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2 years ago

The table shows the potential energy and kinetic energy of a skier at two different positions on a hill.

Physics

2 answers:

Rasek [7]2 years ago

8 0

Answer:

30,000 Units because, It remains unchanged.

Explanation:

Did the test, It's correct

Serggg [28]2 years ago

6 0

Answer:

30,000 units, because total energy remains unchanged

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Where a subducting plate slides beneath the lithosphere, melting takes place and a(n) ____ is created?

11111nata11111 [884]

under ground cave will be created

3 0

2 years ago

A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be

stepan [7]

Answer:

$p_2 = 664081 N/m^{2}$

Explanation:

from the ideal gas law we have

PV = mRT

$P = \rho RT$

$\rho = \frac{P}{RT}$

HERE R is gas constant for dry air = 287 J K^{-1} kg^{-1}

$\rho = \frac{7.00 10^{5}}{287(18+273)}$

$\rho = 8.38 kg/m^{3}$

We know by ideal gas law

$\rho = \frac{m_1}{V_1}$

$m_1 = \rho V_1 = 8.38 *2*10^{-3}$

$m_1 = 0.0167 kg$

for m_2

$m_2 = \rho V_i - V_removed$

$m_2 = 8.38*(.002 - 10^{-4})$

$m_2 = 0.0159 kg$

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

$\frac{p_2}{p_1}=\frac{m_2}{m_1}$

$p_2 = p_1 * \frac{m_2}{m_1}$

$p_2 =7*10^{5} * \frac {.0159}{0.0167}$

$p_2 = 664081 N/m^{2}$

8 0

2 years ago

What kind of symmetry do you have

Keith_Richards [23]

During the daytime, I have mostly line symmetry.

During the night, I often have almost spherical symmetry.

5 0

2 years ago

Read 2 more answers

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does n

Shkiper50 [21]

Answer:

Explanation:

velocity of sound in air at 20⁰C is 343 m /s

velocity of sound in water at 20⁰C is 1481 m /s

The wavelength of the sound is 2.86 m in the air so its frequency

= 343 / 2.86 = 119.93 .

This frequency of 119.93 will remain unchanged in water .

wavelength in water = velocity in water / frequency

= 1481 / 119.93

= 12. 35 m .

7 0

2 years ago

A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int

Inessa05 [86]

Let F be the magnitude of the force applied to the cart, m the mass of the cart, and a the acceleration it undergoes. After time t, the cart accelerates from rest v₀ = 0 to a final velocity v. By Newton's second law, the first push applies an acceleration of

F = m a → a = F / m

so that the cart's final speed is

v = v₀ + a t

v = (F / m) t

If we force is halved, so is the accleration:

a = F / m → a/2 = F / (2m)

So, in order to get the cart up to the same speed v as before, you need to double the time interval t to 2t, since that would give

(F / (2m)) (2t) = (F / m) t = v

3 0

2 years ago