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3 years ago

12

Answer correctly this 14 questions, please and earn 15pts and the brainlist answer!

2 answers:

sammy [17]3 years ago

5 0

1. K 2. P 3. K 4. K 5. P 6. K 7. K 8. P 9. K 10. P

11. A golf ball flying through the air. 12. A skateboard rolling down a hill.
13. A roller coaster car sitting at the top of a drop.
14. Leaning up against a wall.

nikitadnepr [17]3 years ago

3 0

K, P, K, K, P, K, K, P, K, P.
Kinetic:
Walking down the hall.
Dropping a pencil.
Potential:
A book sitting on the edge of a shelf.
A stretched out rubber band.

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The Atomic number tells us the number of ____ in an atom.

yawa3891 [41]

Answer:

Protons

Explanation:

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3 years ago

Read 2 more answers

Finding the spring constant as shown, spring 3, which has an unknown spring constant k3, replaces spring 2. the mass of the weig

nevsk [136]

Replaces spring 2. the mass of the weight and pulley are unchanged: m=5.8 kg and mp=1.7 kg

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4 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

3 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

4 years ago

The attraction will vary directly with the separation between the charges.

Burka [1]

No it won't. It'll vary inversely as the square of the separation.

4 0

3 years ago