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Would it be true that if you double the distance of an astronaut from a planet, the gravitational pull between them would be hal

velikii [3]

Answer:

Yes

Explanation:

Newton's law of universal gravitation is usually stated that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses(m1 and m2) and inversely proportional to the square of the distance between their centers(r).

F = Gm1m2/r²

This is a general physical law derived from

empirical observations by what Isaac Newton called inductive reasoning.

when distance is doubled the gravitational force will be reduced by quarter not half.

5 0

2 years ago

Read 2 more answers

Consider a concave spherical mirr or that has focal length f = +19.5 cm.

lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

f is the focal length of the mirror
v is the object distance
u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

#SPJ1

7 0

1 year ago

A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density

blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

$P = \rho h$

Here,

h = Depth average

$\rho$ = Density

Moreover,

$\text{Density of water}= \rho = 62.4lb/ft^3$

Replacing,

$P = (62.4lb/ft^3)(\frac{35}{2}ft)$

$P = 1092 lb/ft^2$

Finally the force

$\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}$

$F = 1092lb/ft^2(101ft*52ft)$

$F = 5.735*10^6lbf$

6 0

2 years ago

A wire is stretched between two posts. Another wire is stretched between two posts that are four times as far apart. The tension

Elena-2011 [213]

Answer:

Therefore,

The speed of the wave on the longer wire is 95 m/s.

Explanation:

Given:

For Short wire, speed is

$v_{s}=190\ m/s$

Let length of Short and Longer wire be $L_{s}\ and\ L_{l}$ such that

$L_{l}=4\times L_{s}$

To Find:

$v_{l}=?$ Speed on the longer wire

Solution:

The speed of a pulse or wave on a string under tension can be found with the equation,

$v=\sqrt{\dfrac{F_{T}\times L}{m}$

Where,

$F_{T}$ = Tension on the wire

L = Length of Sting

m = mass of String

So here we have,

$F_{T}$ = same

$L_{l}=4\times L_{s}$

Therefore,

$v_{s}=\sqrt{\dfrac{F_{T}\times L_{s}}{m}$ ......equation ( 1 )

And

$v_{l}=\sqrt{\dfrac{F_{T}\times L_{l}}{m}$ .......equation ( 2 )

Dividing equation 1 by equation 2 and on Solving we get

$\dfrac{v_{s}}{v_{l}}=\sqrt{\dfrac{L_{s}}{L_{l}}}$

Therefore,

$v_{l}=v_{s}\sqrt{\dfrac{4\times L_{s}}{L_{s}}}=190\times 2=380\ m/s$

Therefore,

The speed of the wave on the longer wire is 95 m/s.

8 0

3 years ago

A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the

goblinko [34]

Answer:

$\dfrac{K_2}{K_1}=\dfrac{4}{3}$

Explanation:

When an object strikes the ground, the potential energy of the rock gets converted to the kinetic energy.

A rock is dropped from a distance $R_e$ above the surface of the earth and is observed to have kinetic energy $K_1$ when it hits the ground.

For another rock, It is dropped from twice the height $2R_e$ above the earth’s surface and has kinetic energy $K_2$ when it hits.

For the first rock, $K_1=GMm(\dfrac{1}{R_e}-\dfrac{1}{2R_e})=\dfrac{GMm}{2R_e}$ ..............(1)

For the second rock, $K_2=GMm(\dfrac{1}{R_e}-\dfrac{1}{3R_e})=\dfrac{2GMm}{3R_e}$ ..............(2)

Dividing equation (2) by (1) we get :

$\dfrac{K_2}{K_1}=\dfrac{4}{3}$

Hence, this is the required solution.

8 0

2 years ago