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3 years ago

A series circuit has a current of 5 A. The circuit contains 10ohms resistor. What is the voltage of the circuit

Physics

1 answer:

d1i1m1o1n [39]3 years ago

3 0

Answer:

50 Volts

Explanation:

Using Ohm's Law

V= I x R

V= 5 x 10

V = 50 V

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Work is done on a locked door that remains closed while you try to pull it open. True False.

dsp73

Answer:False

Explanation:

Work is being done on a body when it causes displacement of body on the application of force

$Work\ done=Force\times displacement$

When we pull the door by a force it causes zero displacements of the door. So we can say that work done on it is zero.

Thus the above-given statement is false

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The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for

abruzzese [7]

Answer:

The answer is " $\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}$ "

Explanation:

$\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta - \omega_{0}^{r} \\\\\to \omega^{r} = 2 (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}$

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3 years ago

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neonofarm [45]

Answer:

2.55sec

Explanation:

time = distance/speed = (87 m)/(34 m/s) = 2.55sec

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3 years ago

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aleksandr82 [10.1K]

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3 years ago

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

balandron [24]

Answer:

2.78m/s²

Explanation:

Complete question:

A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x\\$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

Hence the acceleration of the box as it slides down the incline is 2.78m/s²

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3 years ago