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3 years ago

A 500 N passenger weighs 24500 N elevator that rises 30 m in exactly 1 minute how much power is needed for the elevators trip

Physics

1 answer:

larisa [96]3 years ago

5 0

Answer:

12500 W (12.5 kW)

Explanation:

The total weight of the passenger + elevator is

$F_g = 500 N + 24500 N=25000 N$

The total work done to rise the elevator + passenger is equal to the product between the total weight and the distance covered during the trip (d = 30 m):

$W=F_g d = (25000 N)(30 m)=750,000 J$

The power needed for the trip is equal to the ratio between the work done (W) and the time taken (t):

$P=\frac{W}{t}$

Since the time taken is t = 1 min = 60 s, the power needed is

$P=\frac{750,000 J}{60 s}=12,500 W=12.5 kW$

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Which material is more closely related to the styrofoam? a. heat exchanger b. diffuser c. insulator d. electrical conductor

nirvana33 [79]

The material that is more closely related to the Styrofoam insulator.

Styrofoam is the term that is used for polystyrene foam in a trademark form. It is a petroleum-based plastic.

Keeping something warm includes the stopping of the transfer of heat from one object to another. This is how insulation works.

Styrofoam is an insulator, which means it'll help keep the heat from the environment out of your cooler. However, you'll still need cooling agents (like ice packs) to make the cooler cold in the first place.

Styrofoam is usually made mostly of air, which means it is a poor conductor of heat, but an excellent convector. It traps the air in small pockets, which blocks the flow of heat energy. This reduces both conduction and convection and makes Styrofoam a good insulator.

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8 0

1 year ago

Please help! Will give brainliest. 10 points. Show work!

Natasha_Volkova [10]

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 396.9 m

Time (t) =.?

t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

5 0

2 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago

PLEASE TRY TO ANSWER AS MANY QUESTIONS AS YOU CAN !

suter [353]

Good luck with solving this

3 0

3 years ago

Read 2 more answers

tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h

Sati [7]

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q is the amount of heat supplied to the object

m is the mass of the object

$\Delta T$ is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

$\Delta T=12^{\circ}$ is the increase in temperature

Substituting into the formula, we find the specific heat:

$C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC$

8 0

3 years ago