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3 years ago

A shortage or a surplus can exist when the current price is equal to the equilibrium price

Physics

1 answer:

Mamont248 [21]3 years ago

5 0

Answer:

The answer is False.

Explanation:

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A 20 kg mass is moving at 10 m/s and collides with a stationary 5 kg mass, transferring all its momentum in the collision, what

Fudgin [204]

Answer:

v = 40 [m/s].

Explanation:

Linear momentum is defined as the product of mass by Velocity. In this way, by means of the following equation, we can calculate the momentum.

$P=m*v\\$

where:

m = mass [kg]

v = velocity [m/s]

$P =20*10\\P =200 [kg*m/s]$

Since all momentum is transferred, we can say that this momentum is equal for the mass of 5 [kg]. In this way, we can determine the speed after the impact.

$v = P/m\\v = 200/5\\v = 40 [m/s]$

3 0

3 years ago

Suppose an automobile has 2000 J of kinetic energy. When it moves at twice the speed, what will be its kinetic energy?

Varvara68 [4.7K]

The answer should be 8000 J

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4 years ago

A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to

o-na [289]

Answer:

angle = 18.40 degree

Explanation:

given data

force = 95 N

distance = 0.50 m

torque = 15 N · m

to find out

angle between the wrench handle and the direction of the applied force

solution

we will apply here torque equation that is express as

torque = distance × force × sin(θ) ...................1

put here value we will get angle that is

15 = 0.50 × 95 × sin(θ)

sin(θ) = 0.315789

θ = 18.40 degree

6 0

3 years ago

Read 2 more answers

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h

vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

$R_A=2R_B$

Also Potential of both sphere is same

$V_A=V_B$

$V=\frac{kQ}{R}$

thus

$k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}$

$\frac{Q_A}{Q_B}=\frac{R_A}{R_B}$

$\frac{Q_A}{Q_B}=\frac{2}{1}=2$

$\frac{Q_B}{Q_A}=\frac{1}{2}$

(b)Ratio of $\frac{E_B}{E_A}$

Electric Field is given by $E=\frac{kQ}{R^2}$

thus $E_A=\frac{kQ_A}{R_A^2}$ ----1

$E_B=\frac{kQ_B}{R_B^2}$ ----2

Divide 2 by 1

$\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}$

$\frac{E_B}{E_A}=\frac{1}{2}\times 4=2$

8 0

4 years ago

How To do this question

777dan777 [17]

Since it is given that their particles have a large difference in size, we can conclude that the particles of small size get embedded into the space between the particles of large size. Thus it doesn't increased it volume as it should have.

6 0

3 years ago