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Two particles have a mass of 8kg and 12kg, respectively. if they are 800mm apart, determine the force of gravity acting between

natima [27]

The magnitude of the force of gravity acting between the particles is:
$F=G\frac{m_1.m_2}{d^2}$
The weight of each particle is:
$P=mg$
Now let's plug in the numbers knowing that $G=6.67\times10^{-11}$ , $g=9.81$ , $d=0.8$ and m1 and m2 are already given in kilograms. We get then:
$P_1=m_1.g=78.48$ N
$P_2=m_2.g=117.72$ N
$F=G\frac{m_1.m_2}{d^2}=1.00\times10^{-8}$ N

This results shows us why we don't often see objects being attracted to each other, their mass is too small compared to the earth gravitational pull.

8 0

3 years ago

If a power lifter raises a 1000N weight a distance of 2 meters in 0.5 seconds, what is his power out put?

Irina18 [472]

Answer:

P = 4000 [W]

Explanation:

In order to solve this problem, we must first determine the work, which is defined as the product of force by distance.

W = F*d

where:

W = work [J] (units in Joules)

F = force = 1000[N]

d = distance = 2 [m]

W = 1000*2

W = 2000 [J]

And power is defined as the relationship between work and the time in which the work is done.

P = W/t

P = power [W] (units of watts)

t = time = 0.5 [s]

P = 2000/0.5

P = 4000 [W]

8 0

3 years ago

1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.

AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Where;

v = The image distance from the lens

We get;

$v = \dfrac{u \times f}{u - f}$

Therefore;

$v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26$

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0

3 years ago

-Cesar the monkey observed the swinging behaviors of his parents. He is now able to swing from branch to branch after much pract

alexgriva [62]

The 3rd one I think . I did something similar a while ago

4 0

4 years ago

Calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30ms

nexus9112 [7]

S= ?
U= 0 $ms^{-1}$
V= 30 $ms^{-1}$
A= 9.8 $ms^{-2}$
T=

$V^2 = U^2 + 2AS$
$900 = 19.6S$
$\frac{900}{19.6} = S$
S = 45.92 $m$

6 0

4 years ago