A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the location shown. A test charge is place at point P, and the charge-sphere system has potential energy U0. If the test charge is then moved to point R, what will the potential energy of the charge-sphere system be? ​

On the moon you weigh 6 times lighter so for this problem you multiply 140n by 6.

Your weight on the Moon is 16.5% what you would experience on Earth. In other words, if you weighed 100 kg on Earth, you would weigh a mere 16.5 kg on the Moon. For you imperial folks, imagine you tipped the scales at 200 pounds.

It’s because of the lower gravity on the Moon. Objects on the surface of the Moon experience only 16.5% of the gravity they would experience on Earth.

The mass of the Moon is only 1.2% the mass of the Earth, so you might expect it to have only 1.2% of the gravity. But it’s only 27% of the size of the Earth, so when you’re standing on the surface of the Moon, you’re much closer to its center of gravity.

Therefore, On the moon you weigh 6 times lighter so for this problem you multiply 140n by 6.

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Answer:

1. heating

2. entertainment

3. cooling

4.educational purpose

5.comfort

Answer:

Approximately \(7.2\; {\rm m\cdot s^{-1}}\) (rounded up), assuming that this circle is vertical and \(g = 9.81\; {\rm m\cdot s^{-2}}\).

Explanation:

Let \(v\) denote the tangential speed of the ball, and let \(r\) denote the radius of the circle. Since the ball is in a circular motion, the acceleration on this ball would be equal to the centripetal acceleration \(a = (v^{2} / r)\). The net force on this ball would be \(F_{\text{net}} = m\, a = (m\, v^{2} / r)\).

The net force on this ball is also the vector sum of the tension \(T\) in the rope and the weight of the ball \(m\, g\):

\(F_{\text{net}} = (\text{weight}) + T\).

\(T = F_{\text{net}} - (\text{weight})\).

Note that:

\(\| T \| = \|F_{\text{net}} - (\text{weight})\| \le \|F_{\text{net}} \| + \| (\text{weight})\|\).

In other words, the magnitude of tension \(T\) is at most equal to \(\|F_{\text{net}} \| + \| (\text{weight})\| = (m\, v^{2} / r) + (m\, g)\), which happens when weight and net force are in opposite directions.

When the speed of the ball is maximized, the magnitude of tension \(T\) would be at the largest possible value of \(130\; {\rm N}\). Rearrange the equation and solve for speed \(v\):

\(\displaystyle \frac{m\, v^{2}}{r} + m\, g = \|T\|\).

\(\begin{aligned}v^{2} = \frac{r}{m}\, (\|T \| - m\, g) = \frac{r\, \|T\|}{m} - r\, g\end{aligned}\).

\(\begin{aligned}v &= \sqrt{\frac{r\, \|T\|}{m} - r\, g} \\ &= \sqrt{\frac{(1.2)\, (130)}{2.5} - (1.2)\, (9.81)}\; {\rm m\cdot s^{-1}} \\ &\approx 7.2\; {\rm m\cdot s^{-1}}\end{aligned}\).

I believe a scale is good

Answer:

C.45 kW

Explanation:

Given parameters:

Initial kinetic energy  = 1.6MJ

Final kinetic energy  = 2.5MJ

Time  = 20s

Unknown:

Minimum average power developed  = ?

Solution:

Power is rate at which work is done.

  Here work done;

Work done  = Final kinetic energy  - Initial kinetic energy

Work done  = 2.5MJ  - 1.6MJ  = 0.9MJ

 Power = \(\frac{work done}{time}\)   = \(\frac{0.9 x 10^{6} J}{20}\)    = 45000W

This is therefore 45kW

Answer:

3504.4 kgm/s

Explanation:

First, convert km/hr to m/s:

19 km/hr x 1000m/1km x 1/60min x 1/60s = 5.28 m/s

p = mv

p = (664 kg)(5.28 m/s) = 3504.4 kgm/s

The velocity at the end of vaneless space(state 3) is  965.92 m/s

It increases kinetic energy to the airstream using a rotating element and then converts it into potential energy in the form of pressure.

Temperature at state 2, T₂ =T₀ + c/2Cp

Substitute T₀ =450K, c=250m/s, Cp =1005, we get

T₂ =418.90K

From the velocity triangle, sinβ₂ =c₂/v₂

v₂ = 250/sin (90°-75°) = 965.92 m/s

Thus, the velocity at the end of vaneless space is 965.92 m/s.

