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4 years ago

The electromagnet in a galavanometer.

Physics

1 answer:

hoa [83]4 years ago

8 0

<span>The electromagnet in a Galvanometer moves a pointer
along a numbered scale in response to a current. </span>

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A 2.37-kg rock is released from rest at a height of 22.2 m. ignore air resistance and determine (a) the kinetic energy at 22.2 m

marta [7]

Yes yes yes yes yes yes

8 0

3 years ago

A person throws a ball with a force of 140 N that accelerates at 15 m/s? What is the mass of the

mr_godi [17]

Newton's 2nd law

$\tt \sum F=m.a$

input the value:

$\tt m=\dfrac{\Sum F}{a}=\dfrac{140~N}{15~m/s^2}=9.3~kg$

6 0

3 years ago

An object-spring system undergoes simple harmonic motion with an amplitude A. Does the total energy change if the mass is double

son4ous [18]

Answer: No, The energy will remain the same

Explanation: Doubling the mass and leaving the amplitude unchanged won't have any effect on the total energy of the system.

At maximum displacement, E=0.5kA^2

Where E = total energy

K = spring constant

A = Amplitude

From the formula above : Total Energy is independent of mass,. Therefore, total energy won't be affected by Doubling the mass value of the object.

Also when the object is at a displacement 'x' from its equilibrium position.

E = Potential Energy(P.E) + Kinetic Energy(K.E)

P.E = 0.5kx^2

Where x = displacement from equilibrium position

E = Total Energy

K. E= E-0.5kx^2

From the relation above, total energy is independent of its mass and therefore has no effect on the total energy.

6 0

4 years ago

A compressor receives air at 290 K, 100 kPa and a shaft work of 5.5 kW from a gasoline engine. It is to deliver a mass flow rate

Sladkaya [172]

Answer:

$P_2=4091\ KPa$

Explanation:

Given that

T₁ = 290 K

P₁ = 100 KPa

Power P =5.5 KW

mass flow rate

$\dot{m}= 0.01\ kg/s$

Lets take the exit temperature = T₂

We know that

$P=\dot{m}\ C_p (T_2-T_1)$

$5.5=0.01\times 1.005(T_2-290})\\T_2=\dfrac{5.5}{0.01\times 1.005}+290\ K\\\\T_2=837.26\ K$

If we assume that process inside the compressor is adiabatic then we can say that

$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{0.285}$

$\dfrac{837.26}{290}=\left(\dfrac{P_2}{100}\right)^{0.285}\\2.88=\left(\dfrac{P_2}{100}\right)^{0.285}\\$

$2.88^{\frac{1}{0.285}}=\dfrac{P_2}{100}$

$P_2=40.91\times 100 \ KPa$

$P_2=4091\ KPa$

That is why the exit pressure will be 4091 KPa.

4 0

3 years ago

during a crash, the acceleration was less than 30g. Calculate the force on a 70 kg person accelerating at this rate.

Igoryamba

Answer:

less than 20580 N

Explanation:

According to the newton's second law of motion

Force = mass * acceleration

(assuming gravitational acceleration =9.8 m/s2 )

acceleration = 30*9.8 = 294 m/s2

acting Force = 70 * 294

= 20580 N

Since the acceleration was less than 30g , acting force should also be less than 20580 N

4 0

3 years ago