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GenaCL600 [577]
3 years ago
9

A 600-N box is pushed up a ramp that is 2 meters high and 5 meters long. The person exerts a force of 300-N. What is the efficie

ncy of the ramp?
Physics
1 answer:
AfilCa [17]3 years ago
6 0
From information given, there are two forces in action, the vertical and the horizontal. We first need to calculate separately work done by these two forces, then determine the efficiency. Work = Force × distance work(a) → 600N × 2m = 1,200 Nm work(b) → 300N × 5m = 1,500 Nm Efficiency = work(a) / work (b) × 100 Efficiency = 1200 / 1500 × 100 Efficiency = 0.8 × 100 Efficiency = 80% Therefore the efficiency of the ramp is 80%
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The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo
Lina20 [59]
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
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K= \frac{1}{2}mv^2
And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
3 0
2 years ago
Par 1/2
BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0
2 years ago
A student uses the right-hand rule as shown.
m_a_m_a [10]

Answer: O:right

Explanation: In this case you place your finger on the current, and your fingers should curl showing the way, you must use your right hand in this case, otherwise that would mean the fingers on your left would bend way back, and snap off, (Not really lol,  just saying)

5 0
2 years ago
A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

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Here \omega is the angular velocity, we replace (2) in (1):

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Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
3 years ago
a car is going around a roundabout.explain why we say that it is accelerating even if it is moving at a constant speed.
kiruha [24]

Because 'acceleration' does NOT mean 'speeding up'. 
It means ANY change in motion ... speeding up, slowing down,
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When traveling a roundabout, or any curved path, the direction
is constantly changing even if the speed is constant, so there is
constant acceleration going on.

4 0
3 years ago
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