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Jobisdone [24]
3 years ago
7

A constant current of I = 15 A exists in a solenoid whose inductance is L = 2.8 H. The current is then reduced to zero in a cert

ain amount of time. (a) If the current goes from 15 to 0 A in a time of 75 ms, what is the emf induced in the solenoid? (b) How much electrical energy is stored in the solenoid? (c) At what rate must the electrical energy be removed from the solenoid when the current is reduced to 0 A in a time of 75 ms? Note that the rate at which energy is removed is the power.
Physics
1 answer:
julia-pushkina [17]3 years ago
7 0

Answer:

Explanation:

Given that

Constant current I=15A

Inductance L=2.8H

a. Current changes from 15A to 0A.

∆I=0-15=-15

For a time of t=75ms=0.075s

EMF=?

EMF is given at

EMF=-L(∆I/t)

EMF=-2.8×-15/0.075

EMF=3.15V

b. Energy stored?

Energy stored is given as

E=½LI²

E=½×2.8×15²

E=315 J

c. Rate of energy? Which is power and the rate of energy is given as

Power=Energy/timetaken

Power=315/0.075

Power=4200W

P=4.2KW

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One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa
Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

<u>ΔP.E = 6.48 x 10⁸ J</u>

7 0
3 years ago
One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

4 0
3 years ago
What gravitational force does the earth exert on a person
lara [203]

The  gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

<h3>What is the gravitational force of the earth on the person?</h3>

The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:

  • F = \frac{Gm^{1}m^{2}}{r^{2}}

where

G = 6.67 * 10⁻¹¹

m¹ = 62 kg

m² = 5.97 * 10²⁷ kg

r = 6.4 * 10⁶ m

F = \frac{5.97*10^{24}*62*6.67*10^{-11}}{(6.4*10^{6}){2}} = 602.74\:N

Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

Learn more about gravitational force at: brainly.com/question/940770

#SPJ1

7 0
2 years ago
The temperature of a 700.96 gram piece of metal falls 120⁰C and in the process releases 2001 Joules of energy. What is the speci
olga nikolaevna [1]

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C  - 100°C   = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

8 0
3 years ago
8. Mark is sitting in a chair, legs out in front of him resting on an ottoman. His knees are.....
anyanavicka [17]
His knees are flat in front of him or 180 degrees, i hope that answered your question   
5 0
3 years ago
Read 2 more answers
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