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Norma-Jean [14]
3 years ago
10

A plane starts from rest and accelerates uniformly over a time of 20 s for a distance of 300 m. Determine the plane's accelerati

on .
Physics
1 answer:
kumpel [21]3 years ago
3 0
Final velocity = 30 m/s
acceleration = 30/20 = 1.5 m/^2
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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>

<u>From (1):</u>

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>

<u>From (3):</u>

\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
Does anyone know what uranium is???
schepotkina [342]
The answer to your question is "Yes".  Great numbers of people ... chemists,
physicists, and common folks alike ... know what Uranium is.

=================================

Uranium is one of the 92 elements that are found to occur in nature.
In the Periodic Table of the Elements, it is Element #92 ...  Its atoms
contain 92 protons in their nucleii.
5 0
3 years ago
Read 2 more answers
3 A jet plane accelerated along a straight runway from rest to a
aleksklad [387]

Explanation:

v=140 m/s

t=50s

u=0

change of velocity = v-u

140-0=140 m/s

for acceleration:

from Newton's first law of motion:

v=u+at

140=0+a50

50a=140

a=140/50

a=2.8 m/s²

8 0
3 years ago
A spring with a spring constant of 50 n/m is compressed 0.2m. if a mass of 0.4kg is connected to the spring, what is the maximum
Trava [24]
Use conservation of energy

\frac{1}{2} k x^{2} =  \frac{1}{2} m v^{2}   \\ \\ v =  \sqrt{\frac{k}{m} }  x
6 0
4 years ago
Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out ov
Mumz [18]

Answer:

The number of excess electrons are on the negative surface is 4.80\times10^{10}\ electrons

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}

C=0.285\times10^{-10}\ F

We need to calculate the potential

Using formula of potential

V=Ed

Put the value into the formula

V=18000\times1.5\times10^{-2}\ V

V=270\ V

We need to calculate the charge

Using formula of charge

Q=CV

Put the value into the formula

Q=0.285\times10^{-10}\times270

Q=76.95\times10^{-10}\ C

Here, the charge on both the positive and negative  plates

Q=+76.95\times10^{-10}\ C

Q=-76.95\times10^{-10}\ C

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

n=\dfrac{q}{e}

Put the value into the formula

n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}

n=4.80\times10^{10}\ electrons

Hence, The number of excess electrons are on the negative surface is 4.80\times10^{10}\ electrons

8 0
3 years ago
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