{1}{2}\times\dfrac{1}{2}\times mr^2\times(\dfrac{v}{r})^2$

$K.E_{r}=\dfrac{1}{4}\times m\times v^2$

$K.E_{r}=\dfrac{1}{4}\times4.1\times(0.79)^2$

$K.E_{r}=0.639\ J$

The total kinetic energy is given by

$K.E=K.E_{l}+K.E_{r}$

$K.E=1.279+0.639$

$K.E=1.918\ J$

Hence, The cylinder’s total kinetic energy is 1.918 J.

8 0

3 years ago