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2 years ago

What velocity is needed for a 2400 kg car to have the same amount of momentum as an 800 kg car?

Physics

1 answer:

maks197457 [2]2 years ago

7 0

Answer:

since velocity is directly proportional to momentum, the heavier car will need 2400/800 = 3 times the speed of the small car.

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Look at the circuit diagram. Which of these components is part of the circuit?

Roman55 [17]

Answer:

b ac power source

Explanation:

ieififuffucjcjcicogore8e8e9e9e9e9ew0ww0w0w

7 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

how long does it take adrian to reach his destination if his displacement is 253 meters and he walks north with a velocity of 1.

klemol [59]

Velocity = displacement/time

1.6 = 253/t

t = 158.125

It takes them about 158 seconds

4 0

3 years ago

A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the s

Andrej [43]

Answer:

$1300 m$

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from $20 s$ to $70 s$ is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.