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Brrunno [24]
3 years ago
8

Arun places a beaker of water on a hotplate and turns the hotplate on. The temperature in the room is 25 ºC. After the water hea

ts to 75 ºC, she adds pieces of ice to the beaker. Which mode of transfer of heat energy occurs in the water during this experiment? Justify your answer
Physics
1 answer:
Rzqust [24]3 years ago
5 0

Answer:

Thermal convection

Explanation:

When ice cubes are placed in water, ice cubes start melting. This is because water transfers its heat to ice cubes until the temperature becomes equal and the process is called "thermal Convection".

Thermal convection is the mode of transfer of heat energy that will occur when ice cubes will be added to the hot water and at the end of the experiment when both are at the same temperature, no ice cube will left.

Hence, the correct answer is "thermal convection".

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A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving.
Anon25 [30]

Answer:

t=7.33 s

Explanation:

According to Newton's second law:

\sum F=m*a

because we don't want the box to slide, the acceleration has to be zero.

\sum F=-F_{friction}+F_{truck}=0\\F_{truck}=F_{friction}\\F_{friction}=\µ*m*g=0.500*9.81*m\\F_{friction}=4.91*m

we know that:

F=m*a\\4.91m=m*a\\a=4.91m/s^2

Now having the acceleration, we can use the following formula.

v_f=a*t\\t=\frac{36.0m/s}{4.91m/s^2}\\\\t=7.33s

7 0
3 years ago
To halt the decline in biodiversity, we must do which of the following?
emmainna [20.7K]
<span>adopt ecological conservation practices </span>
6 0
2 years ago
Water flows through a horizontal pipe. The diameter of the pipe at point b is larger than the diameter of the pipe at point a. W
Natalija [7]

The speed of the water is the greatest at point B

5 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

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4 0
3 years ago
A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t
gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

p_i = (M+m)u

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

p_f = MV+mv

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s

So, the skater is moving at 2.34 m/s (forward) after the shot.

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5 0
3 years ago
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