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3 years ago

Newton’s heliocentric view of the universe meant that the sun was the center of the universe. True False

Physics

2 answers:

Anuta_ua [19.1K]3 years ago

7 0

The answer to this question is true.

Flura [38]3 years ago

5 0

Newton's heliocentric view is true

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5. Gravitational and magnetic forces are examples of _______forces.

Anvisha [2.4K]

Answer:

Explanation:

5.non contact force

6.balanced force

7.unbalanced force

8.net force

9.zero

10.is the difference between two forces

mark me as brainliest please

8 0

3 years ago

Suppose that a certain battery produces a voltage of 1.55V without a load connected (open circuit) and a current of 500mA when s

lubasha [3.4K]

Answer:

Explanation:

Let the internal resistance be r .

Since in open circuit the volt is 1.55 V , this will be the source voltage .

Source voltage = 1.55

If external resistance be R .

1.55 / (R + r ) = .500

R + r = 3.1 ohm

So sum of internal resistance and external resistance will be 3.1 ohm.

7 0

3 years ago

Why in drinking hot water, a thin-bottomed glass is taken?

taurus [48]

Answer:

beacause it's contracts

Explanation:

when using a large bottomed glass the hot water cools that's why is good to use thin bottomed glass

6 0

3 years ago

Help!!Help!!Help!!Help!!

DerKrebs [107]

Answer:

y=8

Explanation:

every time you multiply x by 3 you divide y by 3.

x=2, multiply it by 3: x=6

y=24, divide it by 3: y=8

8 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago