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Alex777 [14]
3 years ago
10

A proton accelerates from rest in a uniform electric field of 630 N/C. At one later moment, its speed is 1.50 Mm/s (nonrelativis

tic because v is much less than the speed of light). (a) Find the acceleration of the proton.
Physics
1 answer:
Vlad1618 [11]3 years ago
8 0

Answer:

the acceleration of the proton is 6.025 x 10¹⁰ m/s².

Explanation:

Given;

magnitude of electric field, E = 630 N/C

final speed of the proton, v = 1.5 M m/s = 1.5 x 10⁶ m/s

charge of proton, Q = 1.6 x 10⁻¹⁹ C

mass of proton, m = 1.673 x 10⁻²⁷ kg

The force experienced by the proton is calculated as;

F = ma = EQ\\\\a = \frac{EQ}{m} \\\\a = \frac{(630)(1.6\times 10^{-19})}{1.673 \times 10^{-27}} \\\\a = 6.025 \times 10^{10} \ m/s^2

Therefore, the acceleration of the proton is 6.025 x 10¹⁰ m/s².

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A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the
Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

W_{r} = |F_{r}|*|d|*cos(\theta_{1})

Where:                

F_{r}: is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

W_{g} = |F_{g}|*|d|*cos(\theta_{2})  

The force of gravity is given by the weight of the bike.

F_{g} = -mgsin(24)     

And the angle between the force of gravity and the direction of motion is 180°.

W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})  

W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J  

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.  

I hope it helps you!                                                                                          

3 0
3 years ago
Which is an example of kinetic energy? A. a stretched rubber band B. wind C. water in a reservoir D. natural gas E. an object su
daser333 [38]

Kinetic energy is energy of motion.

In the cases of a stretched rubber band, water in a reservoir, natural gas, or an object suspended above the ground, everything is just laying there, and nothing is moving. There's nothing there that has kinetic energy.

If there's any wind, then air is moving. The moving air has kinetic energy.

6 0
3 years ago
Pure silicon contains approximately 1.0 X 1016 free electrons per cubic meter. (
Margarita [4]
A. The formula for mean free time is:

t = V/(4π√2 r²vN)
where
N = 1×10¹⁶ molecules (per m³)
V = 1 m³
r = 111×10⁻⁷m (atomic radius of silicon)

Let's solve for v first:
v = √(3RT/M) = √(3(8.314 m³·Pa/mol·K)(25 + 273 K)/28.1 g/mol Si)
v = 16.26 m/s

t = (1 m³)/(4π√2 (111×10⁻⁷m)²(16.26 m/s)(1×10¹⁶ molecules))
<em>t = 2.81×10⁻9 s</em>

<em>Pure silicon has a high resistivity relative to copper because copper is a conductor, while silicon is a semi-conductor. </em>
3 0
3 years ago
A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang
Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

3) E=mgh

Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

8 0
3 years ago
Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t
Alex_Xolod [135]

Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\  T

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

5 0
3 years ago
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