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3 years ago

What is the density of an object whose volume is 100 cubic centimeters that has a mass of 100 grams?

Physics

2 answers:

svp [43]3 years ago

8 0

Answer:

search omni calculator and these problems can be solved easily:) hope this helps! 62.43

Rus_ich [418]3 years ago

3 0

Answer:

1 g/cm3

Explanation:

just had it and this was the answer

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Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric

Goshia [24]

Complete Question:

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:

A. 2F

B. F/4

C. F/2

D. F

E. 4F

Answer:

Explanation:

If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

$F =\frac{kQ*2Q}{r^{2} }$

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.

As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.

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4 years ago

What is a hydraulic system

Simora [160]

Explanation:

Hydraulic systems use the pump to push hydraulic fluid through the system to create fluid power. The fluid passes through the valves and flows to the cylinder where the hydraulic energy converts back into mechanical energy. The valves help to direct the flow of the liquid and relieve pressure when needed

7 0

3 years ago

The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p

aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

$F_{p}$ = $q_{p}$ E

and, we know that, F = ma

So,

$m_{p}$ a = $q_{p}$ E

a = $\frac{q_{p}.E }{m_{p} }$ Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2 $at^{2}$

Here, u = 0.

S = 1/2 $at^{2}$

Put equation 1 into the above equation:

S = 1/2 x ( $\frac{q_{p}.E }{m_{p} }$ ) $t^{2}$ Equation 2

So,

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ( $\frac{q_{e}.E }{m_{e} }$ ) $t^{2}$ Equation 3

We know that the charge of electron is equal to the charge of proton so,

$q_{p}$ = $q_{e}$ = q

By dividing the equation 2 by equation 3, we get:

$\frac{S}{D-S}$ = $\frac{m_{e} }{m_{p} }$

Solve the above equation for S,

S $m_{p}$ = $m_{e}$ D - $m_{e}$ S

So,

S = $\frac{m_{e}.D }{(m_{e} + m_{p}) }$

Plugging in the values,

As we know the mass of electron is 9.1 x $10^{-31}$ and the mass of proton is 1.67 x $10^{-27}$

S = $\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }$

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = $\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }$

As we know the mass of chlorine is 35.5 and of sodium is 23

S = $\frac{35.5 . 4.22}{(35.5 + 23)}$

S = 2.56 cm

7 0

3 years ago

Which is an example of nuclear energy being converted into heat and light energy?

likoan [24]

The answer to the question is A

6 0

3 years ago

Read 2 more answers

A lightning strike can transfer as much as electrons from the cloud to the ground. if the strike lasts 2ms , what is the average

malfutka [58]

The average electric current in the lightning will be 8 × $10^{-17}$ A

<h3>Why Lightning Conductors on top of a tower ?</h3>

The lightning conductors are long metal strips running from the spike end of a conductor on the top of a building to the earth. They are used to prevent buildings from destruction when struck by thunder or lightning.

Given that a lightning strike can transfer as much as electrons from the cloud to the ground. if the strike lasts 2ms, to calculate the average electric current in the lightning, we will first consider the charge released.

one charge = 1.6 × $10^{-19}$ C

Average current I = Q/t

Where

Q = charge = 1.6 × $10^{-19}$ C

t = time = 2ms = 2 × $10^{-3}$ s

I = current = ?

Substitute all the parameters into the formula

I = 1.6 × $10^{-19}$ C ÷ 2 × $10^{-3}$

I = 8 × $10^{-17}$ A

Therefore, the average electric current in the lightning will be 8 × $10^{-17}$ A

Learn more about Lightning here: brainly.com/question/3183045

#SPJ1

5 0

2 years ago