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Lady bird [3.3K]

3 years ago

11

The worked examples of charged-particle motion are relevant to:______. a. a transistor. b. a cathode-ray tube. c. magnetic reson

ance imaging.d. cosmic rays. e. lasers.

1 answer:

Burka [1]3 years ago

3 0

Answer:

B. Cathode Ray Tube

Explanation:

In 1897, British physicist J. J. Thomson showed that cathode rays were composed of a negatively charged particle, which was later named the electron.

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Degger [83]

Answer:

why would you waste points

Explanation:

4 0

3 years ago

Read 2 more answers

What is the mass of a box that accelerates to 10 m/s/s with a 70 Newton push?

Allisa [31]

Answer:

Mass of box = 7 kg

Explanation:

Force = 70 N

Acceleration = 10 m/s²

Force is product of mass and acceleration.

$\longrightarrow$ Force = Mass × Acceleration

$\longrightarrow$ 70 = Mass × 10

$\longrightarrow$ Mass = $\sf \dfrac{70}{10}$

$\longrightarrow$ Mass = 7 kg

8 0

4 years ago

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20°C, and is compressed at steady state to 12 bar, 80°C. The vol

3241004551 [841]

Answer:

The magnitude of the heat transfer rate from the compressor is 87.05 kW

Explanation:

Initial pressure of refrigerant = 4 bar

Final pressure of refrigerant = 12 bar

From steam table,

Internal energy at 4 bar (U1) = 2554 kJ/kg

Internal energy at 12 bar (U2) = 2588 kJ/kg

Change in internal energy (∆U) = U2 - U1 = 2588 - 2554 = 34 kJ

Work input (W) = 120 kJ/kg

Quantity of heat transfer (Q) = ∆U + W = 34 + 120 = 154 kJ/kg

Volumetric flow rate of refrigerant = 8 m^3/min = 8/60 = 0.133 m^3/s

Density of refrigerant = 4.25 kg/m^3

Mass flow rate = density × volumetric flow rate = 4.25 kg/m^3 × 0.133 m^3/s = 0.56525 kg/s

Q = 154 kJ/kg × 0.56525 kg/s = 87.05 kJ/s = 87.05 kW

6 0

3 years ago

A 400 kg satellite is in a circular orbit at an altitude of 500 km above the Earth's surface. Because of air friction, the satel

scoundrel [369]

Answer:

E = 1.45 x 10⁹ J = 1.45 GJ

Explanation:

According to the law of conservation of energy:

Potential Energy Lost by Satellite = Kinetic Energy + Internal Energy

$mgh = \frac{1}{2} mv^2 + E\\\\E = mgh - \frac{1}{2} mv^2$

where,

E = Internal Energy = ?

m = mass = 400 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 500 km = 500000 m

v = speed on ground = 1.6 km/s = 1600 m/s

Therefore,

$E = (400\ kg)(9.81\ m/s^2)(500000\ m)-\frac{1}{2} (400\ kg)(1600\ m/s)^2\\E = 1.962\ x\ 10^9\ J - 0.512\ x\ 10^9\ J$

<u>E = 1.45 x 10⁹ J = 1.45 GJ</u>

8 0

3 years ago

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

user100 [1]

Answer:

A) $\Delta P = 14512.5 Pa = 14.512 kPa$

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as $v_2 = 150 m/s$

since inside air is atmospheric , $v_1 = 0 m/s$

a) By using bernoulli equation between outside and inside of flight

$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh$

$\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$

$\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]$

$\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]$

$\Delta P = 14512.5 Pa = 14.512 kPa$

b) force exerted on window

Area of window $= 25\times 45 = 1125 cm^2 = 0.1125 m^2$

We know that force is given as

$F = P\times A$

$F = 14512.5 \times 0.1125$

F = 1632.65 N

5 0

4 years ago