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To find the speed of the planet in a perfectly circular orbit around the star, we can use the equation v = sqrt(Gm2 / r). Plugging in the given values, we get v = 1843.3 m/s. Therefore, the planet must have a speed of approximately 1843.3 m/s.

Answer:

\(1\times 10^{8} N/C\)

Explanation:

According to the gauss law

As we know that

Electric field is

\(E = -k\frac{q}{r^2}\)

where,

k = column constant = \(9 \times 10 ^{9}\ N. \frac{m^2}{c^2}\)

q = charge

r = distance from the sphere center

For computing the magnitude of e first we have to need to find out the charge outside of sphere which is

\(q = -\frac{Er^2}{k}\)

\(q = -\frac{1 \times 10^{6} \frac{N}{C} (25m)^2}{9 \times 10 ^{9}\ N. \frac{m^2}{c^2}}\)

q = -0.07 C

Now we have to find the electric field

\(E = k\frac{q}{r^2}\)

The r is 2.5m but in question it is given 5m

So,

Electric field is

\(E = 9 \times 10^{9} N . \frac{m^2 \times 0.07 C}{C^2 (2.5 m)^2}\)

\(E = 9 \times 10^{9} N. \frac{m^2 \times 0.07 C} {C^2 (2.5m)^2}\)

\(= 1\times 10^{8} N/C\)

For the given charge the magnitude of electric field at 1 m is  4 times the electric field at 2 m.

Electric charge can create magnetic field as well electric fields.

According to the given question a charge is given and we need to find electric field at 2 different positions that is at 1 m and 2 m.

Electric field at 1 m is E1   E1 = Kq/r₁²

E1 = Kq / 1²

E1 = Kq -------- 1

Electric field at 2 m is E2 , E2 = Kq/ r₂²

E2 = Kq / 2²

E2 = Kq / 4 ---------- 2

After comparing E1 and E2 from the above equation 1 and 2

E1 = 4 E2

Thus electric field at 1 m is  4 times the electric field at 2 m.

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The above question is incomplete. Check complete question below:

The magnitude of the electric field 1 m away from the positive charge is _________ the magnitude of the electric field 2 m away .

1. equal to

2. one-quarter

3. four times

4. two times

5. one-half

Explanation:

The acceleration on an object due to the gravity of any massive body is represented by g (small g). The force of attraction between any two unit masses separated by unit distance is called universal gravitational constant denoted by G(capital g). The relation between G and g is not proportional. That means they are independent entities.

G and g

In physics, G and g can be related mathematically as –

\(g=\frac{GM}{R^{2}}\)

Where,

1=g is the acceleration due to the gravity of any massive body measured in m/s2.

2=G is the universal gravitational constant measured in Nm2/kg2.

3=R is the radius of the massive body measured in km.

4=M is the mass of the massive body measured in Kg.

When two heavy blocks were connected by a uniform rope that has a mass of 4.00 kg, acceleration of the system was found to be 2.7 m/s²; the tension at the top of heavy rope was 136.5 N;  the tension exerted at the midpoint of the rope was112.5N.

Tension is defined as the act of stretching or straining or the condition of  a substance being stretched or strained.

a) Acceleration of the system can be calculated as

a=Fn/m

={200-(16x9.8)}/16

=2.7m/s²

so therefore the acceleration of the system was calculated and found to be=2.7m/s²

b) the tension of the heavy rope at the top can be calculated as

200-50-T1=5(2.7)

T1=136.5N

the tension at the top of heavy rope was 136.5 N

c)the tension at the middle point of the rope was

T2-9g = 9(a)

         =9(g+a)

         =9(9.8+2.7)

         =112.5 N

so here the tension exerted at the midpoint of the rope was 112.5N.

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The final temperature of the water at the given conditions is 2 ⁰C.

The given parameters:

Apply the principle of conservation of energy to determine the final temperature of the water as follows;

\(Q _{cold} = Q _{warm}\\\\mc (t_i - t_f) = mc (t_f - t_i)\\\\mc(1- 0) = mc(t_f - 1)\\\\1 = t_f - 1\\\\t_f = 1+1 \\\\t_f = 2 \ ^0C\)

Thus, the final temperature of the water at the given conditions is 2 ⁰C.

