Mass 2 has a mass of 1.00 kg. Mass 2 rolls

lapo4ka [179]

Answer:

You increase the acceleration of the car.

6 0

3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

So

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

y (x, t) = A*cos(k*x -w*t)

Where k = stiffness and w = angular frequency

Hence,

k = 172 and w = 2730

- Calculate wave speed V:

V = w / k = 2730 / 172 = 13.78 m/s

- Tension in the string T:

T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

T = W = Y*V^2 = (w/L*g)*V^2

W = (0.0126 / 1.8*9.81)*(13.78)^2

W = 0.135 N

4 0

3 years ago

A child stands on the edge of a merry-go-round of radius 1.63 m which is rotating at 2.13 rad/s.

Brilliant_brown [7]

Answer:

9

Explanation:

2.13 rad/s * 26.9 sec

2.13 * 26.9

57.297

3282.88 deg / 360 deg = 9.12

It makes 9 complete revolutoins

7 0

2 years ago

6th grade science I mark as brainliest.

DerKrebs [107]

Answer:

2m 13 $\frac{1}{3}$ s

Explanation:

1.5m = 1s

200m = $\frac{200}{1.5}$ × 1s

= 133 $\frac{1}{3}$ s

= 2m 13 $\frac{1}{3}$ s

3 0

2 years ago