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densk [106]
3 years ago
7

How does the mass of an object affect its inertia?

Physics
1 answer:
Anarel [89]3 years ago
7 0

Answer: If the object has more mass, it will have more inertia because the force of inertia's pull for more massive objects is stronger than a less massive object.

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An air track car with a mass of 0.75 kg and a velocity of 8.5 m/s to the right collides elastically with a 0.65kg car moving to
Sunny_sXe [5.5K]
We can do this with the conservation of momentum. The fact it is elastic means no KE is lost so we don't have to worry about the loss due to sound energy etc.

Firstly, let's calculate the momentum of both objects using p=mv:

Object 1:
p = 0.75 x 8.5 = 6.375 kgm/s

Object 2 (we will make this one negative as it is travelling in the opposite direction):
p = 0.65 x -(7.2) = -4.68 kgm/s

Based on this we know that the momentum is going to be in the direction of object one, and will be 6.375-4.68=1.695 kgm/s

Substituting this into p=mv again:

1.695 = (0.75+0.65) x v
Note I assume here the objects stick together, it doesn't specify - it should!

1.695 = 1.4v
v=1.695/1.4 = 1.2 m/s to the right (to 2sf)
8 0
3 years ago
Newton’s cradle is a contraption where metal balls hang from a frame. When one ball is pulled and released, the collision causes
Marysya12 [62]

Answer: A

Explanation: the principle of friction with swinging and colliding balls.

6 0
2 years ago
Which of these is a measure of how much force is applied to an object from gas particle bouncing into it?
IrinaK [193]

Answer:

Pressure

Explanation:

The measure of how much force is applied to an object from gas particle bouncing into it is called pressure.

Pressure is defined as the force per unit area on a body.

Mathematically;

             Pressure  = \frac{Force}{Area}

The pressure of a gas is the combined force with which gas molecules bombard a unit area of the wall of the container. It is the sum of all tiny pushes on the wall of the container.

Various units of pressure are atm, mmHg, torr, pascal e.t.c

4 0
3 years ago
Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2, đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g,
dusya [7]

Answer:

The angle is 18.3 degree.

Explanation:

A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?

surface charge density, σ = 4 x 10^-5 C/m^2

Charge, q = 10^-8 C

mass, m = 0.008 kg

Let the angle is A and the tension in the string is T.

The electric field due to a plane is

E =\frac{\varepsilon \sigma }{2\varepsilon o}\\\\E =\frac{4\times 10^{-5}}{2\times 8.85\times 10^{-12}}\\\\E = 2.26\times 10^6 V/m \\

Now equate the forces,

T sin A = q E.... (1)\\\\T cos A = m g ..... (2)\\\\divide (1) by (2)\\\\tan A = \frac{10^{-8}\times 2.6\times 10^6}{0.008\times 9.8}\\\\tan A = 0.33\\\\ A = 18.3 degree

5 0
3 years ago
At the same instant that a 0.50 kg ball is dropped from 25 m above Earth, a second ball, with a mass of .25 kg, is thrown straig
Anettt [7]

Answer:

A. 7.1m

B. 3.55m/s

C. 1.775m/s^2

Explanation:

First step is to identify given parameters;

Ball 1: m₁ = 0.5kg, u (initial velocity) =0, t = 2seconds

Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds

<u>Second step:</u> we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;  

y = ut - \frac{gt^2}{2}

y_{1} = 0 X 2 - \frac{9.8 X2^2}{2}

y_{1} = -19.6m

Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)

[tex]y_{1}  = 5.4m[/tex]

<u>Third step</u>: we determine the y-coordinate of ball 2 after 2 seconds

y_{2} = 15 X 2 - \frac{9.8 X2^2}{2}

y_{2} = 10.4m

<u>Fourth step: </u>we determine the y-component of the center mass of the two balls

y = \frac{m_{1}y_{1} +m_{2}y_{2}}{m_{1} +m_{2} }

y = \frac{(0.5)X(5.4) +(0.25) X (10.4)}{(0.5 +0.25) }

y = 7.1m

<u>Fifth step:</u> we solve B part of the question; velocity of the center mass of the two balls

Velocity = \frac{distance of center mass of the two balls (y)}{time}

velocity = \frac{7.1 m}{2 s}

velocity = 3.55m/s

<u>Sixth step:</u> we solve C part of the question; acceleration of the center mass of the two balls

acceleration = \frac{velocity}{time}

acceleration = \frac{3.55}{2}

acceleration = 1.775 m/s^2

7 0
3 years ago
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