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3 years ago

How come when an object has a greater mass, it has a greater Inertia? and the other way around.

1 answer:

4 0

Inertia is a property of a body to resist the change in linear state of motion. It is measured by the mass of the body. An object having greater mass performing linear motion is difficult to stop. Hence greater the mass greater is its inertia. Hope it helps.

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Two long, straight parallel wires are placed 38 cm apart, one above the other. The top and bottom wires are carrying currents 4.

ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

Given;

distance between the two parallel wire, d = 38 cm = 0.38 m

current in the first wire, I₁ = 4.0 kA

current in the second wire, I₂ = 8.0 kA

Force per unit length, between two parallel wires is given as;

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }$

where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

Substitute the given values in the above equation and calculate the force per unit length

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m$

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m

4 0

3 years ago

Two point charges, with charge magnitudes q and ????, are placed a distance r apart. In this arrangement, each point charge expe

sammy [17]

Answer:

1) Q ’= 8 Q , 2) q ’= 16 q , 3) r ’= ¾ r

Explanation:

For this exercise we will use Coulomb's law

F = k q Q / r²

It asks us to calculate the change of any of the parameters so that the force is always F

Original values

q, Q, r

Scenario 1

q ’= 2q

r ’= 4r

F = k q ’Q’ / r’²

we substitute

F = k 2q Q ’/ (4r)²

F = k 2q Q '/ 16r²

we substitute the value of F

k q Q / r² = k q Q '/ 8r²

Q ’= 8 Q

Scenario 2

Q ’= Q

r ’= 4r

we substitute

F = k q ’Q / 16r²

k q Q / r² = k q’ Q / 16 r²

q ’= 16 q

Scenario 3

q ’= 3/2 q

Q ’= ⅜ Q

we substitute

k q Q r² = k (3/2 q) (⅜ Q) / r’²

r’² = 9/16 r²

r ’= ¾ r

6 0

4 years ago

A bicycle pump contains 20 cm3 of air at a pressure of 100 kPa. The air is then pumped in a tyre of volume 100 cm3. Calculate th

Natasha2012 [34]

Answer:

The pressure of the air in the tyre is 20 kPa

Explanation:

The parameters for the bicycle pump and tyre are;

The volume of air contained in the bicycle pump, V₁ = 20 cm³

The pressure of the air contained in the bicycle pump, P₁ = 100 kPa

The volume (available) of the tyre, where the air is pumped, V₂ = 100 cm³

Let P₂ represent the pressure in the tyre after the air is pumped

By Boyle's law, we have that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure;

Mathematically, Boyle's law gives the following equation;

P₁ × V₁ = P₂ × V₂

∴ P₂ = (P₁ × V₁)/V₂

Substituting the known values gives;

P₂ = (100 kPa × 20 cm³)/(100 cm³)

∴ P₂ = 100 kPa × 1/5 = 20 kPa

P₂ = 20 kPa

The pressure of the air in the tyre = P₂ = 20 kPa.

7 0

3 years ago

A block of mass 10 kg slides down an inclined plane that has an angle of 30. If the inclined plane has no friction and the block

Helga [31]

No friction present means: Ek = Ep

So Ek = mgh = 10 * 9.8 * 2 = 196 J

4 0

3 years ago

Read 2 more answers

Why is friction like applied force but different from gravity?

kirill115 [55]

Answer:

Friction is when a force is applied or done by weight dragging onto something.

Explanation:

Gravity is when an object is getting pulled toward the center of what is attracting it. And applied force is when someone/sommething is applying force.

3 0

4 years ago

Read 2 more answers