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2 years ago

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A tank in the shape of an inverted cone has a height of 11 meters and a base radius of 6 meters. The tank is full of water. How

much work is required to pump the water over the upper edge of the tank until the height of the water remaining in the tank is 9 meters

1 answer:

SOVA2 [1]2 years ago

7 0

Explanation:

Below are attachments containing the solution.

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How much heat is evolved when 1234 g of water condenses to a liquid at 100.°c?

lbvjy [14]

During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to
$Q=mL_e$
where
m is the mass of the water
$L_e$ is the latent heat of evaporation.

For water, the latent heat of evaporation is $L=2257 kJ/kg$ , while the mass of the water is
$m=1234 g=1.234 kg$

so, the amount of heat released in the process is
$Q=mL_e = (1.234 kg)(2257 kJ/kg)=2785 kJ$

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3 years ago

A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0

3 years ago

(b) A piece of wood of volume 0.6 m² floats in water. Find the volume

enot [183]

Answer:

Explanation:

Let the volume below water be v . Then

buoyant force = v d g where d is density of water , g is acceleration due to gravity

= v x 1000 x g

weight of wood piece = volume x density of wood x g

= .6 x 600 x g

for equilibrium while floating

buoyant force = weight

= v x 1000 x g = .6 x 600 x g

v = .36 m²

volume above water or volume exposed = .6 - .36

= .24 m²

When immersed completely ,

buoyant force = .6 x 1000 x 9.8

= 5880 N

weight of wood

= .6 x 600 x g

= 3528 N

buoyant force is more than the weight . In order to equalise them for floating with full volume in water

weight required = 5880 - 3528

= 2352 N.

6 0

3 years ago

Luba_88 [7]

<u>T</u><u>he area w</u><u>h</u><u>ich touches the sand floor of 4 feet of the bull is less than the man.</u>

because the area of surface of contact is more in bull than in man. More is the area of contact, leads is the force or pressure affected.

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3 years ago

You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive

Irina-Kira [14]

Answer:

Explanation:

In the first case you can use the expression for the Doppler effect when the source is getting closer and getting away

$f'=f(\frac{v}{v-v_{s}})$ ( 1 )

$f''=f(\frac{v}{v+v_s})$ ( 2 )

f' = perceived frequency when the source is getting closer

f'' = perceived frequency when the source is getting away

f = source frequency

v = relative speed

vs = sound speed

by dividing (1) and (2) you have

$\frac{f'}{f''}=\frac{f}{f}\frac{\frac{v}{v-v_s}}{\frac{v}{v+v_s}}=\frac{v+v_s}{v-v_s}\\\\f'v-f'v_s=f''v+f''v_s\\\\v(f'-f'')=v_s(f''+f')\\\\v=v_s\frac{f''+f'}{f'-f''}=(340\frac{m}{s})\frac{1370Hz+1330Hz}{1370Hz-1330Hz}=67.5\frac{m}{s}$

but this is the relative velocity, you have that

$v=v_{sir}-v_{car}\\v_{sir}=v+v_{car}=67.5\frac{m}{s}+35\frac{m}{s}=102.5\frac{m}{s}$

a. hence, the speed of the police car is 102.5m/s

6 0

3 years ago