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3 years ago

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A freezer has a coefficient of performance of 4.1. You place 0.45 kg of water at 16°C in the freezer, which maintains its temper

ature of -18°C. In this problem you can take the specific heat of water to be 4190 J/kg/K, the specific heat of ice to be 2100 J/kg/K, and the latent heat of fusion for water to be 3.34 ×105 J/kg. How much additional energy, in joules, does the freezer use to cool the water to ice at -18°C?

1 answer:

velikii [3]3 years ago

5 0

Ohh this is quite a lot umm i’ll be back hold on

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Kids are pelting a window with snowballs. On average, two snowballs of roughly 300-g mass hit the window each second, moving hor

hram777 [196]

Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs $m = 300 \times 10^{-3}$ Kg

Velocity of snowball $v = 10 \frac{m}{s}$

For finding the average force,

Force is equal to the change in momentum,

$F = \frac{dP}{dt}$

Here, final velocity is zero so we write,

$F = \frac{mv}{1}$

Where $dt = 1$ sec

$F = 300 \times 10^{-30} \times 10$

$F = 3$ N

Above value of force is due to one ball, but here given in question there are two ball,

$F = 3 \times 2$

$F = 6$ N

Therefore, the average force exerted on the window due to two snowballs is 6 N

5 0

3 years ago

On a nice summer day,Kim takes her niece Madison for a walk in her stroller.If they start from rest and accelerate at a rate of

11111nata11111 [884]

2.5m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 0.5m/s²

time of travel = 5s

Solution:

Final velocity = ?

Solution:

Acceleration can be defined as the change in velocity with time:

Acceleration = $\frac{Final velocity - Initial velocity}{time}$

From the equation above, the unknown is final velocity:

Final velocity - initial velocity = Acceleration x time

since initial velocity = 0

Final velocity = 0.5 x 5 = 2.5m/s

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

6 0

3 years ago

the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range

Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

$v_y(t)=vo*sin(A)-g*t$

the velocity is Zero when the projectile reach in the maximum altitude:

$0=vo-gt\\t=\frac{vo}{g}$

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

$d_x(t)=vo*cos(A)*t\\$

R=Range

$R=d_x(t=2*\frac{vo}{g})$

$R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}$

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

$\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\$

2B=60°

B=30°

7 0

3 years ago

Compare and contrast electric potential energy and electric potential difference? Explain.

forsale [732]

Answer:

<u><em>Electric Potential Energy:</em></u>

The energy that is needed to move a charge against an electric firld is called Electric Potential Energy

<u><em>Electric Potential Difference:</em></u>

The amount of work done in carrying a unit charge from one point to an other in an electric field is called Electric Potential Difference.

<u><em>Relation:</em></u>

Relation between Electric potential and electrical potential energy is given by

$\delta V=\frac{PE}{q}$

Here PE represents Electric potential energy

and $\delta V$ is Electric potential difference

it means electric potential difference is the difference in electric potential energy divided by the charge.

6 0

3 years ago

When the distance between two stars decreases by one-third, the force between them

pashok25 [27]

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

$r'=\frac{2}{3}r$

So the new force will be

$F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F$

So, the force will be 2.25 times the previous value.

4 0

3 years ago