All categories

2 years ago

PLEASEEE HELPPPPPPP:

Physics

2 answers:

tatuchka [14]2 years ago

7 0

V(voltage) = I(current)R(resistance)
substitute in the values

V = 15 * 0.10
V = 1.5 volts

Sophie [7]2 years ago

4 0

Voltage is an electromotive force or potential difference expressed in volts.

You might be interested in

HALP MEH! 50 PTS!!!!

PilotLPTM [1.2K]

A free body diagram shows the vectors for all the forces acting in the body

5 0

3 years ago

Read 2 more answers

One problem that fireplace designers try to solve is how to get more of the fire's heat into the room. Where does the rest of th

Eva8 [605]

Answer:

the rest of the heat enters the holes in the house

4 0

2 years ago

Which formula is used to calculate average velocity?

Flauer [41]

Answer:

Explanation:

<h3>for average velocity we use this formula V Vavg =V1+V2<u><em>÷</em></u><u>2</u></h3>

5 0

3 years ago

A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

Kisachek [45]

Answer:

$B_2 = 1.71 mT$

Explanation:

As we know that the magnetic field near the center of solenoid is given as

$B = \frac{\mu_0 N i}{L}$

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

$\frac{B_1}{B_2} = \frac{L_2}{L_1}$

now plug in all values in it

$\frac{2.0 mT}{B_2} = \frac{21}{18}$

$B_2 = 1.71 mT$

3 0

2 years ago

A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.

pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

$h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t = 6.55 \ s$

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

$X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m$

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0

2 years ago