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Ahat [919]
3 years ago
15

The force supplied by a spring depends on what?

Physics
1 answer:
Morgarella [4.7K]3 years ago
8 0
The force and motion
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A 900 kg car is traveling at 20 m/s along the road. What force must be applied to the car to stop it in a distance of 30 m2 Assu
serg [7]

Answer:

A. 6000 N

Explanation:

v²=u²+2as

0²=20²+2x30xa

-400=60a

a=-400/60

a =-6.667m/s²

f =ma

f = 900 x 6.667 = 6003N

F = 6000N

5 0
3 years ago
Why do all planets rotate around the sun in the ssame direction
Nutka1998 [239]
This is because of the conservation of angular momentum, which allows most planets to spin in the same direction, but Venus and Uranus have seem to ignore this as they spin in different directions, as Venus spins clockwise and Uranus is on its side
3 0
3 years ago
g If the x-component of a force vector is 5.69 newtons and its y-component is 8.00 newtons, then what is its magnitude?
beks73 [17]

Answer:

F = 9.82 N

Explanation:

given,

Force x-component = 5.69 N

Force y-component = 8 N

magnitude of force = ?

Resultant of force

F = \sqrt{F_x^2 + F_y^2}

F = \sqrt{5.69^2 + 8^2}

F = \sqrt{32.3761 + 64}

F = \sqrt{96.3761}

F = 9.82 N

Hence, the magnitude of force is equal to 9.82 N

4 0
3 years ago
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Which of the following is a scalar quantity?
neonofarm [45]

Answer:

55

Explanation:

I hope this answer help u

4 0
2 years ago
Read 2 more answers
A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

3 0
3 years ago
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