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3 years ago

Help me do this question

Physics

1 answer:

Zepler [3.9K]3 years ago

3 0

Answer:

c20800

Explanation:

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Numerical problems:

Hatshy [7]

Answer:

this is answer please mark me as brainlest

Explanation:

A force of 500N acts on the area of 2.5m2; calculate the pressure

exerted,

(Ans: 200 Pa]

b. Calculate the force applied when 400 Pa of pressure is exerted by a box

of surface area 2m2.

(Ans: 800 N]

The weight of a stone is 1400N. If it exerts a pressure of 700 Pa, calculate

the area.

(Ans: 2m]

d. A force of 2000 N acts on 5m² surface area. Calculate the pressure

exerted.

(Ans: 400 Pa]

7 0

3 years ago

a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially

CaHeK987 [17]

Answer:

Velocity of the helium nuleus = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

$P_{1i} + P_{2i} = P_{1f} + P_{2f]$ (1)

<em>where $P_{1i}$ : is the proton linear momentum initial, $P_{2i}$ : is the helium nucleus linear momentum initial, $P_{1f}$ : is the proton linear momentum final, $P_{2f}$ : is the helium nucleus linear momentum final </em>

$m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f}$ (2)

<em>where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$ : are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$ : are the proton and helium nucleus velocities, respectively, after the collision </em>

By conservation of energy, we have:

$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

<em>where $K_{1i}$ and $K_{2i}$ : are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and $K_{2f}$ : are the kinetic energy for the proton and helium, respectively, after the colission </em>

$\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$ (4)

<u>Now we have two equations: (2) ad (4), and two incognits: $v_{1f}$ and $v_{2f}$ . </u>

Solving equation (2) for $v_{1f}$ , we have:

$v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f}$ (5)

<u>From getting (5) into (4) we can obtain the $v_{2f}$ :</u>

$v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0$

$v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0$

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

$v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s}$ (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

$v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}$

$v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}$

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

7 0

4 years ago

How do you find the rest mass (kg) of a 3.1 eV electron?

Scilla [17]

Answer:

Explanation:

The rest energy of any substance is defined by the Einstein's mass energy equivalence relation. Thus the rest mass of a electron is 9.11x10^-31 kg. The speed of light is 299,792,458 m/s. Thus multiplying the square of speed of light with the rest mass of electron gives the rest energy of the electron.

3 0

3 years ago

What would MOST LIKELY happen if the amount of incoming solar energy decreased without a change in the amount of reflection or o

777dan777 [17]

<h2><u><em>Well, you see, that depends. </em></u></h2><h2><u><em>The firsy thing we have to tak intp account is the angle at witch the sun's rays hit the earth, and that fact can make all the difference, seeing as it does discriminate against seasons. It's more likely that i the winter, a more drastic effect would talk.</em></u></h2><h2 /><h2 /><h2 /><h2>oωo</h2>

8 0

3 years ago

Read 2 more answers

If you are standing at Earth’s North Pole, which of the following will be directly overhead?

Sidana [21]

The north celestial pole (and also the south) are described as two imaginary points that project on the axis of rotation of the earth, indefinitely towards space and that intersects with the celestial sphere (Remember that the earth has a slight inclination with respect to to its axis of rotation). The north and south celestial poles are directly above the head of someone located at the north or south pole of the earth.

Correct Answer is A.

7 0

3 years ago

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