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wariber [46]
2 years ago
7

If you are going 60 mph what is your speed in m per second

Physics
1 answer:
LenKa [72]2 years ago
5 0

Answer:

1/60 mps

Explanation:

We would first have to divide 60 by 60 because there is 60mins per hour to get 1mpm. After that we would have to divide 1 by 60 because there are 60 secs in a min. So our final answer after doing 1/60 would be a fraction.

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Astronomers estimate that a 2
tia_tia [17]
M)³ / 6 = 4.2e9 m³ 
<span>so its mass is </span>
<span>M = 3300kg/m³ * 4.2e9m³ = 1.4e13 kg </span>
<span>and so its KE at 16 km/s = 16000 m/s is </span>
<span>KE = ½ * 1.4e13kg * (16000m/s)² = 1.8e21 J 

</span># of bombs N = 1.8e21J / 4.0e16J/bomb = 44 234 bombs 
<span>give or take. 
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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

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Any four difference between velocity and acceleration​
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https://physicsabout.com/acceleration-and-velcoity/

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Two cars collide and come to a complete stop. where did all of their energy go?
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Read 2 more answers
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
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