A proton is held at rest in a uniform electric field. When it is released, the proton will gain:_________
a) electrical potential energy.
b) kinetic energy.
c) both kinetic energy and electric potential energy.
d) either kinetic energy or electric potential energy.

Answer:

[tex]M.E=41J[/tex]

Explanation:

From the question we are told that:

Amplitude [tex]a=4.20cm[/tex]

Spring Constant [tex]K=262N/m[/tex]

Mass [tex]m=560g[/tex]

Generally the equation for mechanical energy is mathematically given by

[tex]M.E=\frac{1}{2}km^2[/tex]

[tex]M.E=0.5*262*0.56^2[/tex]

[tex]M.E=41J[/tex]

Answer:

Wheel and axle

Explanation:

Which simple machine is shown in the diagram?

a wheel and axle

From the given diagram, the machine shown is actually a wheel and axle

Description of wheel and axle

The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.

Answer:

Wheel and axle

Explanation:

Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.

Hope this helps!!!!

Answer:

electric current

Explanation:

The answer is electric current

Answer:

True

Explanation:

With the gravitational pull that our planets have, we are able to remain in orbit. This demonstrates how every object in the galaxy is attracted to every other object. Every object in the universe that has mass exerts a gravitational pull on every other mass. We as humans do it too, but since our force isn't strong, we don't have much of an effect. I hope this helped and please don't hesitate to reach out with more questions!

Answer:

8m/s

Explanation:

Vavg= 16-0/2=8m/s

Answer:

What is (a) the x component and (b) the y component of the net electric field at the square's center

Answer:

Explanation:

From the given information:

mass = 64 kg

speed = 3.2 m/s

coefficient of friction [tex]\mu =[/tex] 0.70

The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:

[tex]W = \Delta K.E[/tex]

[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]

[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]

Thus, the mechanical energy touted = 327.68 J

According to the formula used in calculating the frictional force

[tex]F_r = \mu mg[/tex]

= 0.70 × 64  kg× 9.8 m/s²

= 439.04 N

The distance covered now can be determined as follows:

d = W/F

d = 327.68 J/  439.04 N

d = 0.746 m

Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).

At point A, the block has total energy

E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²

E (A) = 686 J + 1/2 (10.0 kg) v₀²

At point B, the block's potential energy is converted into kinetic energy, so that its total energy is

E (B) = 1/2 (10.0 kg) v₁²

The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,

E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J

Throughout this whole process, energy is conserved, so

E (A) = E (B) = E (C) = E (D)

(a) Solve for v₀ :

686 J + 1/2 (10.0 kg) v₀² = 2548 J

==>   v₀ ≈ 19.3 m/s

(b) Solve for v₁ :

1/2 (10.0 kg) v₁² = 2548 J

==>   v₁ ≈ 22.6 m/s

Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:

• net horizontal force:

∑ F = -f = ma

• net vertical force:

∑ F = n - mg = 0

where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :

n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N

f = µn = 0.500 (98.0 N) = 49.0 N

==>   - (49.0 N) = (10.0 kg) a

==>   a = - 4.90 m/s²

The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that

v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)

==>   v₂² = 490 m²/s²

and thus the block has total/kinetic energy

E (C) = 1/2 (10.0 kg) v₂² = 2450 J

(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so

2450 J = (10.0 kg) (9.80 m/s²) h

==>   h = 25.0 m

(d) At half the maximum height, the block has speed v₃ such that

2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²

==>   v₃ ≈ 15.7 m/s

The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by

v = v₁ + at = 22.6 m/s - (4.90 m/s²) t

The block comes to a rest when v = 0 :

0 = 22.6 m/s - (4.90 m/s²) t

==>   t ≈ 4.61 s

It covers a distance x after time t of

x = v₁t + 1/2 at ²

so when it comes to a complete stop, it will have moved a distance of

x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m

(e) The block crosses the rough region

(52.0 m) / (2.00 m) = 26 times

Explanation:

For engine 1,

Energy removed = 239 J

Energy added = 567 J

[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]

For engine 2,

Energy removed = 457 J

Energy added = 789 J

[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]

For engine 3,

Energy removed = 422 J

Energy added = 1038 J

[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]

So, the engine 2 has the highest thermal efficiency.

Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

Answer:

KE = 2800 J

Explanation:

Usually a velocity is expressed as m/s. Then the energy units are joules.

