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horrorfan [7]
2 years ago
10

Approximately how much electrical energy does a 5-W lightbulb convert to radiant and thermal energy in one hour?​

Physics
1 answer:
k0ka [10]2 years ago
8 0

Answer:

18,000 j

Explanation:

the lightbulb dissipates 5W of power

P = ΔE / Δt

rearrange to solve for energy

ΔE = PΔt

P = 5W

t =  1 hour = 60 minutes = 3600 seconds

ΔE = 5 * 3600

ΔE = 18000 J

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The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given  

N = W Cos30

N = (500) Cos30

N = 433 N

3 0
3 years ago
HELP PLEASE!!
aliya0001 [1]

Answer:

m=41 kg

v=0,02 ms

R=2,1 m

F-?

a=v²/R

a=(0,02) ²/2,1=0,0002

F=m*a

F=41*0,0002=0,0082 H

F=0,0082 H

3 0
3 years ago
A car with a velocity of 22 m/s is accelerated at a rate of 1.6m/s2 for 6.8s. determine the final velocity
Ivahew [28]

A car with a velocity of 22 m/s is accelerated at a rate of 1.6 m/s^2 for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

time, t = 6.8 s

acceleration, a = 1.6 m/s^2

We will be using the equation of motion:

v = u + at

\therefore v=22+1.(6.8)

\Rightarrow v=22+10.88

\Rightarrow v=32.88 \ m/s

The final velocity become 32.88 m/s.

To learn more about Attention here:

https://brainly.in/question/10557838

#SPJ4

3 0
2 years ago
Use the right-hand rule for magnetic force to determine the charge on the moving particle. This is a charge.
german

Answer:

Use the right-hand rule for magnetic force to determine the charge on the moving particle.

This is a  

negative

charge

Explanation:

4 0
3 years ago
Read 2 more answers
An object rotates about a fixed axis, and the angular position of a reference line on the object is given by θ = 0.220e3t, where
OLga [1]

Answer:a_{t}=3.96

a_{c}=0.8712

Explanation:

Given

\theta =0.220e^{3t}

r=2cm

Now angular velocity is given by \omega =\frac{\mathrm{d}\theta}{\mathrm{d}t}

\omega =0.66e^{3t}

Now linear velocity(v) is given =\omega r

v=1.32e^{3t}

Now tangential component of acceleration is given by

a_{t}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=3.96e^{3t}

at t=0

a_{t}=3.96cm/s^2

radial component of acceleration is given by

a_{c}=\omega ^{2}r

a_{c}=0.4356e^{6t}\times 2

a_{c}=0.8712e^{6t} cm/s^{2}

at t=0

a_c=0.8712 cm/s^{2}

5 0
3 years ago
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