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3 years ago

5

PLEASE HEP FAST PLEASE!!!!! IM SO SCARED!!!! An elevator has the mass of 3 tons of power needed to raise the elevator 50m in 15

s?

2 answers:

gayaneshka [121]3 years ago

6 0

Answer:

Mass=3,height=50 and time=15,g=10,P=?

P=mgh/t....which is 3×10×50=1500/15=100Watts

krok68 [10]3 years ago

5 0

Answer:

$\huge \boxed{\mathrm{100 \ Watts}}$

Explanation:

$\sf \displaystyle P=\frac{mgh }{t} \\\\\\ P=power \ (W) \\\\ m=mass \ (kg) \\\\ g=gravity \ acceleration \ (m/s^2) \\\\ h=height \ (m) \\\\ t=time \ taken \ (s)$

$\sf \displaystyle P=\frac{3 \times 10 \times 50 }{15}$

$\sf \displaystyle P=\frac{1500 }{15}$

$\sf \displaystyle P=100$

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Speed = Distance/ Time

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Speed = 100 km/hr.

100 km per hour.

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3 years ago

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N

irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

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Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

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3 years ago

A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at

VARVARA [1.3K]

Answer:

μ = 0.18

Explanation:

Let's use Newton's second Law, the coordinate system is horizontal and vertical

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N-W = 0

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F = fr

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fr = μ N

fr = μ m g

At the limit point just before starting the movement

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μ = F / m g

calculate

μ = 34.8 / (19.8 9.8)

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7 0

3 years ago

Read 2 more answers

A washing machine heats 10kg of water in each wash cycle. How much energy is saved by washing at 30'c instead of 50'c if the sta

Volgvan

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3 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

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Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

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On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

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$S = \Delta S_{source}+\Delta S_{sink}$

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PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

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3 years ago