When a candle burns which forms of energy does the chemical energy in the candle change to

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgl$

$v=\sqrt{2gl}$

Put the value into the formula

$u=\sqrt{2\times9.8\times50.0\times10^{-2}}$

$u=3.13\ m/s$

The initial speed of the ball $u_{1}=3.13\ m/s$

The initial speed of the block $u_{2}=0$

(a). We need to calculate the speed of the ball after collision

Using formula of collision

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13$

$v_{1}=-2.01\ m/s$

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

$v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0$

$v_{2}=1.11\ m/s$

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0

4 years ago

I need help quickly: another name for car battery ?

Alekssandra [29.7K]

Another name is voltaic battery

3 0

4 years ago

Read 2 more answers

The dome of a Van de Graaff generator receives a charge of 0.00011 C. The radius of the dome is 5.2 m. Find the strength of the

wel

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

F=kq1/r²

F=kQ/r²

F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

Then Electric field is given as,

E=kQ/r²

E=8.98755E9 ×0.00011/12.6²

E=6227.34N/C

5 0

3 years ago

An elevator is rising up a vertical shaft in a skyscraper. Between the 12th and 20th floors, it travels at a constant speed. Dur

zaharov [31]

3. The upward force is greater
.................,

8 0

3 years ago

An object is moving to the right in a straight line. The net force acting on the object is also directed to the right, but the m

Shtirlitz [24]

Answer:

A) continue to move to the right, with its speed increasing with time.

Explanation:

As long as force is positive , even when it is decreasing , it will create positive increase in velocity . Hence the body will keep moving with increasing velocity towards the right . The moment the force becomes zero on continuously decreasing , the increase in velocity stops and the body will be moving with the last velocity uniformly towards right . When the force acting on it becomes negative , even then the body will keep on going to the right till negative force makes its velocity zero . D uring this period , the body will keep moving towards right with decreasing velocity .

Hence in the present case A , is the right choice.

6 0

3 years ago