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BARSIC [14]
3 years ago
10

A 24-V battery is powering a light bulb with a resistance of 3.0 ohms. What is the current flowing through the bulb? A) 7.20 A B

) 8.0 A C) 12.0 A D) 72.0 A
Physics
1 answer:
Usimov [2.4K]3 years ago
3 0

According to Ohm's law for a portion of the circuit we have:

U=RI=>I=U/R=24/3=8 A

The correct answer is  B


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Jill applies a force of 15 N to a wrench. If
Ivenika [448]
Jill is the input, as she creates the force. The wrench is the output because it gives the force to the finish peace of the chain.
6 0
3 years ago
A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
GREYUIT [131]

Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

6 0
3 years ago
amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i
Harman [31]

Answer:

Work done = 422.45 kJ

Explanation:

given,                                  

weight of equipment = 6 kN      

coefficient of kinetic friction = 0.05          

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²                    

Work done by the camper = ?                

actual acceleration acting a'      

m a = m a' - μ mg            

a' = a + μ g                      

a' = 0.2 + 0.05 x 9.8                

a' = 0.69 m/s²                              

Work done = Force x distance

F = m a'                                                            

F = \dfrac{6000}{9.8} \times 0.69

F = 422.44897 N                            

Work done = F x d                          

Work done = 422.44897 x 1000

Work done = 422449 J                  

Work done = 422.45 kJ            

4 0
3 years ago
A 1.5-kg block slides at rest starts sliding down a snow-covered hill Point A, which has an altitude of 10 m. There is no fricti
damaskus [11]

Answer:

the speed of the block when it reaches point B is 14 m/s

Explanation:

Given that:

mass of the block slides = 1.5 - kg

height = 10 m

Force constant  = 200 N/m

distance of rough surface patch = 20 m

coefficient of kinetic friction = 0.15

In order to determine the speed of the block when it reaches point B.

We consider the equation for the energy conservation in the system which can be represented by:

\dfrac{1}{2}mv^2=mgh

\dfrac{1}{2}v^2=gh

v^2=2 \times g \times   h

v^2=2 \times 9.8 \times   10

v=\sqrt{2 \times 9.8 \times   10

v=\sqrt{196

v = 14 m/s

Thus; the speed of the block when it reaches point B is 14 m/s

7 0
3 years ago
Which mixture is most likely to be homogeneous?
BlackZzzverrR [31]
C would be the answer
4 0
2 years ago
Read 2 more answers
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