What would be the difference in your welght if you were four times farther from the center of Earth than you are now?

Hi guys. Help with any of these questions! Pressure in physics year 7 :) Thanks!

SVEN [57.7K]

Answer:

2 a) it is less dense than the water

2 b) it is more dense than the water

3 a ping pong ball is hollow and less dense than the water so it quickly bounces up to the surface of the water

5 0

3 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

3 years ago

And I need help with seven and eight only I will appreciate it

Mumz [18]

Answer:

7] Force = mass × acceleration

Force = 2 × 5

Force = 10 N

8] Velocity = acceleration due to gravity × time taken

Velocity = 9.8 × 12

Velocity = 117.6 m/s

8 0

2 years ago

At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear

Mandarinka [93]

We apply the gravity calculation expressed in the formula: g=GM/r2
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1) Radius = √6.674e-11*5.972e24/8 = 7058 kms Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2 Radius=√6.674e-11*5.972e24/4.9 = 9019 kms Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.

4 0

3 years ago

The length of the adult mosquito is typically between 3.0 mm and 6.0 mm. The smallest known mosquitoes are around 2.5 mm, and th

FrozenT [24]

0.74in

Explanation:

Given parameters:

length range of adult mosquito = 3.0 - 6.0mm

Smallest mosquito = 2.5mm

Largest mosquito = 19mm

Average mass range = 3.0 - 5.0mg

Unknown:

Length of largest known mosquito in inches = ?

Solution;

The length is the longest dimension. It is how long a body is.

The problem here is converting from mm to inches;

The length of the longest mosquito which is the largest is 19mm

19mm to inches;

1mm = 0.039inches

19mm = 19mm x $\frac{0.039in}{1mm}$ = 0.74in

learn more:

Scale brainly.com/question/570757

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4 0

3 years ago