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2 years ago

What iz inertia ? -,-

Physics

2 answers:

natali 33 [55]2 years ago

6 0

A tendency to do nothing or to remain unchanged.

kipiarov [429]2 years ago

4 0

Inertia refers to the tendency of a body to resist changes.

<h3>Inertia</h3>

It is the property of a body to resist change.

A moving body will continue moving until a force acts on it. Unless the acting force is big enough to overcome the inertia of the body, the body will continue moving.

Also, a body at rest will continue resting unless a force big enough to overcome the inertia force is used to displace it.

More on inertia can be found here: brainly.com/question/1358512

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Two ping pong balls have been painted with metallic paint and charged by contact with an Van de Graaff generator. The charge on

andre [41]

Answer:

0.035 N

Explanation:

Parameters given:

Charge q1 = -3.31x10^(-7) C

Charge q2 = -5.7x10^(-7) C.

Distance between them, R = 22 cm = 0.22 m

Electrostatic force between to particles is given as:

F = (k* q1 * q2) / R²

F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²

F = 0.035 N

6 0

3 years ago

Read 2 more answers

Technician A says a short to voltage may or may not blow a fuse and may or may not affect more than one circuit. Technician B sa

Tomtit [17]

Answer:

C. Both A and B

Explanation:

Fuses are rated by the amperage they can carry before heat melts the element. The fuse is ideal for protection against short circuits. Short circuits produce enough amperage to vaporize a fuse element and break connection in one cycle of a 60-cycle system.

Specifically, the voltage rating determines the ability of the fuse to suppress the internal arcing that occurs after a fuse link melts and an arc is produced.

6 0

4 years ago

Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle

Zepler [3.9K]

Answer:

1/9 E

Explanation:

The electric field produced by a charged sphere outside the sphere is equal to that produced by a single point charge:

$E=k\frac{Q}{r^2}$

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

In this problem, we have a sphere of radius R (smaller sphere) with a charge Q that produces an electric field of magnitude E at r=R.

For the larger sphere,

R' = 3R

Therefore, the electric field at r=R' will be

$E'=k\frac{Q}{R'^2}=k\frac{Q}{(3R)^2}=\frac{1}{9}(k\frac{Q}{R^2})=\frac{E}{9}$

So, the electric field just outside the larger sphere will be 1/9 E.

8 0

3 years ago

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

4 years ago

A 1500-kg car accelerates from 0 to 30 m/s in 6.0 s. What is the minimum average power delivered by the engine?

nadezda [96]

Answer:

Power, P = 112500 watts

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Initial velocity of the car, u = 0

Final velocity of the car, v = 30 m/s

Time taken, t = 6 s

The minimum average power delivered by the engine is given by :

P = W/t

Where

W = work done

t = time taken

Work done = change in kinetic energy

So, $P=\dfrac{\Delta K}{t}=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{t}$

$P=\dfrac{\dfrac{1}{2}\times 1500\ kg\times (30\ m/s)^2}{6\ s}$

P = 112500 watts

So, the minimum average power delivered by the engine is 112500 watts. Hence, this is the required solution.

7 0

3 years ago

What iz inertia ? -,-​

What iz inertia ? -,-