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kramer
3 years ago
7

Gravity anything with mass as gravity we know the earth has gravity because you and I are standing on earth and not floating off

into space when we jump we come right back to earth which model best represents the force of gravity pulling us all down to earth?
Physics
1 answer:
Paraphin [41]3 years ago
8 0

Answer:

F=mg

Explanation:

Close to Earth's surface, the force of gravity that pulls an object towards the ground is

F=mg (2)

where

m is the mass of the object

g is the acceleration due to gravity, which is 9.81 m/s^2 close to Earth's surface

This is an approximation of the general formula of gravity valid only close to Earth's surface. The more general formula is

F=G\frac{Mm}{r^2} (1)

where

G is the gravitational constant

M is the Earth's mass

m is the object's mass

r is the distance of the object from Earth's center

At the Earth's surface,

r = R (Earth's radius), and by calling the following factor

g=\frac{GM}{R^2}

we see that eq.(1) becomes eq.(2).

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A glass capillary tube with a diameter of 8.5 mm and length 8 cm is filled with a salt solution with a resistivity of 2.5 ?m. Wh
MA_775_DIABLO [31]

Answer:

The resistance is 3.5\times10^{-4}\ \Omega

Explanation:

Given that,

Diameter of tube = 8.5 mm

Length = 8 cm

Resistivity = 2.5 m

We need to calculate the resistance

The resistance is equal to the product of the resistivity and length divided by the area of cross section .

In mathematical form,

R = \dfrac{\rho\times l}{A}

Where, \rho=resistivity

l = length

A = area of cross section

Put the value into the formula

R = \dfrac{2.5\times8\times10^{-2}}{3.14\times(\dfrac{8.5}{2}\times10^{-3})^2}

R=3526.32\ \Omega

R=3.5\times10^{-4}\ \Omega

Hence, The resistance is 3.5\times10^{-4}\ \Omega

6 0
3 years ago
By what factor would your weight be multiplied if the earth were1/2 as massavise and the diameter was unchanged
Nutka1998 [239]
<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then: 

F = G(Mm / (4(pi)*r^2)) 

The above expression gives the force that you feel on the earth's surface, as it is today! 

Let us now double the mass of the earth and decrease its diameter to half its original size. 

This is the same as replacing M with 2M and r with r/2. 

Now the gravitational force (F' ) on the new earth's surface is given by: 

F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F 

So: 

F' = 8F 

This implies that the force that you would feel pulling you down (your weight) would increase by 800%! 

You would be 8 times heavier on this "new" earth!</span>
4 0
3 years ago
What is the process of conduction in terms of particle movement
ivann1987 [24]
<span>What is the process of conduction in terms of particle movement chocies

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8 0
3 years ago
If a 25 kg lawnmower produces 347 w and does 9514 J of work, for
Igoryamba

Steps 1 and 2)

The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.

The goal is to solve for the unknown time t.

-----------------------

Step 3)

Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P

-----------------------

Step 4)

t = W/P

t = 9514/347

t = 27.4178674351586

t = 27.4 seconds

-----------------------

Step 5)

The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.

-----------------------

Note: we don't use the mass at all

6 0
3 years ago
A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen
IgorLugansk [536]

Answer:

Approximately 0.077\; {\rm m\cdot s^{-1}} (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

Momentum of the t-shirt:

\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}.

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if p(\text{cannon}) denote the momentum of this cannon:

p(\text{t-shirt}) + p(\text{cannon}) = 0.

p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}.

Rewrite p = m\, v to obtain v = (p / m). Since the mass of this cannon is m(\text{cannon}) = 33\; {\rm kg}, the velocity of this cannon would be:

\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}.

8 0
2 years ago
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