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Neko [114]
3 years ago
11

A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac s

ource. What is the current I1 in the primary coil
Physics
1 answer:
Nataly [62]3 years ago
4 0

Answer:

I = 0.5 A

Explanation:

Given: P=60 Watts, Voltage supply V = 120 Volts (for primary coil)

Solution:

we have P = V I

⇒ I = P /V = 60 Watts / 120 Volts

I = 0.5 A

You might be interested in
PART ONE
Helga [31]

Answer:

1.129×10⁻⁵ N

1.295 m

Explanation:

Take right to be positive.  Sum of forces on the 31.8 kg mass:

∑F = GM₁m / r₁² − GM₂m / r₂²

∑F = G (M₁ − M₂) m / r²

∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²

∑F = 1.129×10⁻⁵ N

Repeating the same steps, but this time ∑F = 0 and we're solving for r.

∑F = GM₁m / r₁² − GM₂m / r₂²

0 = GM₁m / r₁² − GM₂m / r₂²

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

516 / r² = 207 / (0.482 − r)²

516 (0.482 − r)² = 207 r²

516 (0.232 − 0.964 r + r²) = 207 r²

119.9 − 497.4 r + 516 r² = 207 r²

119.9 − 497.4 r + 309 r² = 0

r = 0.295 or 1.315

r can't be greater than 0.482, so r = 0.295 m.

5 0
4 years ago
3.
nikitadnepr [17]
The answer is
B:because
8 0
3 years ago
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

4 0
3 years ago
Read 2 more answers
If you wanted to detect x rays coming from the sun, where would you place the detector? Why?
Damm [24]
You would have to place your sensor above earth's atmosphere because it blocks out nearly all x-rays. this is why we have the Chandra observatory

hope this helps
4 0
3 years ago
Read 2 more answers
Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro
luda_lava [24]

Answer:

c. 4.582\times10^{21} kg

Explanation:

r_{i} = Initial distance between asteroid and rock = 7514 km = 7514000 m

r_{f} = Final distance between asteroid and rock = 2823 km = 2823000 m

v_{i} = Initial speed of rock = 136 ms⁻¹

v_{f} = Final speed of rock = 392 ms⁻¹

m = mass of the rock

M = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg

8 0
3 years ago
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