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ycow [4]
3 years ago
6

What do you mean by acceleration due to gravity? ​

Physics
1 answer:
likoan [24]3 years ago
4 0

It is the acceleration of an object in free fall

Explanation:

When an object is in free fall, it is subjected only to one force: the force of gravity, which pulls the object downward, with a magnitude (near the Earth's surface) which is given by

F=mg

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration due to gravity

We can apply Newton's second law to the object in free fall:

F=ma

where

F is the net force on the object

a is the acceleration of the object

m is the mass

However, since there is only the force of gravity acting on the object, the net force is equal to the force of gravity: so we can equate the two equations, obtaining that

mg = ma\\\rightarrow a = g

Which means that the acceleration of an object in free fall (acted upon the force of gravity only) is equal to the acceleration due to gravity, g=9.8 m/s^2.

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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Identify two fields where physical quantities are used in motion calculations​
larisa86 [58]

The two fields were physical quantities are used in motion calculations are length and mass with time.

The physical quantity in a field is referred as every point in a particular space time.

<h3>How physical quantities are used in motion calculations?</h3>

 If we consider an object, the physical property of the object is considered as physical quantity and to measure that object is known as units. The Physical quantity can be classified as elemental physical quantity and derived physical quantity. Length, mass, time, etc.. are elemental physical quantity, momentum, density, acceleration, etc... are derived physical quantity. Only for charge and temperature the physical quantity will be less than zero.

Length, mass and time  are the physical quantities used in motion calculations.

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4 0
2 years ago
What unit is used to measure force?
jarptica [38.1K]

Answer:

a

b

a

d

9

Explanation:

6 0
3 years ago
Read 2 more answers
Find the westward component of a resultant vector 85.42 unit, 23 degrees W of N
Sindrei [870]

Since the angle is West of North, therefore to find for the westward component (horizontal component) of the vector, we use the sin function:

sin θ = opposite side / hypotenuse = westward component / resultant vector

So the westward component (x) is:

x = 85.42 sin 23

<span>x = 33.38 unit</span>

5 0
3 years ago
Dos coches, uno de 1.000 kg que circula a 100 km/h, y otro de 1.200 kg que circula a 90
Kobotan [32]

The collision of car 2 is more violent (because more impulse is exerted)

Explanation:

The collision which is more violent is the one in which more impulse is exchanged.

The impulse exerted by each car on the wall is equal to the change in momentum of the car:

I=m\Delta v

where

m is the mass of the car

\Delta v is the change in velocity

For the car 1,

m = 1000 kg

\Delta v = -100 km/h \cdot \frac{1000 m/km}{3600 s/h}=-27.8 m/s (the sign is negative because the velocity of the car has changed from 100 km/h to 0 km/h)

So the magnitude of the impulse of car 1 is

I_1 = (1000)(27.8)=27,800 kg m/s

For the car 2,

m = 1200 kg

\Delta v = -90 km/h \cdot \frac{1000 m/km}{3600 s/h}=-25 m/s

So the magnitude of the impulse of car 2 is

I_2 = (1200)(25)=30,000 kg m/s

So, car 2 exerts a larger impulse, therefore its impact is more violent.

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr
DiKsa [7]

Answer:

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

x_A = v_0 t + \frac{1}{2} at^2

v_0* t= velocity ( time ) = distance

since ball is at height, the above equation changes as

x_A = H - \frac{1}{2} gt^2

for ball B

xB = v0 t - \frac{1}{2} gt^2

the condition of collision is

xA = xB

vA = - 2vB (given)

from the kinematic equation

the speed of the ball A is

v_A = u- gt

since initial speed of the ball A is zero

, so

v_A = -gt

the speed of the ball B is

v_B = v_0 - gt

sincev_A = - 2v_B

   -gt = -2 ( v_0 - gt)

-gt = -2 v_0 +2gt

3gt =2 v_0

t = \frac{2v_0}{3g}

since x_A = x_B

H -  \frac{1}{2} gt^2= v_0 t - \frac{1}{2} gt^2

H = v_0 t

= v_0 (2v_0/3g)

= \frac{2 v_0^2}{ 3g}

x_A = 2 (\frac{v_0^2}{ 3g})- \frac{1}{2} gt^2

  = 4 \frac{v_0^2}{9g} = (2/3) H

so, (2/3) times height collision occur

3 0
3 years ago
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