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Oxana
12 hours ago
10

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

itude 950 \text{ N/C}950 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x
Physics
1 answer:
Ostrovityanka [2.2K]12 hours ago
7 0
The alteration in potential energy is  \Delta PE = - 3.8*10^{-16} \ J

In the query, it is stated that

  The intensity of the uniform electric field equals E = 950 \ N/C

     The distance the electron covers is  x = 2.50 \ m

Typically, the force exerted on this electron is expressed mathematically as

     F = qE

Where F signifies the force and  q represents the charge of the electron, which is a fixed value of q = 1.60*10^{-19} \ C

    Thus  

      F = 950 * 1.60 **10^{-19}

      F = 1.52 *10^{-16} \ N

Generally, the work-energy theorem is mathematically framed as

          W = \Delta KE

Where W denotes the work done on the electron by the electric field and  \Delta KE  is the change in kinetic energy

Additionally, work done on the electron can also be described as

        W = F* x *cos( \theta )

Where  \theta = 0 ^o assuming that the electron's movement aligns with the x-axis  

        So

             \Delta KE = F * x cos (0)

Inserting values

         \Delta KE = 1.52 *10^{-16} * 2.50 cos (0)

          \Delta KE = 3.8*10^{-16} J

According to the conservation of energy

       \Delta PE = - \Delta KE

Where \Delta PE signifies the change  in  potential energy  

Thus  

        \Delta PE = - 3.8*10^{-16} \ J

               

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A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total
Ostrovityanka [2214]

Answer:

0.000047N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

Thus, the power absorbed can be calculated as:

Power = Intensity x Area

This equals = 1.4 x 10^3 x(10)

Thus,

14000 Watts = 14 kWatt

However, the radiation pressure can be defined as

time-averaged intensity divided by the speed of light in a vacuum

So,

P = (1.4 x 1000)/c

Also,

F= P x A

Thus,

((1.4 x 1000)/(3 x10^8)) x 10

This results in

=0.000046666N

Rounded to two significant figures gives us

=0.000047 N

3 0
14 days ago
d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s
Yuliya22 [2446]
Assuming that the mass of the empty wagon is "M," according to Newton's second law, we can derive the following relationships. Given that the empty wagon accelerates at 1.4 m/s², we proceed with this information. If a child weighing three times the mass of the wagon is on it, we can establish the relevant equations.
8 0
8 days ago
A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
ValentinkaMS [2433]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is \frac{4}{3} \pi (\frac{D}{2})^3

Thus \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

The radius of the disk is 1.15(\frac{ D}{2} )

Applying angular momentum conservation;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disk = m*\frac{ \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
25 days ago
Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi
Keith_Richards [2268]

The answer is:

V=14m/s

Details are as follows:

According to the problem, we have

The combined mass of A and B is 60kg

A's speed is 2m/s

B's speed is 1m/s

The mass of the bag is 5kg

Typically, the momentum of astronaut A along with the bag is defined by

M_A=(60+5)*2

M_A=130kgm/s

To prevent a collision, astronaut A should maintain a speed that is either equal to or less than astronaut B's speed

Thus, the minimum speed astronaut A should achieve corresponds to that of astronaut B, which is 1

Consequently,

130=(60*1)=(5*v)

V=14m/s

7 0
11 days ago
You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A
kicyunya [2264]

Answer:

The initial speed at which the baseball is thrown is 20.58 m/s

Explanation:

To determine the time taken to reach the peak height in projectile motion, the formula is:

t = (u sin θ)/g

Where u = initial speed of the baseball =?

θ = angle of launch above the horizontal

g = gravitational acceleration = 9.8 m/s²

1.05 = (u sin 30)/9.8

u can be calculated as (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0
14 days ago
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