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frosja888 [35]
1 year ago
9

Which diagram best represents the relationship between these terms?

Physics
1 answer:
shepuryov [24]1 year ago
6 0
Quantum and state physics
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The lift does 3000 J of work in 5 seconds. What is the power of the lift?
diamong [38]

Answer:

600

Explanation:

p=Work/time

3000/5=600 is power

8 0
3 years ago
g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and
cupoosta [38]

Answer:

= 236N

Explanation:

tension T = mg + ma

Given that,

m = 20kg

g = 9.8 m/s²

a = 2.0 m/s²

T = m(g + a)

T = 20( 9.8 + 2.0)

  = 20(11.8)

  = 236N

4 0
3 years ago
Veronica claims that she can throw a dart at a dartboard from a distance of 2 m and hit the 5 cm wide bullseye if she throws the
wariber [46]
First, find the amount of time for the dart to hit the board using this equation: t = d/v

t = 2 m/ 15 m/s = 0.133 s

Then, find the height the dart has fallen from its initial point using this equation: h = 0.5gt²

h = 0.5(9.81 m/s²)(0.133 s)² = 0.0872 m or 8.72 cm

Since the diameter of the bull's eye is only 5 cm, and you started at the same level of the top of the bull's eye, that means the maximum allowance would only be 5 cm. Since it exceeded to 8.72 cm, it means that <em>Veronica will not hit the bull's eye.</em>
3 0
3 years ago
An electric motor converts electrical energy into (blank) energy
Delicious77 [7]

Answer:

Mechanical

Explanation:

5 0
2 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
2 years ago
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