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3 years ago

Andrew skis down a hill.

Physics

1 answer:

IceJOKER [234]3 years ago

5 0

Explanation:

gravitational potential energy at the top of the hill, which transforms into kinetic energy as he moves bottom of the hill

that's mean potential energy transfoms into kinetic energy

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Which country’s data center industry taps latest in energy-efficient technology?.

SCORPION-xisa [38]

Answer:

N i g e r i a

Explanation:

Hope that helps!!!

Have A good day/night!!

3 0

2 years ago

A car is moving in uniform circular motion. If the car's speed were to double to keep the car moving with the same radius, the a

Anestetic [448]

Answer:

Increase by a factor of 4.

Explanation:

The acceleration of a car moving with speed v in a circle of radius R is given by:

$a=\frac{v^2}{R}.$

Now if we double the speed $v$ in the equation above, it becomes $2v$ . Thus:

$a=\frac{v^2}{R}\:\: {\rightarrow}\:\: a_n=\frac{(2v)^2}{R}=4\frac{v^2}{R}=4a.$

Therefore the acceleration is increased by a factor of 4.

7 0

3 years ago

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

kramer

Answer:

$v_x = -6\ m/s$

Explanation:

initial velocity

magnitude of velocity, v = 12 m/s

angle made of velocity with negative x-axis,θ = 60°

We need to calculate x- component of v

$v_x = v cos \theta$

velocity is in negative x-direction, v = -12 m/s

now,

$v_x = -12\times cos 60^0$

$v_x = -12\times 0.5$

$v_x = -6\ m/s$

Hence, the velocity x-component is equal to -6 m/s.

5 0

3 years ago

What is the approximate energy required to raise the temperature of 1.00 L of hydrogen by 90 °C? The pressure is held constant a

Oliga [24]

Answer:

Q = 116.8 J

Explanation:

Here given that the temperature of 1 L hydrogen is increased by 90 degree C at constant pressure condition.

So here we will have

$Q = n C_p \Delta T$

here we know that

n = number of moles

$n = \frac{1}{22.4}$

$n = 0.0446$

for ideal diatomic gas molar specific heat capacity at constant pressure is given as

$C_p = \frac{7}{2}R$

now we have

$Q = (0.0446)(\frac{7}{2}R)(90)$

$Q = 116.8 J$

3 0

3 years ago

Suppose that a meter stick is balanced at its center. A 0.24 kg mass is then positioned at the 6-cm mark. At what cm mark mus

Scilla [17]

Answer:

80.17 cm

Explanation:

Taking moments of forces about the center, the total clockwise moments is equal to the total counter clockwise moment:

Force * distance (counter clockwise) = force * distance (clockwise)

0.24 * 9.8 * (50 - 6) = 0.35 * 9.8 * (x - 50)

0.24 * 44 = 3.43x - 171.5

103.5 = 3.43x - 171.5

=> 3.43x = 103.5 + 171.5

3.43x = 275

x = 275/3.43 = 80.17 cm

8 0

3 years ago