Question

Acceleration is defined as the rate of change for which characteristic?
A. displacement
B. position
C. velocity
D. time
What acceleration does the force of Earth's gravity produce?

A. 9.8m/s/s

B. it varies depending on how long it has been falling

C. it changes direction every 9.8 meters

D. gravity does not produce an acceleration

PLEASE HURRY NEED THIS

Answer 1

1) C. velocity

Acceleration is defined as the rate of change of velocity per unit time. In formulas:

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in velocity

$\Delta t$ is the time interval

Therefore, the correct answer is C. velocity.

2) A. 9.8m/s/s

Earth's gravity is a force, so it produces an acceleration on every object with mass located on the Earth's surface. This acceleration can be calculated, as it is given by the formula

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

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

$r=6.37\cdot 10^6 m$ is the Earth's radius

By substituting these numbers into the formula, one can find that the acceleration due to Earth's gravity is $g=9.81 m/s^2$.

Answer 2

1) The acceleration is defined as the rate of change of velocity.

2) The value of acceleration due to the Earth’s gravity is  $\boxed{{\text{9}}{\text{.8}}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$.

Further Explanation:

Part (1):

An object is under acceleration if its velocity changes with the time and the velocity of object is the rate of change of displacement of the object.

$v =\dfrac{{dx}}{{dt}}$  

If the velocity of the object continues to change along with the time, the object is said to be accelerating. It may be decreasing or increasing. The acceleration of the object is defined as the rate of change of velocity of the object.

 $\boxed{a=\frac{{dv}}{{dt}}}$

Thus, the acceleration is defined as the rate of change of velocity.

Part (2):

The value of the acceleration due to gravity on the surface of the planet is defined as the acceleration experienced by the body due to the Earth’s gravitational pull.

The acceleration due to gravity on the Earth is given by:

$\boxed{g=\frac{{GM}}{{{R^2}}}}$  

Here, $g$ is the acceleration due to gravity, $G$ is the gravitational constant, $M$ is mass of the Earth and $R$ is radius of Earth.

The mass of the Earth is $6 \times {10^{24}}\,{\text{Kg}}$.

The radius of the Earth is $6400\,{\text{km}}$.

The value of the gravitational constant is  $6.67 \times {10^{ - 11}}\,{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}\mathord{\left/{\vphantom {{{\text{N}} \cdot {{\text{m}}^{\text{2}}}} {{\text{k}}{{\text{g}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{\text{k}}{{\text{g}}^{\text{2}}}}}$.

Substitute the values in above expression:

$\begin{aligned}g&=\dfrac{\left(6.67\times 10^{-11}\right)\left(6\times 10^{24}\right)}{\left(6.4\times ^{6}\right)^{2}}\\&=\dfrac{4.002\times 10^{14}}{4.096\times 10^{13}}\\&=9.77\text{m}/ \text{s}^{2}\\&\approx 9.8\text{m}/ \text{s}^{2}\end{aligned}$

Thus, the value of acceleration due to the Earth’s gravity is $\boxed{{\text{9}}{\text{.8}}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

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Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Kinematics

Keywords:  Acceleration, rate of change, velocity, time, position, displacement, acceleration due to gravity, force on a body, rate of change of displacement, gravitational constant.