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Nady [450]
3 years ago
12

What is the minimum work needed to push a 920-kg car 310 m up along a 6.5 ∘incline? Ignore friction.Express your answer using tw

o significant figures.
Physics
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

2800000J

Explanation:

Parameters given:

Mass = 920kg, weight = 920 * 9.8 = 9016N

Distance = 310m

Angle of inclination = 6.5°

Work done is given as :

W = F*d*cosA

Where A = angle of inclination

W = (9016 * 310 * cos6.5)

W = 2776993.59J

In 2 significant figures, W = 2800000J

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Answer:

\boxed{q = 1.2 \times  {10}^{ - 6} C}

Explanation:

f_e =   \frac{{q}^{2}k }{ {r}^{2} }  \\ q =  \sqrt{ \frac{f_e( {r}^{2} )}{k} }  =  \sqrt{ \frac{5.2(5 \times  {10}^{ - 2} )^{2} }{9 \times  {10}^{9} } }  \\ q =\sqrt{ \frac{5.2(5 \times  {10}^{ - 2} )^{2} }{9 \times  {10}^{9} } }  =  \sqrt{ \frac{0.013}{9 \times  {10}^{9} } }  \\ q = 1.2 \times  {10}^{ - 6}

3 0
3 years ago
A fisherman noticed that a wave strikes the boat side every 5 seconds. The distance between two consecutive crest is 1.5m. What
GrogVix [38]
5/1.5=7 I think that is the answer
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Find the frequency of a wave with a period of 0.5 seconds
strojnjashka [21]

Frequency of the wave is 2 per second

Explanation:

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f = 1/T, where f is the frequency and T is the time period

⇒ f = 1/0.5 = 2 per second

8 0
3 years ago
A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have
Doss [256]

Answer:

2000\; {\rm cm^{3}}.

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let m(\text{ball}) denote the mass of this ball. Let m(\text{water}) denote the mass of water that this ball has displaced.

Let g denote the gravitational field strength. The weight of this ball would be m(\text{ball}) \, g. Likewise, the weight of water displaced would be m(\text{water})\, g.

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

\text{buoyancy} \ge m(\text{ball})\, g.

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

\text{buoyancy} = m(\text{water}) \, g.

Therefore:

m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g.

m(\text{water}) \ge m(\text{ball}).

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let \rho(\text{water}) denote the density of water. The volume of water that this ball should displace would be:

\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})}  \end{aligned}.

Given that m(\text{ball}) = 2000\; {\rm g} while \rho = 1.00\; {\rm g\cdot cm^{-3}}:

\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})}  \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}.

In other words, for this ball to stay afloat, at least 2000\; {\rm cm^{3}} of the volume of this ball should be under water. Therefore, the volume of this ball should be at least 2000\; {\rm cm^{3}}\!.

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2 years ago
What is something that has a lot of thermal energy?
Artemon [7]
a heater that produces lots of warm air which would include thermal energy or a kitchen stove
8 0
3 years ago
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