[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]

v = 30 m / sec

KE = 1/2 * 4 * (30)^2

KE =2800 kg m^2/sec^2

KE = 2800 Joules

Answer:

30J

Explanation:

Given data

The total quantity of heat recieved= 50J

Quantity of heat used to do work= 20J

Hence the net change is

ΔU= Total Heat - Net work

ΔU= 50-20

ΔU= 30J

Hence the change in the internal energy is 30J

Answer:

Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.

Explanation:

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Complete Question

A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall​ (above the fire​ truck) if the ladder makes an angle of  

40° 16' with the horizontal.

Answer:

 [tex]d=8.01m[/tex]

Explanation:

From the question we are told that:

Length of ladder [tex]l=12.5m[/tex]

Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]

Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by

 [tex]d=l sin \theta[/tex]

 [tex]d=12.5 sin 40.26[/tex]

 [tex]d=8.01m[/tex]

Answer:

A) nm, um, mm, cm

B) 1mm, 1cm, 1m, 1km

A) 3500g, B) 2500m, C) 7200 seconds

Answer:

Explanation:

Speed= distance/time

Or time = distance/speed

According to your question

Speed=15m/s

and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s

D= 1.2km = 1.2×1000m =1200meter

Time = distance/ speed

1200/15 =80second

Or. 1min and 20 sec will be your answer.

Answer:

They can be mixed together to make almost any other color.

Explanation:

All the three primary colors can mix to form white color.

Blue and red mix to form a black color.

Answer: masses

Explanation:

Trust me

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) g

• the upward normal force of the surface, magnitude n₁

• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction

• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction

• the applied force, mag. F, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) g, pointing downward

• normal force, mag. n₂, pointing upward

• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction

• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

n₂ - (3.0 kg) g = 0   ==>   n₂ = (3.0 kg) g   ==>   f₂ = 0.30 (3.0 kg) g

• net horizontal force:

c₂ - f₂ = 0   ==>   c₂ = 0.30 (3.0 kg) g ≈ 8.8 N

The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.

Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.

Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Explanation:

The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.

[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.

                   [tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Initial:            a                0          0

Equilibrium:  (a - x)          0          2x

Total pressure = (a - x) + 2x = a + x

As it is given that the total pressure is 1.1 atm.

So, a + x = 1.1

a = 1.1 - x

Now, expression for equilibrium constant for this equation is as follows.

[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]

Hence, the value of 'a' is calculated as follows.

a + x = 1.1 atm

a = 1.1 atm - x

  = 1.1 atm - 0.31 atm

  = 0.79 atm

Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Answer:

Look at work

Explanation:

Θ= 4t^3+10t-40

a) In order to find ω, we need to find displacement so plug in t=2 to find Θ.

Θ= 4*8+20-40=12

use ω=Θ/t

Plug in values

ω=6 rad/s

b) In order to find α we use ω/t.

Plug in values

α=6/2= 3 rad/s^2

Answer:

The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.

Weight can be calculated using the equation:

weight = mass × gravitational field strength

This is when:

weight (W) is measured in newtons (N)

mass (m) is measured in kilograms (kg)

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

The total net force acting on the balloon will be 24498 Newtons

Given that

Volume of the balloon = 2210 cubic meter

Density of the air inside the balloon = 1.13  kg/m3

Here force on the balloon will be equal to the weight of the air displaced by balloon

[tex]F= mass of air displaced\times gravity[/tex]

[tex]F= Density \times volume \times gravity[/tex]

[tex]F=1.13 \times 2210 \times 9.81[/tex]

[tex]F=24498 N[/tex]

The total net force acting on the balloon will be 24498 Newtons

To know more about buoyancy force follow

https://brainly.com/question/117714

Answer:

A will attract

B will repare

Answer:

See the answer below

Explanation:

After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem

1. Drop a few drops of immersion oil on the slide and view again under high the power objective.

2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.

Answer:

13.33 K

Explanation:

Given that,

Heat absorbed, Q = 30 J

Mass of nail, m = 5 g = 0.005 kg

The specific heat capacity of the nail is 450 J/kg-°C.

We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:

[tex]Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K[/tex]

So, the increase in temperature is 13.33 K.

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]

             a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

Answer "a" would be correct.

Answer:

d

Explanation:

There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D

Brainliest please~