A 1.5m wire carries a 2 A current when a potential difference of 55 V is applied. What is the resistance of the wire?

What is the kinetic energy of a ball that has a mass of 3kg and Is moving at 5m/s?

Arada [10]

Answer:

E kin = 1/2 • m • v^2

E kin: kinetic energy

m: mass

v: velocity

Explanation:

6 0

3 years ago

3. When two things are the same temperature:

JulijaS [17]

The answer is definitely C.) their molecules move at the same average speed

3 0

3 years ago

You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the

lana [24]

Answer:

The fifth option is the correct answer: mv; 2 mv

Explanation:

Change of Momentum

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

$\Delta p=p_2-p_1$

The momentum is computed as

$p=mv$

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

$p_1=mv$

When it hits the wall, it sticks, thus its final speed is 0 and

$p_2=0$

The change of momentum is

$\Delta p=0-mv=-mv$

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

$p_2=-mv$

The change of momentum is

$\Delta p=-mv-mv=-2mv$

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0

3 years ago

Make the following conversion. 120 mL = _____ cm³ 12.0 1200 120 1.20

omeli [17]

120 is your answer. 120 mL = 120 cm^3

7 0

3 years ago

Read 2 more answers

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

3 years ago