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masya89 [10]
3 years ago
8

if the resistance of a car headlight is 15 ohm and the current through it is 0.60, what is the voltage across the headlight?

Physics
2 answers:
alex41 [277]3 years ago
5 0
Answer is 9 V you use Ohm’s Law to solve the problem
Strike441 [17]3 years ago
4 0

Answer:

9 volts (assuming 0.60 is in Amperes)

Explanation:

Recall that Ohms law can be expressed as

V = IR, where

V = voltage,

I = current (given as 0.6. I'm going to assume that the units is Amperes because it is not given)

R = resistance (given as 15 ohm)

substituting the above values into the formula

V = IR

V = (0.6)(15)

V = 9 Volts

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A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface. The surface of the ramp is the x-
kifflom [539]

Magnitude of the force  of tension: 139 N

Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

  • The force of tension, F_T, acting up along the plane
  • The force of friction, F_f=14.8 N, acting down along the plane
  • The component of the weight in the x-direction, F_{gx}, acting down along the plane

We know that the magnitude of the weight is

F_g=70.0 N

So its x-component is

F_{gx}=F_g sin \theta =(70.0)(sin 22^{\circ})=26.2 N

The net force along the x-direction can be written as

F_x = F_T-F_f-F_{gx}

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

F_T=F_x+F_f+F_{gx}=98+14.8+26.2=139 N

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0
3 years ago
Read 2 more answers
A bus is moving at a speed of 150km/hr. Begins to slow at a constant rate of 3.0m/s each second. Find how far it goes before sto
ladessa [460]

Answer:

Distance = 13.9 meters

Explanation:

Given the following data;

Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s

Decelerating speed = 3m/s

To find the distance travelled with this speed;

Distance = maximum speed/decelerating speed

Distance = 41.67/3

Distance = 13.9 meters

Therefore, the bus would travel a distance of 13.9 meters before stopping.

4 0
2 years ago
Two small nonconducting spheres have a total charge of 93.0 μC . Part A
Travka [436]

Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

5 0
2 years ago
PLEASE HELP ASAP!!!
Hatshy [7]
The distance the spring stretches is the answer. 

I hope this helps. 

Have a nice day. 
5 0
3 years ago
3. The shrillness of sound is known as_________​
swat32

Answer:

pitched sound

Explanation:

The shrillness of sound is known as   Pitched Sound

Please mark me as brilliant

plz plz plz

3 0
2 years ago
Read 2 more answers
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