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3 years ago

What did Blaise Pascal discover, and how did he impact Physics?

Physics

1 answer:

yan [13]3 years ago

6 0

Answer:

Well it really depends because he found out about a lot of things in a lot of scientific fields actually. He is known for contributing Pascal's triangle and probability theory. He also invented an early digital calculator and a roulette machine. In the field of physics, Blaise contributed to the study of atmospheric pressure by discovering that vacuums are real and exist in the real world.

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The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago

A 12.0-g bullet is fi red horizontally into a 100-g wooden block initially at rest on a horizontal surface. After impact, the bl

allochka39001 [22]

Answer:

$v_1 = 91.3 m/s$

Explanation:

By energy conservation and work energy theorem we can say that after bullet hits the block, it will move on the rough floor and comes to rest

so here work done by frictional force = change in kinetic energy

so we know that

$W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$-\mu mg d = \frac{1}{2}m(v_f^2 - v_i^2)$

$-0.650(9.81)(7.5) = \frac{1}{2}(0 - v_i^2)$

$47.8 = \frac{1}{2}v^2$

$v = 9.78 m/s$

now by momentum conservation we have

$m_1 v_1 = (m_1 + m_2) v$

$12 v_1 = (100 + 12) 9.78$

$v_1 = 91.3 m/s$

3 0

3 years ago

a hammer is dropped on the moon. it reaches the ground 5 seconds later. if the distance it fell was 5 meters, calculate the acce

Artyom0805 [142]

It is one meters per second

8 0

3 years ago

How many balance electrons does sulfur and silicon has?

marusya05 [52]

I believe you meant valence electrons?

Sulfur has 6 valence e-, and Silicon has 4.

Hope this helps! :)

4 0

4 years ago

a suction cup works by creating a vacuum between the rubber cup and surface its stuck to. if the total area of the rubber cups i

Licemer1 [7]

<span>1.0x10^3 newtons of force 1.0x10^2 kilograms of weight (assuming local gravity of 9.8 m/s^2) I will assume that the external pressure is 1 atmosphere (101325 Pascals) since it wasn't specified in the problem. So let's simply multiply the area of the suction cup by the pressure and see what we get: 101325 P * 0.01 m^2 = 101325 kg/(m*s^2) * 0.01 m^2 = 1013.25 kg*m/s^2 = 1013.25 N So the suction cup can support 1013.25 Newtons of force. Assuming the local gravitational acceleration is 9.8 m/s^2, let's see how many kilograms that is 1013.25 N / 9.8 m/s^2 = 1013.25 kg*m/s^2 / 9.8 m/s^2 = 103.3928571 kg Rounding to 2 significant figures gives 1.0x10^3 newtons, or 1.0x10^2 kilograms.</span>

5 0

3 years ago