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Brilliant_brown [7]

3 years ago

11

1. What is the gravitational force of attraction between two 800 kg spherical objects that are 5 m apart? What would the gravita

tional force of attraction between these two objects be if the distance between them was changed to 10 m?

1 answer:

Katyanochek1 [597]3 years ago

4 0

Answer:

the force of gravity between them is quadrupled .

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

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Object 1 has a momentum of 10 kg m/s and Object 2 has a momentum of 25 kg m/s. Will it be easier to change the direction of move

svetoff [14.1K]

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

5 0

3 years ago

A spherical helium filled balloon (B) with a hanging passenger cage being held by a single vertical cable (C) attached to Earth

Gre4nikov [31]

Answer:

The tension is $T = 4326.7 \ N$

Explanation:

From the question we are told that

The total mass is $m = 200 \ kg$

The radius is $r = 5 \ m$

The density of air is $\rho_a = 1.225 \ kg/m^3$

Generally the upward force acting on the balloon is mathematically represented as

$F_N = T + mg$

=> $(\rho_a * V * g ) = T + mg$

=> $T = (\rho_a * V * g ) - mg$

Here V is the volume of the spherical helium filled balloon which is mathematically represented as

$V = \frac{4}{3} * \pi r^3$

=> $V = \frac{4}{3} * 3.142 *(5)^3$

=> $V = 523.67\ m^3$

So

$T = (1.225 * 523.67* 9.8 ) - 200 * 9.8$

$T = 4326.7 \ N$

5 0

3 years ago

Help me please!!!

lilavasa [31]

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7 0

2 years ago

Read 2 more answers

What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate

Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

The diameter is $d = 12 \ cm = 0.12 \ m$

The charge is $Q = 4.4 nC = 4.4 *10^{-9} \ C$

The distance from the center is $k = 0.1 \ mm = 1*10^{-4} \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.12}{2}$

=> $r = 0.06 \ m$

Generally electric field is mathematically represented as

$E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]$

substituting values

$E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]$

$E =248.2 \ N/C$

4 0

3 years ago

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

Nina [5.8K]

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

4 0

3 years ago