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Nady [450]
3 years ago
6

8. A steel ball rolls from rest down a ramp with an acceleration of -3 m/s/s. What is its velocity after rolling for 5 s?

Physics
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

C. -15m/s

Explanation:

using formula, v = u + at

where v = final velocity

u = initial velocity = 0

a = acceleration = -3m/s/s

t = time = 5sec

=> v = 0 + -3×5 = -15m/s

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A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. Approximately how m
Alex73 [517]

Amount of work done is zero and so power = 0 watts.

<u>Explanation:</u>

Power is the rate at which work is done, or W divided by delta t. Since the barbell is not moving, the weightlifter is not doing work on the barbell.Therefore, if the work done is zero, then the power is also zero.It may seem unusual that the data given in question is versatile i.e. A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. But, still the answer is zero watts , this was a tricky question although conceptual basis of question was good! Power is dependent on amount of work done which is further related to displacement and here the net displacement is zero ! Hence, amount of work done is zero and so power = 0 watts.

6 0
3 years ago
Gauss's law: Group of answer choices can always be used to calculate the electric field. relates the electric field throughout s
kvasek [131]

Answer:

relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss Law states that overall electric flux of a closed surface is equivalent right to charge enclosed which is divided by the permittivity. In other words Gauss Law stress that

net electric flux that pass through an hypothetical closed surface is equivalent to overall electric charge present within that closed surface.

The Gauss law can be expressed mathematically as

ϕ = (Q/ϵ0)

Q = total charge within the surface,

ε0 = the electric constant

5 0
3 years ago
A gardener uses a 60-N wheelbarrow to transport two bags of fertilizer weighing W = 252-N. Determine the maximum allowable horiz
Butoxors [25]

Explanation:

Let us assume that the maximum allowable horizontal distance be represented by "d".  

Therefore, torque equation about A will be as follows.

   60 \times 0.15 + 252 \times 0.15 \times 2 + 252 \times d = 2 \times 75 \times (0.7 + 0.15 + 0.15)

      d = \frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}

       d = 0.409 m

Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.

6 0
3 years ago
What would happen if both the radius and mass of a planet were cut in half
Shtirlitz [24]

Explanation:

According to formula

g = GM/R^2

when mass is halved the value of g becomes half but when radius is halved the value of g increases 4 times.

As a result of both value of g becomes twice.

5 0
3 years ago
What 2 events increase power output
OverLord2011 [107]
P=I*E
Power (P)
Voltage (E)
Amps (I)

7 0
4 years ago
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