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2 years ago

Why does water’s high specific heat make it a good coolant for car radiators?.

Physics

2 answers:

Alexeev081 [22]2 years ago

7 0

Answer:

Car radiators: Water is used as coolant car radiators. Due to its high specific heat capacity, it can absorb a large amount of heat energy from the engine of the car, but its temperature does not rise too high.

Explanation:

i hope this answer your question if it s wrong please let know

Nataly [62]2 years ago

6 0

Answer:

Water has a high specific heat, meaning it takes more energy to increase the temperature of water compared to other substances. This is why water is valuable to industries and in your car's radiator as a coolant.

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Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.

Tcecarenko [31]

The energy of a photon is given by
$E=hf$
where
$h=6.6 \cdot 10^{-34} Js$ is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is
$f=5.49 \cdot 10^{14}Hz$
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
$E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J$

8 0

3 years ago

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

Mathematical expression for the Newton's second law of motion is given as:

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

∵mass is constant at speeds v << c (speed of light)

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

& Impulse:

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0

3 years ago

The speed of a boat in still water is 20 mph. it travels from one pier to another with the current in 4 hours and back against t

saw5 [17]

The answer is, "the speed of the current is 5 miles per hour."

To calculate the speed of the current,
let's assume speed of current = xmph. Time taken to travel from one pier to another with the current = 100/(20+x)h

But the time taken to travel from one pier to another with the current, which is given is = 4 hours. So, 4=100/(20+x) 80+4x = 100

4x = 20

x = 5 Thus, the speed of the current is 5 miles per hour.

3 0

3 years ago

The normal is a line perpendicular to the reflecting surface at the point of incidence.

ladessa [460]

Answer:

True

Explanation:

The normal line is defined as the line which is perpendicular to the reflecting surface at the point where the incident ray meet with the reflecting surface.

The angle of incident is defined as the angle which is subtended by the incident ray with respect to the normal ray by consider the normal ray as the base line and angle is measured from the point where incident ray is incident on the reflecting surface of the mirror.

Similarly reflecting ray can be defined as the ray which is reflected after the incident of a ray and the angle subtended by the reflecting ray is measure with respect to normal ray by considering normal ray as a base line.

Therefore, the normal ray is the perpendicular line to the reflecting surface at the point of incidence.