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ladessa [460]
3 years ago
12

What does a line on a graph look like when an object is maintaining average speed?

Physics
1 answer:
Brut [27]3 years ago
5 0

The line on the graph would be constant if the speed stays the same throughout the entire time

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Answer:

I think first rock is right answer

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If a resistance is added in parallel to a circuit resistance is than?
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Answer: The total resistance in a parallel circuit is always less than any of the branch resistances. Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease. As you add more and more branches to the circuit the total current will increase because Ohm's Law states that the lower the resistance, the higher the current.

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4 years ago
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John rides his motorcycle with a constant speed of 40 miles per hour. How far can he travel in 1/2 an hour?
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The answer would be 60 miles.
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3 years ago
An open bed truck is moving at 20 m/s and comes to a stop. A 100 kg crate sitting in the back of the truck remains at rest in th
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The work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.

<h3>Work done by friction to prevent the crate from moving</h3>

The work done is determined by applying the principle of conservation of energy.

W = K.E

W = ¹/₂mv²

where;

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  • v is speed of the crate with respect to the truck = 20 m/s

W = ¹/₂(100)(20²)

W = 20,000 J

Thus, the work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

3 0
2 years ago
A ball whirls around in a vertical circle at the end of a string. The other end of the string is fixed at the center of the circ
never [62]

Answer:

T_b=m(g+\frac{v_b^2}{R})

Explanation:

At the bottom the tension would be upwards and the weight downwards, their difference being the centripetal force. Taking the upwards direction as positive we then have:

T_b-mg=F_{cp}=ma_{cp}=m\frac{v_b^2}{R}

where we have used the equation for centripetal acceleration. Thus we have:

T_b=m(g+\frac{v_b^2}{R})

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3 years ago
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