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3 years ago

Does gravity always include the Earth’s pull or can it be between other objects?

2 answers:

7 0

Well other planets have a gravitational pull that's what keep the planet's orbiting the sun. But the amount of gravity differs by mass.

Hope this helps :)

sammy [17]3 years ago

4 0

What a great question !

There is a pair of forces of gravity (one in each direction) between EVERY two specks of mass in all of creation.

-- Between you and the Earth.

-- Between the Moon and the Earth.

-- Between the Moon and you.

-- Between you and your dog.

-- Between your dog and your goldfish.

-- Between every pair of corn flakes in your cereal bowl.

-- Between the lint in your pocket and every grain of sand on the beach on the far side of the farthest planet in orbit around the farthest star in the farthest galaxy on the other side of the Universe. You don't feel it, and it's too small to measure, but its strength can be calculated.

You might be interested in

How can I solve the following statement?

Firlakuza [10]

Answer:

The net electric field at the midpoint is 6.85 x 10^7 N/C.

Explanation:

q = − 8.3 μC

q' = + 7.8 μC

d = 9.2 cm

d/2 = 4.6 cm

The electric field due to the charge q at midpoint is

$E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C$ leftwards

The electric field due to the charge q' at midpoint is

$E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C$

The resultant electric field at mid point is

E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C

7 0

2 years ago

All objects are either ___________ or ___________. Charged objects can have a ____________ or _____________ charge. Uncharged ob

KATRIN_1 [288]

All objects are either charged or uncharged. Charged objects can have a positive or negative charge. Uncharged objects will have a no charge. Charged objects that have the same charges will repel each other. Charged objects with opposite charges will attract each other. Uncharged objects can become charged. The charge an object has gives it electric energy. The charged object’s ability to attract (pull) or repel (push) other objects is called electrostatic force.

5 0

2 years ago

Two convex lenses are placed 15 cm apart. The left lens has a focal length of 10 cm, and the right lens a focal length of 5 cm.

Minchanka [31]

Answer:

The image distance from right lens is 2.86 cm and image is real.

Explanation:

Given that,

Focal length of left lens = 10 cm

Focal length of right lens = 5 cm

Distance between the lenses d= 15 cm

Object distance = 50 cm

We need to calculate the image distance from left lens

Using formula of lens

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

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{-50}$

$\dfrac{1}{v}=\dfrac{3}{25}$

$v=8.33\ cm$

We need to calculate the image distance from right lens

The object distance will be

$u = 15-8.33 = 6.67\ cm$

Using formula of lens

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

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{-6.67}$

$\dfrac{1}{v}=\dfrac{1167}{3335}$

$v=2.86\ cm$

The image is real.

Hence, The image distance from right lens is 2.86 cm and image is real.

4 0

3 years ago

Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used

natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is $T =8391.6 N$

Explanation:

From the question we are told that

The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

The contact area is $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$

Looking at the first diagram

At 600 MPa of stress

The strain is $0.3mm/mm$

At 450 MPa of stress

The strain is $0.0015 mm/mm$

To find the stress at $\Delta l$ we use the interpolation method

$\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

$\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

$\sigma _{\Delta l} -450 = 49.50$

$\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

$F = \sigma_{\Delta l} * A$

Substituting values

$F = 499.50*10^{6} * 2.8*10^{-6}$

$=1398.6N$

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

$T = 6 * F$

$= 6* 1398.6$

$T =8391.6 N$

6 0

3 years ago

The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

The gage pressure,

$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

3 0

3 years ago