The average value of the load between A and B is 6.0 N. The spring has an unstretched length

A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc

Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

6 0

3 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33

Anuta_ua [19.1K]

Answer:

$\frac{v_{2}}{v_{1}}=2$ .

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

$\bar{K}=\frac{3k_{B}T}{2}$

Now, we know that $\bar{K} = (1/2)m\bar{v}^{2}$

Before the absorption we have:

$(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2}$ (1)

After the absorption,

$(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2}$ (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

$\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}$

$\frac{v_{2}}{v_{1}}=\sqrt{4}$

Therefore the ratio will be $\frac{v_{2}}{v_{1}}=2$

I hope it helps you!

4 0

3 years ago

Read 2 more answers

A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm

bekas [8.4K]

Answer:

$f = 8.89 cm$

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

$-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}$

$f = 8.89 cm$

3 0

3 years ago

A future space explorer has been kidnapped and is held prisoner on a planet in our solar system. With nothing else to do, our pr

DIA [1.3K]

To develop this problem we will apply the linear motion kinematic equations. Specifically, the second law that describes the position of a body as a function of its initial velocity, time and acceleration.

$y = ut+\frac{1}{2}gt^2$

Here,

u = Initial velocity

t = Time

g = Acceleration due to gravitation

If we replace the values to find the gravitational acceleration we have then,

$1.7m = 0+\frac{1}{2} g*0.36^2$

$g= 26.2346m/s^2$

Recall that the force of gravity on the planet Jupiter is 24.79 m / s² so the measure is closer to this planet. It is likely that you are in Jupiter.

4 0

3 years ago

Ceres has an orbital semi-major axis = 2.768 AU. What is Ceres’ orbital period?

Doss [256]

Answer: 4.6 years

Explanation:

According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period $T$ of a planet is proportional to the cube of the semi-major axis $a$ (size) of its orbit”: </em>

<em />

$T^{2}\propto a^{3}$ (1)

However, if $T$ is measured in Earth years, and $a$ is measured in astronomical units (unit equivalent to the distance between the Sun and the Earth), equation (1) becomes:

$T^{2}=a^{3}$ (2)

Knowing $a=2.768 AU$ and isolating $T$ from (2):

$T=\sqrt{a^{3}}$ (3)

$T=\sqrt{(2.768 AU)^{3}}$ (4)

Finally:

$T=4.6 years$ This is Ceres' orbital period

6 0

3 years ago