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3 years ago

7. DRAW A PICTURE TO SHOW WORK.

Physics

1 answer:

Degger [83]3 years ago

8 0

Answer:
Distance covered 28 km
displacement is 22.8 km North-East
Explanation;

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Se um corpo 1 está em movimento em relação a um corpo 2, e se o corpo 2 está em repouso em relação a um corpo 3, pode-se afirmar

stealth61 [152]

Yo habla ingles no espanol

8 0

3 years ago

Read 2 more answers

When a wire is made thicker its resistance what?

NeTakaya

Making a wire thicker has the same effect as making a road wider. It makes it easier for the electron traffic to flow. The resistance decreases, and the current (traffic) increases.

7 0

3 years ago

A secret agent is locked in a room. He pushes against the door but cannot open it. Finally, he falls to the floor exhausted. Has

Vitek1552 [10]

Answer: There is not work done at the door because the door did not move.

Explanation: Work is defined as the movement done by a force.

So if you move to apply a force F in an object and you move it a distance D, the work applied on the object is

W = F*D

In this case, the secret agent pushes against the door, so there is a force, but the agent does not move the door, so D = 0, so there is no motion of the door, which implies that there is no work done at the door.

8 0

3 years ago

A 10 µf capacitor is charged to 108 v and is then connected across a 328 ω resistor. what is the initial charge on the capacitor

valina [46]

The capacitance is defined as the maximum charge stored in a capacitor, Q, divided by the voltage applied, V:
$C= \frac{Q}{V}$

The capacitor is initially charged with the battery of 108 V, so the the initial charge on the capacitor can be found by re-arranging the previous formula:
$Q=CV=(10 \mu F)(108 V)=1080 \mu C$

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4 years ago

A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o

dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?

we know $p = \frac{v^2}{R}$ .....................(i)

we need to find change in power

$\Delta P = \frac{\Delta (V^2)}{R}$

$\Delta P = \frac{2 V \Delta V}{R}$ ..............(ii)

from equations we get

$\frac{\Delta P}{P} = \frac{2 \Delta V}{V}$

$\frac{\Delta P}{P} = 2 \frac{110 -120}{120}$

$\frac{\Delta P}{P} = -2(\frac{10}{120})$

$\frac{\Delta P}{P} = - 0.16 %$

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller

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3 years ago