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2 years ago

Which of these is NOT a reason why the geocentric model of the solar system was once commonly accepted as the correct model?

Physics

2 answers:

IrinaK [193]2 years ago

5 0

I'm not really sure but I think the answer is D

Serhud [2]2 years ago

5 0

Answer:

Explanation:

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A device that converts electrical energy into kinetic energy to turn an axle?

Advocard [28]

A motor is the devise

8 0

3 years ago

a runner covers the last straigjt stretch of a race in 4 s. during that time, he speeds up from 5m/s to 9m/s.

PtichkaEL [24]

During that final period of time,
his acceleration is
(9 m/s - 5 m/s) / (4 sec) = 1 m/s² .

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8 0

3 years ago

How much voltage is required to run 0.42 A of current through a 150

igomit [66]

The voltage in the resistor is 63 V

Explanation:

We can solve the problem by applying Ohm's law, which states the relationship between voltage, current and resistance in a resistor:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

For the resistor in this problem, we have:

I = 0.42 A is the current

$R=150 \Omega$ is the resistance

Substituting into the equation, we find the voltage needed:

$V=(0.42)(150)=63 V$

Learn more about voltage and current:

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4 0

3 years ago

Read 2 more answers

Two equally charged insulating balls each weigh 0.16 g and hang from a common point by identical threads 35 cm long. The balls r

Dafna11 [192]

Answer:

Q₁ = Q₂ = 8.84 x 10⁻⁹ C

Explanation:

given,

mass of ball, m = 0.16 g = 1.6 x 10⁻⁴ Kg

ball each other, r = 6.8 cm

Weight of the ball

F_w = m g

F_w = 1.6 x 10⁻⁴ x 9.8

F_w = 1.56 x 10⁻³ N

The tension in each string is a force directed along the length of the string and is the hypotenuse of a right triangle.

we have to find the horizontal component of the forces.

The length of the string,L is 35 cm so, it will be the hypotenuse.

θ be the angle made with imaginary vertical line and the string.

now,

$sin \theta = \dfrac{r\2}{L}$

$sin \theta = \dfrac{3.4}{35}$

θ = 5.57°

horizontal component of the force = ?

vertical component of force,F_v = 1.56 x 10⁻³ N

$tan\theta = \dfrac{F_H}{F_v}$

$tan(5.57^0) = \dfrac{F_H}{1.56\times 10^{-3}}$

F_h = 1.52 x 10⁻⁴ N

now, each ball will be repelled by

F = 1.52 x 10⁻⁴ N

now calculation of charges

$F = \dfrac{kQ_1Q_2}{r^2}$

Q₁ = Q₂ because both charge are same

$1.52\times 10^{-4} = \dfrac{9\times10^9Q^2}{0.068^2}$

Q² = 7.809 x 10⁻¹⁷

Q = 8.84 x 10⁻⁹ C

hence the change on the balls were Q₁ = Q₂ = 8.84 x 10⁻⁹ C

5 0

3 years ago

A box (m = 20 kg) is sliding on a horizontal surface. it is connected to a massless hook by a light string passing over a massle

Varvara68 [4.7K]

mass of the box = 20 kg

force of friction on the box due to surface

$F_s = \mu_s N$

$F_s = 0.80 * 20 * 9.8$

$F_s = 156.8 N$

similarly kinetic friction on it

$F_k = \mu_k N$

$F_k = 0.30* 20 * 9.8$

$F_k = 58.8 N$

now the weight of the suspended block will be

$W = mg = 15*9.8 = 147 N$

so here the weight of the suspended block is less than the limiting friction on it

So here we will say that friction will counter balance the weight of the suspended block and it will not move at all

So acceleration of the box will be zero

5 0

3 years ago