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zvonat [6]
3 years ago
13

As you found in Part A, your weight will be greater than normal when the elevator is moving upward with increasing speed. For wh

at other motion would your weight also be greater than your normal weight?
A) The elevator moves upward with constant velocity.
B) The elevator moves downward with constant velocity.
C) The elevator moves upward while slowing in speed.
D) The elevator moves downward while slowing in speed.
E) The elevator moves downward while increasing in speed.
Physics
1 answer:
shepuryov [24]3 years ago
4 0

Answer:

D) The elevator moves downward while slowing in speed.

Explanation:

The elevator is moving downward with decreasing speed , that means a force is acting on it in upward direction . So it has  an acceleration in upward direction . This type of movement is similar to motion with acceleration in upward direction.

In this case apparent weight will be greater.

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In rooms where there are multiple lights, a parallel circuit is better.

In a series circuit, if one light broke, all of the lights would turn off, as the circuit would be broken.

However, in parallel, if one bulb broke, the circuit could still be complete through the other bulbs, so they will stay on.
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Who was the first chemist to organize elements by atomic number? Johann Wolfgang Döbereiner Antoine Lavoisier Henry Moseley John
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If the volume of a cylinder is reduced from 8.0 liters to 4.0 liters, the pressure of the gas in the cylinder wil change from 70
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4 years ago
(30 Points) A piston–cylinder arrangement containing Nitrogen (N2), initially at 4 bar and 438 K, undergoes an expansion to a fi
BartSMP [9]

Answer:

A. Final pressure P2

P2/P1 = (T2/T1)^n/n-1

P1 = 4bar

T1 = 438K

T2 = 300K

Polytropic index, n, = 1.3

P2 = 4 (300/438)^1.3/1.3-1

P2 = 4 (300/438)^4.333

P2 = 4 * 0.19400

P2 = 0.776bar.

B. The work done is;

W = mR/ n-1 (T1 -T2)

Where, R = 0.1889kJ/kg.K, m = 1

W = 1 * 0.1889/ 1.3-1 * (438-300)

W = 86.89kJ/kg.

C. The heat transfer, Q

Q = W + ΔU

Q = W + mCv(T2-T1), where Cv of nitrogen is 0.743kj/kgk

Q = 86.89 + 1 * 0.743 (300-438)

Q = 86.89 + (-102.534)

Q = -15.644kJ/K

Q = 15.64kJ/K

3 0
3 years ago
An 18g bullet is shot vertically into a 10kg block. The block lifts upward 9mm. The bullet penetrates the block in a time interv
stealth61 [152]

Answer:

The initial kinetic energy of the bullet is closest to 491.87 J

Explanation:

Given;

mass of bullet, m₁ = 18g = 0.018kg

mass of block, m₂ = 10kg

height moved by the block, h = 9 mm = 0.009 m

time taken for the bullet to travel through the block, t = 0.001s

let the initial velocity of the bullet = v₁

let the final velocity of the bullet = v₂

Apply the principle of conservation of linear momentum;

initial momentum = final momentum

0.018v₁ = v₂(0.018 + 10)

0.018v₁ = 10.018v₂ -----equation (1)

Apply the law of conservation of energy when the bullet lifts the block through 9mm

mgh = ¹/₂mv₂²

gh = ¹/₂v₂²

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.009)

v₂ = 0.42 m/s

Substitute in v₂ in equation 1, to determine the initial velocity of the bullet;

0.018v₁ = 10.018v₂

0.018v₁  =  10.018(0.42)

0.018v₁  = 4.208

v₁ = 4.208 / 0.018

v₁ = 233.78 m/s

Now, determine the initial kinetic energy of the bullet;

K.E₁ = ¹/₂m₁v₁²

K.E₁ = ¹/₂(0.018)(233.78)²

K.E₁ = 491.87 J

Therefore, the initial kinetic energy of the bullet is closest to 491.87 J

6 0
3 years ago
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