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MAVERICK [17]
1 year ago
12

A 6V radio with a current of 2A is turned on for 5 minutes. Calculate the energy transferred in joules

Physics
1 answer:
iogann1982 [59]1 year ago
7 0

Answer:

R=V/I=6/2=3ohm

time =5minutes =5*60=300seconds

I=2A

Heat =I^2Rt=(2)^2*3*300=4*900=3600J

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The length of a 60 W, 240 Ω light bulb filament is 60 cm Remembering that the current in the filament is proportional to the ele
faust18 [17]

Answer:

Finally current will be

i = 0.35 A

Explanation:

As we know that power of the bulb is given by the formula

P = \frac{V^2}{R}

now we have

P = 60 W

R = 240 ohm

so we have

60 = \frac{V^2}{240}

V = 120 Volts

now the current in the bulb is given as

i = \frac{V}{R}

i = \frac{120}{240} = 0.5 A

now when length of the filament is double

so the resistance of the wire also gets double

so we have

P = \frac{V^2}{R}

60 = \frac{V^2}{480}

V = 169.7 volts

now the current in the bulb is given as

V = i R

169.7 = i(480)

i = 0.35 A

8 0
2 years ago
A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol
qaws [65]

Answer:

Part a)

v = \sqrt{v_o^2 + g^2t^2}

Part b)

d = \frac{1}{2}gt^2

Part c)

v_f = v_o - gt

Part d)

d = \frac{1}{2}gt^2

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

v_y = -gt

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{v_o^2 + g^2t^2}

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

d = \frac{1}{2}gt^2

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

v_f = v_i + at

v_f = v_o - gt

Part d)

distance from helicopter

d = \frac{1}{2}gt^2

8 0
2 years ago
Read 2 more answers
Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth a
Natasha_Volkova [10]

Answer:

I. The horizontal distance traveled by the bullet is greater for the Moon.

II. The flight time is less for the bullet on the Earth.

Explanation:

Horizontal distance depends on the initial speed, height and gravity. Bullets have the same initial speed and are shot from the same height. In these conditions horizontal distance only depends on gravity, which is inversely proportional. Therefore, the less gravity the greater the horizontal distance. Gravity slows bullet and causes its impact on the ground. Since gravity is greater in Earth, the bullet hits faster on the earth.

3 0
1 year ago
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Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi
Doss [256]

Answer:

Explanation:

Given that,

Basket ball is drop from height

H=10m

It is dropped on planet mass

And the acceleration due to gravity on Mars is given as

g= 3.7m/s²

Time taken for the ball to reach the ground

Initial velocity of the body is zero

u=0m/s

Using equation of motion: free fall

H = ut + ½gt²

10 = 0•t + ½ × 3.7 ×t²

10 = 0 + 1.85t²

10 = 1.85t²

Then, t² =10/1.85

t² = 5.405

t = √ 5.405

t = 2.325seconds

So the time the ball spend on the air before reaching the ground is 2.325 seconds

5 0
2 years ago
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
netineya [11]

Answer:

E=\frac{\lambda}{2\pi r\epsilon_0}

Explanation:

We are given that

Linear charge density of wire=\lambda

Radius of hollow cylinder=R

Net linear charge density of cylinder=2\lambda

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

\oint E.dS=\frac{q}{\epsilon_0}

q=\lambda L

E(2\pi rL)=\frac{L\lambda}{\epsilon_0}

Where surface area of cylinder=2\pi rL

E=\frac{\lambda}{2\pi r\epsilon_0}

8 0
2 years ago
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