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3 years ago

What is acceleration due to gravity

Physics

2 answers:

rjkz [21]3 years ago

7 0

The acceleration developed between two bodies containing mass as a result of the gravitational force between them.

Studentka2010 [4]3 years ago

5 0

If you stay on the same planet and drop a lot of objects one at a time,
it turns out that every object you drop falls from your hand to the ground
with the same acceleration, and hits the ground with the same speed,
no matter whether the object is light, heavy, or anything in between.

That particular value of acceleration is the "acceleration due to gravity".
On Earth, it's 9.81 meters per second². On the moon, it's 1.62 meters
per second². On Jupiter, it's 25.89 meters per second².

Why we don't generally notice it: The previous description is true if the
ONLY force on the object is the force of gravity. If it has to fall through
air on the way down, then the air can have a great effect on it. Many
museums have an exhibit where they drop things in a long tube with
all the air removed from it, and there you can see some pretty weird
stuff ... like a bowling ball, a rock, a sheet of paper, and a feather, all
falling together, with nothing fluttering.

Why everything falls with the same acceleration ? That's a separate question.

You might be interested in

When the temperature of the air is 50°C, the velocity of a sound wave traveling through the air is approximately?

noname [10]

The answer is:

C. 361 m/s

The explanation:

To calculate the speed of sound at a given temperature (50°C) we are going to use this formula:

v = 331 + 0.6T

when V is the velocity

and T is the temperature = 50°C

by substitution:

v = 331 + 0.6(50)

v = 361 m/s

So, The correct answer is C.

because of the variation of the motion of the molecules of air with change of temperature so, the velocity (V) of the sound in the air is change with temperature.

5 0

3 years ago

Read 2 more answers

Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area

cluponka [151]

Answer:

5.09 x 10⁵ Nm²/C

Explanation:

The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e

φ = E A

From the question;

E = (8.0j + 2.0k) ✕ 10³ N/C

r = radius of the circular area = 9.0m

A = area of a circle = π r² [Take π = 3.142]

A = 3.142 x 9² = 254.502m²

Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.

Therefore;

φ = (2.0) x 10³ x 254.502

φ = 5.09 x 10⁵ Nm²/C

The electric flux is 5.09 x 10⁵ Nm²/C

4 0

2 years ago

Is the image formed by a plane mirror always real? True or false?

mafiozo [28]

This statement is false

6 0

2 years ago

While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish

Bingel [31]

Answer:

the smallest separation between two objects is 0.8067 m

Explanation:

Given the data in the question;

Altitude h = 5.75 km = 5750 m

Diameter D = 4.0 mm = 0.004 m

λ = 460 nm = 4.6 × 10⁻⁷ m

Now, Using Rayleigh criterion for Airy disks resolution.

we know that, Minimum angular separation for resolving two points is;

θ = 1.22λ / D

so we substitute

θ = (1.22 × 4.6 × 10⁻⁷) / 0.004

θ = 5.612 × 10⁻⁷ / 0.004

θ = 1.403 × 10⁻⁴ rad

so minimum separation $d_{min$ = θh

so we substitute

$d_{min$ = (1.403 × 10⁻⁴) × 5750 m

$d_{min$ = 0.8067 m

Therefore, the smallest separation between two objects is 0.8067 m

6 0

3 years ago

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik

Anon25 [30]

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$