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mars1129 [50]
3 years ago
6

A push or a pull on an object is known as a(n)

Physics
2 answers:
balu736 [363]3 years ago
6 0
Force is a push or a pull of an object that causes the object to speed up, slow down, or stay in one place. In other words, a force is what causes an object to move. Friction and gravity are two types of forces that influence how an object moves.

Hope I can help you!
Reptile [31]3 years ago
5 0

A push or a pull on an object is known as a force.

Option D.

<u>Explanation:</u>

For example, if we kick a ball the ball moves in the direction of the kick. Here on kicking the ball we apply a force on the ball that made the ball move. So, the motion (push/pull) of an object is due to the force applied on it.

Force is a vector quantity defined by the product mass and acceleration. That is, F=ma. The SI unit is Newton (N).

Velocity is the rate of change of speed of an object.

Acceleration of an object is the rate of change of velocity.

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If I want to calculate the normal force of an object on two legs but the weight of the object is not uniform throughout, how do
RoseWind [281]

Use the right equation. To calculate the normal force of an object at an angle, you need to use the formula: N = m * g * cos (x) For this equation, N refers to the normal force, m refers to the object's mass, g refers to the acceleration of gravity, and x refers to the angle of incline.

6 0
1 year ago
List inner planets and outer planets
elena-s [515]

Answer:

The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars .  the outer planets, Jupiter, Saturn, Uranus and Neptune

Explanation:

7 0
3 years ago
Read 2 more answers
Determine the change in electric potential energy of a system of two charged objects when a -2.1-C charged object and a -5.0-C c
elena55 [62]

Answer:

Change in electric potential energy ∆E = 365.72 kJ

Explanation:

Electric potential energy can be defined mathematically as:

E = kq1q2/r ....1

k = coulomb's constant = 9.0×10^9 N m^2/C^2

q1 = charge 1 = -2.1C

q2 = charge 2 = -5.0C

∆r = change in distance between the charges

r1 = 420km = 420000m

r2 = 160km = 160000m

From equation 1

∆E = kq1q2 (1/r2 -1/r1) ......2

Substituting the given values

∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)

∆E = 94.5 × 10^9 (3.87 × 10^-6) J

∆E = 365.72 × 10^3 J

∆E = 365.72 kJ

6 0
3 years ago
Identify the energy transformations that take place in an electrical fan heater
Sergio [31]
The energy goes from electric energy and gets converted into thermal energy.
3 0
3 years ago
Read 2 more answers
An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c
Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

                      \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

                      \frac{1}{10}=\frac{1}{v}-\frac{1}{-5}

                      \frac{1}{v}=\frac{1}{10}-\frac{1}{5}

                      v = -10 cm

The magnification factor of lens

                     m=\frac{-10}{-5}

                     m = 2

                     m=\frac{h'}{h}

                     h'=h\times m

                     h'=3\times 2

                     h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0
3 years ago
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