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Gwar [14]
3 years ago
5

Can someone please help me with this question thank you!

Physics
1 answer:
katovenus [111]3 years ago
7 0
<h2>at the highest point that is B</h2><h2>because PE is directly proportional to height </h2><h2>hope it helped</h2>

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
3 years ago
Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
Nookie1986 [14]

Explanation:

We have,

The initial position of an object is zero.

The starting velocity is 3 m/s and the final velocity was 10 m/s.

The object moves with constant acceleration..

The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".

8 0
3 years ago
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

= 365 d²

This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

= 1/2 x 1.2 x 2.3²

= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

d = .097 m

= 9.7 cm

If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

5 0
3 years ago
How can we magnetise a piece of iron.
liq [111]
Place the magnet at one end of the piece of metal. The magnet must make as much contact with the metal as possible. Place light pressure on the magnet and rub the metal in one direction only. Magnetization will take some time to accomplish so continue rubbing until the iron or steel attracts other pieces of metal.
4 0
3 years ago
80N force ;
lutik1710 [3]

The power of the engine is 320 W.

<u>Explanation:</u>

Power may be defined as the rate of doing work (or) work done per unit time. One unit of energy is used to do the one unit of work.

                          Power = Work done / Time taken

Given, Force = 80 N,    height = 5 m , final velocity = 4 m/s

To calculate the power, we must know the time taken.

To find the time, use the distance and speed formula which is given by

                              Time = Distance / speed

Here distance = 5 m and speed = 4 m/s

                              Time = 5 / 4 = 1.25 s.

Now,          Power = work done / time

                              = (F * d) / t  = (80 * 5) / 1.25

                  Power = 320 W.

The standard unit of power is watt (W) which is joule per second.

                                                               

4 0
3 years ago
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