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PSYCHO15rus [73]
3 years ago
13

A space explorer is moving through space far from any planet or star. He notices a large rock, taken as a specimen from an alien

planet, floating around the cabin of the ship. Should he push it gently, or should he kick it toward the storage compartment? Explain.
Physics
1 answer:
yaroslaw [1]3 years ago
3 0
He should push it gently.
This is because the forces of resistance this situation are minimal, so the rock will not slow as it would on Earth. Kicking the rock may result in it travelling too fast and hitting something else, causing damage. Moreover, the rock could start rebounding off of surfaces and create havoc. 
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if an object is moving with a velocity of 24m/s and has an acceleration of -4m/s how long will it take to stop
ololo11 [35]
The negative sign on the acceleration is only a vector quantity that means the object is accelerating to the left. Hence, we can only focus on it magnitude which is 4 m/s^2. Acceleration is the change in velocity over time. The change in velocity must be 24 m/s - 0 m/s, if you want the object to stop. Therefore,

a = (v2 - v1)/t
4 = (24 - 0)t
t = 6 seconds

The object will stop after 6 seconds.
7 0
3 years ago
Which description best matches the image below of a hand that is using the right-hand palm rule?
otez555 [7]

Answer:

When reviewing the results, the correct one is C

Explanation:

The right hand rule is widely useful in knowing the direction of force in a maganto field,

The ruler sets the thumb in the direction of the positive particle, the fingers extended in the direction of the magnetic field, and the palm in the direction of the force.

Let's apply this to our exercise.

The thumb that is the speed goes in the negative direction of the axis,

The two extended that the magnetic field look negative x,

The span points entered the dear sheet the negative the Z axis

When reviewing the results, the correct one is C

8 0
3 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
A 3250 N car is pushed a distance of 35 m the power was 11375 J, how long did it take?
irakobra [83]

Answer:

10.8s

Explanation:

Given parameters:

Force on the car  = 3250N

Distance  = 35m

Power  = 11375W

Unknown:

Time taken = ?

Solution:

To solve this problem;

 Power is the rate at which work is done

         Power = \frac{work done }{time}  

  Work done  = force x distance  = 3250 x 35  = 123200J

Now;

          11375  = \frac{123200}{t}  

           11375t  = 123200  

                   t  = 10.8s

5 0
3 years ago
40 POINTS!!! PLEASE HELPP!!!
NeX [460]

Answer:

A

Explanation:

if it was B it would say from one to another to another

7 0
2 years ago
Read 2 more answers
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