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4 years ago

Ques

Physics

1 answer:

Amanda [17]4 years ago

4 0

Answer:

When a particle or a system of particles move in a system where no external force acts, then the total linear momentum of the particle system remains constant.

Explanation:

Given data:

Total mass of the skateboarder, $m_{1} = 70 \ kg$

Mass of the friend, $m_{2} = 50 \ kg$

Initial velocity of the skateboarder, $u_{1} = 3 \ m/s$

Initial velocity of the the friend, $u_{2} = 0$

Let the new velocity of the skateboarder when his friend jumps be $v$ .

From the conservation law of linear momentum,

$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

$70 \times 3 + 50 \times 0 = (70 + 50)v$

$\Rightarrow \ v &= 1.75 \ m/s.$

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I’m hope this help

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3 years ago

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

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4 years ago

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igor_vitrenko [27]

Answer:

Explanation:

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3 years ago

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4 years ago

The unit of area is called a derived unit.why?

oksano4ka [1.4K]

Explanation:

the unit of area is called a derived unit because it is made of two fundamental unit metre and metre.

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3 years ago