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3 years ago

Two objects each with a mass of 5x10^15 kg have a gravitational

Physics

1 answer:

kipiarov [429]3 years ago

6 0

Answer:

Distance, r = 700.31 m

Explanation:

Mass of two objects, $m_1=m_2=5\times 10^{15}\ kg$

Force between the objects , $F=3.4\times 10^{15}\ N$

We need to find the distance between the two objects. The force of gravitational between two objects is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\r=\sqrt{\dfrac{Gm_1m_2}{F}} \\\\r=\sqrt{\dfrac{6.67\times 10^{-11}\times (5\times 10^{15})^2}{3.4\times 10^{15}}}\\\\=700.31\ m$

So, the distance between the objects is 700.31 m.

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Take for example driving by with a cake in your hand, then dropping it while going 30 mph. It will not drop directly down, it will gradually go in the direction you were driving while falling.
This is true I believe, if I'm interpreting correctly.

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3 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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5 0

3 years ago

If a light wave is reflected with a 27° angle of reflection, what was the angle of incidence of which it was transmitted?

Leni [432]

Answer:

it is 27 degree.

Explanation:

because angle of reflection is equal to angle of Inc dence

8 0

3 years ago

One major difference between plant and animal cells is that plant cells are the only ones with A) cellulose. B) lysosomes. C) nu

stira [4]

The correct answer is A .

5 0

3 years ago

Read 2 more answers

The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a

brilliants [131]

Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

vector (AB) = 2 i + j - 5 k

mag (AB) = sqrt (2^2 + 1^2 + 5^2)

mag (AB) = sqrt(30)

unit (AB) = ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

vector (T) = unit(AB)* 2.3 KN

= 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

vector (AC) = 2 i - 2 j - 5 k

mag (AC) = sqrt (2^2 + 2^2 + 5^2)

mag (AC) = sqrt(33)

unit (AC) = ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k) . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

T_ac = 0.2923947572 - 0.146973786 - 1.827467232

T_ac = 1.972888 KN

4 0

3 years ago