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3 years ago

Why does saturn and jupiter have more moons than the other planets in our solar system ?

Physics

1 answer:

Vikentia [17]3 years ago

5 0

Because they are larger planets the gravitational pull is heavier and they attract more celestial bodies

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When the current in a toroidal solenoid is changing at a rate of 0.0275 a/s, the magnitude of the induced emf is 12.4 mv. when t

slavikrds [6]

Answer:

41 turns

Explanation:

$\dfrac{d\phi}{dt}$ = Induced emf = 12.4 mV

$\dfrac{dI}{dt}$ = Current changing rate = 0.0275 A/s

L= Inductance

$\phi$ = Average flux = 0.00285 wb

N = Number of turns

Change in flux is given by

$\dfrac{d\phi}{dt}=L\dfrac{dI}{dt}\\\Rightarrow 12.4\times 10^{-3}=L0.0275\\\Rightarrow L=\dfrac{2.4\times 10^{-3}}{0.0275}\\\Rightarrow L=0.08727\ H$

Flux through each turn is given by

$\dfrac{\phi}{N}=L\dfrac{I}{N}\\\Rightarrow 0.00285=0.08727\times \dfrac{1.34}{N}\\\Rightarrow N=\dfrac{0.08727\times 1.34}{0.00285}\\\Rightarrow N=41.03221\ turns$

The number of turns is 41

3 0

3 years ago

A sports car accelerates at a constant rate from rest to a speed of 90 km/hr in 8 s. What is its acceleration?

AfilCa [17]

First convert 90km/hr to m/s.

Initiate velocity = 0m/s (car was at rest)

Final velocity is 25m/s (90km/hr converted)

25m/s - 0m/s / 8s = 3.125 m/s^s

Therefore the answer is option A (3.13m/s^2)

3 0

3 years ago

If an object is dropped from a tall building and hits the ground 3.0 s. later, what is the magnitude of the object's displacemen

marin [14]

Find the velocity of the object after one second.

v = vo + at

v = (0 m/s) + (9.8 m/s^2)(1 s)

v = 9.8 m/s

Now, using that, you can find the displacement in that one second between 1 and 2.

d = vot + (1/2)at^2

d = (9.8 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)^2

d = 14.7 m

5 0

3 years ago

Read 2 more answers

A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t

Talja [164]

Answer:

$P = 133.13 Watt$

Explanation:

Initial angular speed of the ferris wheel is given as

$\omega_i = 2\pi f$

$\omega_i = 2\pi(8.5/3600)$

$\omega_i = 0.015 rad/s$

final angular speed after friction is given as

$\omega_f = 2\pi f$

$\omega_f = 2\pi(7.5/3600)$

$\omega_f = 0.013 rad/s$

now angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

$\alpha = \frac{0.015 - 0.013}{15}$

$\alpha = 1.27 \times 10^{-4} rad/s^2$

now torque due to friction on the wheel is given as

$\tau = I \alpha$

$\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})$

$\tau = 8875.3 N m$

Now the power required to rotate it with initial given speed is

$P = \tau \omega$

$P = 8875.3 \times 0.015$

$P = 133.13 Watt$

8 0

3 years ago

A 60 kg student is standing motionless. Approximately how much force is the floor exerting on the student?

Solnce55 [7]

The normal force is equal in magnitude and opposite in direction to the weight of the student.
W = m · g = 60 kg · 10 m/s² = 600 N ( downward )
N = - 600 N
Answer:
B ) 600 N upward.

7 0

3 years ago