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3 years ago

What happens as water evaporates from a lake

Physics

1 answer:

NARA [144]3 years ago

4 0

Answer:

As that liquid water is further heated, it evaporates and becomes a gas—water vapor. So C... Need branliest pls

Explanation:

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A force of 9 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then imme

mart [117]

Answer:

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$

Explanation:

let $m$ be the mass attached, let $k$ be the spring constant and let $\beta$ be the positive damping constant.

-By Newton's second law:

$m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}$

where $x(t)$ is the displacement from equilibrium position. The equation can be transformed into:

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$ shich is the equation of motion.

7 0

3 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

3 years ago

Read 2 more answers

A 1,650 kg SUV comes uniformly to a stop. If the vehicle is accelerating at -1.3 m/s^2,

lions [1.4K]

Answer:b

Explanation:

4 0

3 years ago

Why is it important that two different metals are used in building an electrochemic<br> cell?

Mnenie [13.5K]

Answer:

Explanation:

They need a galvanic difference. Or saying that less technically, they need to have different electron attraction, so that one can collect electrons (oxidation/reduction) and flow current from the other. :)

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3 years ago

Best exercises for these topics. <br> (have the amount of exercises, have 4 exercises for each.)

Kipish [7]

Answer:

what is the question?

Explanation:

i dont understand

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2 years ago

Read 2 more answers