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2 years ago

Parallel Circuits:

Physics

1 answer:

Ber [7]2 years ago

8 0

Answer:

D.)

Explanation:

the current separates on each branch according to the resistance it experience.

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A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

2 years ago

What is the magnitude of the gravitational force acting on the earth due to the sun?

expeople1 [14]

Answer: $3.524(10)^{22}N$

Explanation:

According to Newton's law of Gravitation, the force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m1=1.99(10)^{30}kg$ is the mass of the Sun

$m2=5.972(10)^{24}kg$ is the mass of the Earth

$r=1.50(10)^{11}m$ is the distance between the Sun and the Earth

Substituting the values in (1):

$F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.99(10)^{30}kg)(5.972(10)^{24}kg)}{(1.50(10)^{11}m)^2}$ (2)

Finally:

$F=3.524(10)^{22}N$ This is the gravitational force acting on the earth due to the sun

3 0

2 years ago

Read 2 more answers

At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi

Nataliya [291]

Answer:

The distance from the charge is 3.35 m.

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

8 0

2 years ago

Is Racism Good or Bad? Why? (Explain)

k0ka [10]

Bad, it's bring more hatred to the world

3 0

2 years ago

Read 2 more answers

(3) What is the weight of a 50-kg astronaut (a) on Earth (b) On the Moon ,(g=1.7m/s2), (c) on Mars (g=3.7m/s2) (d)in outer space

artcher [175]

Answer:

a) On Earth

490N

b) On the Moon

85N

c) On Mars

185N

d)in outer space traveling with constant velocity.

Explanation:

The weight is defined as:

$W = mg$ (1)

Where m is the mass and g is the gravity

a) On Earth $g = 9.8m/s^{2}$

Then, equation 1 can be used:

$W = (50Kg)(9.8m/s^{2})$

$W = 490Kg.m/s^{2}$

but 1N = Kg.m/s^{2}

$W = 490N$

Hence, the weight of the astronaut on Earth is $490N$

b) On the Moon $g = 1.7m/s^{2}$

$W = (50Kg)(1.7m/s^{2})$

$W = 85N$

Hence, the weight of the astronaut on the Moon is $85N$

c) On Mars $g = 3.7m/s^{2}$

$W = (50Kg)(3.7m/s^{2})$

$W = 185N$

Hence, the weight of the astronaut on Mars is $185N$

(d) in outer space traveling with constant velocity.

Tanking into consideration that the astronaut is traveling in outer space at a constant velocity, it can be concluded that the acceleration will be zero.

Remember that the acceleration is defined as:

$a = \frac{v_{f} - v_{i}}{t}$

Since the acceleration is the variation of the velocity in a unit of time.

Therefore, from equation 1 is gotten.

$W = (50kg)(0)$

Remember that g is the acceleration that a body experience as a consequence of the gravitational field.

$W = 0$

5 0

2 years ago