All categories

2 years ago

How does the value of gravity vary as you go:

Physics

1 answer:

iragen [17]2 years ago

6 0

Answer:

Gravity changes with altitude. as we know The gravitational force is proportional to 1/R2, where R is your distance from the center of the Earth.

eg. The radius of the Earth at the equator is 6400 kilometers.

Let's say you were in a jet at the equator that was 40 kilometers high above the earth's surface.

may be helpfull

You might be interested in

A fisherman casts his lure at an angle of 33 degrees above the horizontal. The lure reaches a maximum height of 2.3 m. Assuming

omeli [17]

Answer:

12.32 m/s

Explanation:

Using the formula of maximum height of a projectile,

H = u²sin²Ф/2g................... Equation 1

Where H = maximum height, u = initial velocity, Ф = angle of projection, g = acceleration due to gravity

make u the subject of the equation

u = √(2Hg/sin²Ф)............ Equation 2

Given: H = 2.3 m, Ф = 33°, g = 9.8 m/s²

Substitute into equation 2

u = √[(2×2.3×9.8)/sin²33°]

u =√ [45.08/(0.545)²]

u = 45.08/0.297

u = √(151.785)

u = 12.32 m/s

3 0

3 years ago

Ifa truck starts from rest and it has acceleration of 4 m/s for 5 second

yan [13]

Answer:

Distance 50m

final velocity 20ms^-1

$s = ut + \frac{1}{2} a {t}^{2} \\ = 0 \times t + \frac{1}{2} \times 4 \times {5}^{2} = 50m$

$v = u + at \\ v = 0 + 4 \times 5 \\ v = 20ms^{ - 1}$

6 0

3 years ago

Explain what the ionsphere is and how it interacts with some radio waves

mrs_skeptik [129]

Answer:

not sure

Explanation:

8 0

3 years ago

When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration d

Mice21 [21]

The gravitational acceleration at any distance r is given by

$g= \frac{GM}{r^2}$

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is $r_e=6.37 \cdot 10^6 m$ , so the meteoroid is located at a distance of:

$r=r_e+2.60 r_e =3.60 r_e = 2.29 \cdot 10^7 m$

And by substituting this value into the previous formula, we can find the value of g at that altitude:

$g= \frac{GM}{r^2} = \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2$

5 0

3 years ago

Read 2 more answers

Please help i would really appreciate it

Verdich [7]

Answer:

THE MINIONSSSSSSS AYEEEEE

Explanation:

4 0

2 years ago

Read 2 more answers