3.

The gravitational force between two spheres in outer space is 1000 n. how large would this force be if the mass of each sphere w

kipiarov [429]

The force will remain same if the mass of each sphere is doubled and the distance between them is doubled as well.

The force of gravitation between two spheres is given by

$f=\frac{G m1 m2}{r^{2}}$

m1= mass of first object

m2= mass of second object

r= distance between the masses

Now if each mass is doubled and the distance r is doubled as well

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

$F'= \frac{4 G m1 m2}{4r^{2}}$

F'= $\frac{G m1 m2 }{r^{2}}$

so F'= F

so the force remains same and is equal to 1000 N.

3 0

3 years ago

Find the voltage drop (in volt) along a 93.4 meter long 10 gauge copper wire carrying acurrent of 72.5 A. The diameter of a 10 g

Sophie [7]

Answer: 5.41 V

Explanation:in order to explain this result we have to use the Ohm law given by:

ΔV=R*I where R is the resistance which is equal R= ρ*L/A . ρ is the resistivity, L the length of the wire and A is the cross section. I is the current.

Then we have

ΔV=ρ*L*I/A= 1.68 * 10^-8 Ωm*93.4 m*72.5A/2.1* 10^-5 m^2=5.41 V

3 0

3 years ago

A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplishe

melamori03 [73]

Answer:

$a_r=389\ rev.s^{-2}$

$n=58350 rev$

Explanation:

Given:

time of constant deceleration, $t=10\ s$

A.

initial angular speed, $N_i=3890\ rpm\$

<u>Using equation of motion:</u>

$N_f=N_i+a_r.t$

$0=3890+a_r\times 10$

$a_r=389\ rev.s^{-2}$

B.

Using eq. of motion for no. of revolutions, we have:

$n=N_i.t+\frac{1}{2} a_r.t^2$

$n=3890\times 10+0.5\times 389\times 100$

$n=58350\ rev$

8 0

3 years ago

JUST PLZ HELP!!! Why does the lightbulb in the right electrical circuit turn on but not the one on the left?

jekas [21]

In the open circuit the current can not flow from one end of the power source to the other. Because of this there is no current flow, and therefore the light does not turn on.

8 0

3 years ago

Read 2 more answers

A crane lifts a 425 kg steel beam vertically upward a distance of 66 m. How much work does the crane do on the beam if the beam

iogann1982 [59]

Answer:

W = 311074.5 [J]

Explanation:

In order to solve this problem we must analyze two parts, in the first part by means of Newton's second law we can determine the acceleration of the beam, remembering that the sum of the forces is equal to the product of mass by acceleration.

∑F = m*a

F = forces acting on the beam [N]

m = mass = 425 [kg]

a = acceleration = 1.8 [m/s²]

The forces acting on the beam are the force of the crane up (positive) and the weight of the beam down (negative)

$F_{crane}-(425*9.81)= 425*1.8\\F_{crane}=4713.25 [N]$

Now in the second part, we use the definition of work, which is equal to the product of the force applied in the direction of displacement, that is, the product of force by distance.

$W=F*d$

where:

W = work [J]

F = force = 4713.25 [N]

d = distance = 66 [m]

$W=4713.25*66\\W=311074.5[J]$

5 0

3 years ago