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tamaranim1 [39]
3 years ago
11

During a certain period of time, the speedometer of a car reads a constant

Physics
1 answer:
kogti [31]3 years ago
7 0
No it just means that at one point you reached 60 km/h how long you remain there was dependent on how long u kept your foot on the gas
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As a football player moves in a straight line [displacement (3.00 mm)i^i^ - (6.50 mm)j^j^], an opponent exerts a constant force
Brilliant_brown [7]

Answer:

Work done, W=(0.378i-1.092j)\ J

Explanation:

Displacement,

d=(3i-6.5j)\ mm\\\\d=(0.003i-0.0065j)\ m

Force, F=(126i+168j)\ N

Work done by the opponent do on the football player is given by :

W=F{\cdot} d\\\\W=(126i+168j){\cdot} (0.003i-0.0065j)\ m\\\\W=(0.378i-1.092j)\ J

So, the work done by the opponent do on the football player is  (0.378i-1.092j)\ J.

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3 years ago
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Whats mechanical energy mean
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Mechanical energy<span> is the sum of kinetic and potential </span>energy<span> in an object that is used to do work. In other words, it is </span>energy<span> in an object due to its motion or position, or both.

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3 years ago
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The shortest path from a starting point to an endpoint, regardless of the path
liubo4ka [24]

Answer:

answer is C shortest vector

8 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
Luden [163]

Answer:

The ball will be attracted to the negatively charged plate. It'll touch and pick up some electrons from the plate so that the ball becomes negatively charged. Immediately the ball is repelled by the negative plate and is attracted to the positive plate. The ball gives up electrons to the positive plate so that it is positively charged and suddenly attracts to the negative plate again, flies over to it and picks up enough electrons to be repulsed by negative plate and again to the positive plate and that continues.

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3 years ago
Driving 30.7 m/s in your car you see a dog up ahead and slam on your brakes. You stop just before hitting the dog after skidding
Nitella [24]

Answer: 2.74

Explanation:

We can solve this problem using the stopping distance formula:

d=\frac{(V_{o})^{2}}{2 \mu g}

Where:

d=132.1 m is the distance traveled by the car before it stops

V_{o}=30.7 m/s is the car's initial velocity

\mu is the coefficient of friction between the road and the tires

g=9.8 m/s^{2} is the acceleration due gravity

Isolating \mu:

\mu=\frac{2dg}{(V_{o})^{2}}

Solving:

\mu=\frac{2(132.1 m)(9.8 m/s^{2})}{(30.7 m/s)^{2}}

\mu=2.74 This is the coefficient of friction

7 0
3 years ago
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