All categories

3 years ago

what is the magnitude of the gravitational force between two 5.0-kilogram masses separated by a distance of 5.0 meters?

Physics

2 answers:

Vladimir79 [104]3 years ago

8 0

6.7x10^-11n is your answer

satela [25.4K]3 years ago

8 0

Answer:

$6.67\times 10^{-11}$ N

Explanation:

Given,

Mass of the two objects (m₁ and m₂) = 5 kg

Distance between the two objects (r) = 5 m

Also, we know that Universal Gravitational Constant (G) = $6.67\times 10^{-11}$ Nm²/kg²

Gravitational Force (F) between the objects is,

$F = \frac{G\times m_{1}\times m_{2}}{r^{2}}$

$F = \frac{6.67\times 10^{-11}\times 5\times 5}{5^{2}}$

$F = 6.67\times 10^{-11}$

Thus the magnitude of the gravitational force is $6.67\times 10^{-11}$ N

You might be interested in

Significant digits show the _____ of measurements and calculations.

SashulF [63]

Significant digits show the precision of measurements and calculations

7 0

4 years ago

___ is the combination of all colors of light

TiliK225 [7]

The color white is what you'd see when every color of light is combined.

5 0

3 years ago

Read 2 more answers

If 710- nm and 660- nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wa

alexira [117]

0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.

<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.5m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3276.92 x 10⁻⁶ m

= 3.276x 10⁻³ m

= 3.276mm .

For λ = 660 nm

position = 2 λ D / d

λ = 660 nm , D = 1.5 m

d = .65 x 10⁻³

position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3046.15 x 10⁻⁶ m

= 3.046 x 10⁻³ m

= 3.046 mm .

Difference between their position

= 3.276mm ₋ 3.046 mm

= 0.23 mm .

To know more about Fringes refer to: brainly.com/question/15649748

#SPJ4

7 0

2 years ago

One baked potato provides an average of 30 mg of vitamin C. If 70 potatoes weigh 20 lb., how many milligrams of vitamin C are pr

JulsSmile [24]

Answer:

105 mg

Explanation:

Given that:

1 baked potato provides 30 mg of vitamin C.

So,

70 baked potatoes provide $30\times 70$ mg of vitamin C

Also,

70 potatoes = 20 lb

So,

20 lb potatoes provide $30\times 70$ mg of vitamin C

Thus,

1 lb potatoes provide $\frac {30\times 70}{20}$ mg of vitamin C

<u>Thus, 105 mg of Vitamin C are provided per pound of the potatoes.</u>

5 0

3 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

8 0

3 years ago