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3 years ago

1.

Physics

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

The rider experiences weight or does not experience weight depends upon the direction of the elevator in which it moves and its acceleration.

Explanation:

Consider the mass of a rider as m. So its actual weight is mg and it acts vertically downward. The apparent weight of the rider is depended on the acceleration of the elevator and its direction of movement.

When elevator moves with a constant speed i.e. its acceleration is zero, the apparent weight of the rider is equal to the actual weight. Thus the rider's sensation is normal.

If the elevator is moving upward with an acceleration a, then the apparent weight of the rider will be more and the rider will experience an increase in weight or the sensation is heavy.

Or when the elevator moves downward with an acceleration a, the apparent weight of the rider is less. And the rider sensation is lighter.

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The heat loss from a boiler is to be held at a maximum of 900Btu/h ft2 of wall area. What thickness of asbestos (k= 0.10 Btu/h f

zmey [24]

Answer:

a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F

Explanation:

a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.

Now [dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft

b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A = -70 Btu/h ft²

c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.

Substituting these values into the equation, we have

-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft

-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂ - 1312°F + 70 °F - 0.28T₂

collecting like terms, we have

-900 °F + 1312°F - 70 °F = 0.82T₂ - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

8 0

3 years ago

What happens when Earth rotates on its axis and how long does it take

faust18 [17]

Answer:

You get Day and Night

It takes 24 hour

8 0

3 years ago

Read 2 more answers

Show that the speed with which a projectile leaves the ground is equal to its speed just before it strikes the ground at the end

Rina8888 [55]

Answer:

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

Explanation:

Let a projectile is projected vertically upwards with a speed of u and reaches to the maximum height H.

At maximum height , the speed is zero and then the projective comes back on the ground.

Use the third equation of motion

$v^2 = u^2 + 2 g h \\\\0 = u^2 - 2 g H\\\\\u =\sqrt{2gH}$

Now let the velocity at the time of strike is v'.

Use third equation of motion, here initial velocity is zero.

$v'^2 = 0 + 2 g H \\\\v = \sqrt{2gH}$

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

8 0

3 years ago

Charlie Brown kicks a football at 24.5 m/s at 35.0. What is the maximum height of the ball?

lozanna [386]

Answer:

d = 10.076 m

Explanation:

We need to obtain the velocity of the ball in the y direction

Vy = 24.5m/s * sin(35) = 14.053 m/s

To obtain the distance, we use the formula

vf^2 = v0^2 -2*g*d

but vf = 0

d = -vo^2/2g

d = (14.053)^2/2*(9.8) = 10.076 m

5 0

4 years ago

Please help on these two thanks

Lisa [10]

Explanation:

Number 29

= 10 N (from rocket) + 10 N (from wheels) - 7 N (from parachute 'cause force in other direction)

= 20 N - 7 N

= 13 N ✔

Number 30

= 100 N - 85 N

= 15 N to the left ✔

4 0

3 years ago