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3 years ago

What provides the centripetal force needed to keep Earth in orbit?

Physics

2 answers:

anyanavicka [17]3 years ago

8 0

The centripetal force is the gravitational attraction
between the Earth and the Sun.

kkurt [141]3 years ago

3 0

The centripetal force needed to keep earth in orbit is gravity.

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A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res

Luden [163]

The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW

8 0

3 years ago

Inga [223]

Answer is c, they are equal:

Explanation:

4 0

3 years ago

Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp

erma4kov [3.2K]

Answer:

c. 981 watts

$P=981\ W$

Explanation:

Given:

horizontal speed of treadmill, $v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}$

weight carried, $w=588.6\ N$

grade of the treadmill, $G\%=10\%$

<u>Now the power can be given by:</u>

$P=v.w$

$P=588.6\times\frac{5}{3}$ (where grade is the rise of the front edge per 100 m of the horizontal length)

$P=981\ W$

6 0

3 years ago

Which branch of physics deals with the study of force, energy, and motion?

dexar [7]

The correct answer is Mechanics

4 0

3 years ago

Read 2 more answers

A 0.260 m radius, 525 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

Sophie [7]

Answer:

B = 0.37T

Explanation:

In order to calculate the needed magnitude of the magnetic force you use the following formula, which calculate the induced emf of the solenoid when there is a change in the magnetic flux:

$emf=-N\frac{\Delta \Phi_B}{\Delta t}=-N\frac{\Delta (BAcos\theta)}{\Delta t}$ (1)

emf: induced voltage in the solenoid = 10,000V

N: turns of the solenoid = 525

ФB: magnetic flux

B: magnitude of the magnetic field = ?

A: cross-sectional area of the solenoid = π*r^2

r: radius of the cross-sectional area = 0.260m

Δt: interval time of the change of the magnetic flux = 4.17ms = 4.17*10^-3s

First, you have the magnetic field direction perpendicular to the plane of the solenoid, after, the angle between them is 90° (quarter of a revolution)

In the equation (1) the only parameter that changes on time is the angle, then, you can solve for B from the equation (1):

$emf=-NBA\frac{cos(90\°)-cos(0\°)}{\Delta t}=\frac{NBA}{\Delta t}\\\\B=\frac{(\Delta t)(emf)}{NA}=\frac{(\Delta t)(emf)}{N(\pi r^2)}\\\\$

Finally, you replace the values of the parameters to calculate B:

$B=\frac{(4.17*10^{-3}s)(10000V)}{(525)(\pi(0.260m)^2)}=0.37T$

The strength of the magnetic field is 0.37T

7 0

3 years ago