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natima [27]
4 years ago
15

1. 3 V is applied across a 6Ω resistor. What is the current?

Physics
1 answer:
pav-90 [236]4 years ago
4 0

The current is 0.5A

<u>Explanation:</u>

Given:

Voltage, V = 3 V

Resistance, R = 6Ω

Current, I = ?

We know:

Voltage = current X resistance

Substituting the value we get:

3V = I X 6

I = \frac{3}{6} A

I = 0.5A

Therefore, the current is 0.5A

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Solar energy is the most abundant energy resource on earth.
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What a medium-mass star becomes at the end of its life?
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Answer:

What a medium-mass star becomes after a planetary nebula; a very bright, dense mass about the size of the planet Earth. ... The process that generates all of the energy that a star produces. Supernova. A Red Super Giant explodes into this when it runs out of elements to fuse together.

Explanation:

7 0
3 years ago
A 800-gram grinding wheel 27.0 cm in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the cent
SSSSS [86.1K]

Explanation:

d = Diameter of wheel = 27 cm

r = Radius = \frac{d}{2}=\frac{27}{2}=13.5\ cm

m = Mass of wheel = 800 g

\omega_i = Initial angular velocity = 245\times \frac{2\pi}{60}\ rad/s

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-245\times \frac{2\pi}{60}}{50}\\\Rightarrow \alpha=-0.51312\ rad/s^2

Moment of inertia is given by

M=\frac{1}{2}mr^2\\\Rightarrow M=\frac{1}{2}\times 0.8\times 0.135^2\\\Rightarrow M=0.00729\ kgm^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=0.00729\times -0.51312\\\Rightarrow \tau=-0.0037406448\ Nm

The torque the friction exerts is -0.0037406448 Nm

For more information on torque and moment of inertia refer

brainly.com/question/13936874

brainly.com/question/3406242

7 0
3 years ago
A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera
AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel r=0.3\ m

Initial angular velocity \omega _0=0

It rotates 3 revolution in 5 s therefore

\omega =2\pi 3=\6\pi =18.85\ rad/s

using \omega =\omega _0+\alpha t

where \alpha =angular\ acceleration

\omega =Final\ angular\ velocity

t=time

\alpha =\frac{18.85}{5}=3.77 rad/s^2

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

\omega at t=1

\omega =0+3.77\times 1=3.77 rad/s

a_c=\omega ^2\cdot r

a_c=(3.77)^2\cdot 0.3=4.26 m/s^2

Tangential acceleration a_t=\alpha \times r

a_t=3.77\times 0.3=1.13 m/s^2

a_{net}=\sqrt{a_t^2+a_c^2}

a_{net}=\sqrt{(1.13)^2+(4.26)^2}

a_{net}=4.41 m/s^2

                       

7 0
4 years ago
Anyone know the answer to this? Please help.. NO LINKS
Harlamova29_29 [7]
Option C homie. 0.00001 C at 2 meters is 0.225 N

4 0
3 years ago
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