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3 years ago

15

What must be true about two objects if heat is flowing between them? the objects must be different temperatures. the objects mus

t be the same temperature. the objects must both be over room temperature. the objects must both be under room temperature.

2 answers:

tester [92]3 years ago

7 0

Heat always flows from high temperature to low temperature and flows until the temperature between the two locations become equal. hence for the two objects having heat flowing between them, the temperature must be different. one object must have higher temperature than the other object.

hence the correct choice is

the objects must be different temperatures

VladimirAG [237]3 years ago

4 0

They must have different temperatures.

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A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op

Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

$F cos \theta$

where

F is the magnitude of the pull

$\theta=30^{\circ}$ is the angle

- Backward: the force of friction, which is

$F_f = 30 N$

So, the equation of motion is

$F cos \theta - F_f = ma$

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

$F cos \theta - F_f = 0$

Therefore, we can find the pulling force:

$F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N$

(b) 43.9 N

In this case, the acceleration is

$a=0.40 m/s^2$

So, the equation of motion in this case is

$F cos \theta - F_f = ma$

So this time we have to take into account the term (ma).

Using the same data as before:

m = 20 kg

$\theta=30^{\circ}$

$F_f = 30 N$

We find the new magnitude of F:

$F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N$

6 0

3 years ago

yulyashka [42]

The force applied to lift the crate is 171 N

Explanation:

The lever works on the principle of equilibrium of moments, so we can write:

$F_i d_i = F_o d_o$

where

$F_i$ is the force in input

$d_i$ is the arm of the input force

$F_o$ is the output force

$d_o$ is the arm of the output force

For the lever in this problem, we have:

$d_i = 0.25 m$

$d_o = 0.19 m$

$F_i = 130 N$ (force applied)

Solving the equation for $F_o$ , we find the force applied to lift the crate:

$F_o = \frac{F_i d_i}{d_o}=\frac{(130)(0.25)}{0.19}=171 N$

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

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3 years ago

Two drums of the same size and same height are taken.

Gelneren [198K]

Answer:

i) The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) The pressure acting on the base of <em>B</em> will be the same as the pressure acting on the base of <em>A</em>

iii) The pressure on the base of drum <em>A</em> will be slightly less than the pressure on the base of drum <em>B</em>

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

$h_{max}$ = The height of the drums

ρ = The density of the liquid in the drum

g = The acceleration due to gravity ≈ 9.81 m/s²

i) If <em>A</em> is completely filled, we have $h_A$ = $h_{max}$

Therefore, $P_A$ = $h_{max}$ × $\rho_{liquid}$ ×g

If <em>B</em> is half filled, we have, $h_B$ = (1/2)· $h_{max}$

$P_B$ = (1/2) × $h_{max}$ × $\rho_{liquid}$ ×g

Therefore, $P_B$ = (1/2) × $P_A$

The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) If both <em>A</em> and <em>B</em> are each filled with water (the same liquid), then the pressure on their bases will be $P_A$ = $h_{max}$ × $\rho_{water}$ ×g = $P_B$ , the same, given that the acceleration due to gravity, <em>g</em>, is constant and the same in Nepal and India

iii) If <em>A</em> is filled with water, and <em>B</em> is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

$P_A$ = $h_{max}$ × $\rho_{water}$ ×g < $P_B$ =

The pressure on the base of drum <em>A</em> will be less than the pressure on the base of drum <em>B.</em>

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