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The radar range equation is used to compute the range distance of the aircraft. The radar range equation is given by: are the transmitting and receiving antenna gain respectively,

Therefore, the range distance of discovering this airplane is approximately $41.1$ km.Solutions to extend the range distance of the radar can be:Use a more directional antenna on the radar to increase gain and reduce sidelobes

Use higher transmitting powerUse a lower operating frequencyUse pulse compression or frequency modulationUse a larger aperture antennaUse a more sensitive receiver

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According to Newton's first law, an unbelted victim in a car accident will continue to move in the same direction and with the same speed until the dashboard causes a change in motion.

Inertia is the tendency of an object to remain in motion in the absence of an unbalanced force. It is the property of an object to resist any change in motion unless acted upon by an external force.

The dashboard applies an external force that changes the direction and speed of the victim. This is because the person has no external forces acting on them to cause them to stop. Since they were in motion at the time of the accident, they will continue in that motion unless acted upon by another force, such as the dashboard, until they come to a stop or another force acts upon them.

Therefore, the best exemplifies the law of inertia. The law of inertia states that an object at rest will remain at rest, and an object in motion will remain in motion at a constant velocity unless acted upon by an external unbalanced force.

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The resultant momentum of the object is 13 meters per second.

A moving body's amount of motion is referred to as momentum. In general, it is more difficult to stop a moving object the more momentum it has. For this reason, the term is frequently used in a metaphorical sense, as in the sports team example.

It indicates that the team is having success (usually a winning streak) and is growing as a result. The other teams will find it more difficult to prevent the team from picking up steam.

Given that an object has an x–the momentum of 8.5 kilograms meters/second and a y–the momentum of 9.8 kilograms ·meters/second.

The resultant momentum will be calculated as:-

R = √ ( 9.8² + 8.5²)

R = √168.29

R = 13 meters per second

Therefore, the resultant momentum of the object will be 13 meters per second.

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48% is percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb.

Define resistance

A circuit's opposition to current flow is measured by its resistance. The unit of resistance is one ohm.

A resistor is an electrical component with two terminals that implements electrical resistance as a circuit element. Restricting current flow, adjusting signal levels, dividing voltages, biasing active parts, and terminating transmission lines are just a few of the functions for resistors in electronic circuits.

P = VI

I = V/R

P =  V²/R

R= r + R

P =  12*12(26+3.5)

P = 4.8W

48% is percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb.

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The lack of impact craters, evidence of ice flows, and geysers supplying material for the E-ring suggest ongoing geological activity on Enceladus. Option B is the correct answer.

Lack of impact craters on much of its surface, evidence of ice flows from possible volcanic activity and geysers that supply material for the E-ring.

The evidence suggesting that Enceladus has ongoing geological activity includes the lack of impact craters on much of its surface, which indicates that the surface is relatively young and has been resurfaced. This suggests ongoing geological processes.

Additionally, evidence of ice flows indicates the presence of possible volcanic activity, which contributes to the geological activity on Enceladus. Furthermore, geysers on Enceladus supply material for Saturn's E-ring, indicating ongoing geologic activity.

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The momentum of a car of mass 2,200 kg at ​21 ms−1 is 46,200 kg m/s, the impulse produced by a force of 5 N acting for 2.5 s is 12.5 Ns and a mass of 1.3 kg is initially at rest. The change in velocity when subject to an impulse of 52 Ns is 40 m/s.

(a) The momentum of a car is given by the product of its mass and velocity.

Momentum, p = mv

Where,m = mass of the car

= 2200 kg

v = velocity of the car

= 21 m/s

p = 2200 × 21

= 46,200 kg m/s

(b)  Impulse is the product of force and time.

Impulse, J = Ft

Where,

F = Force

= 5 N

t = time

= 2.5 s

J = 5 × 2.5

= 12.5 Ns.

(c) The change in velocity of an object is given by the ratio of impulse to mass.

Δv = J/m

Where, m = mass of the object

= 1.3 kgJ

= Impulse

= 52 Ns

Δv = J/m

= 52 / 1.3

= 40 m/s.

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Answer:

a) λ = 1.12 m

b) f = 5.41 Hz

c) v = 154.54 m/s

d) A = 0.22m

e)

\(v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\\)

Explanation:

You have the following equation for a wave traveling on a cord:

\(D=0.22sin(5.6x+34t)\)     (1)

The general expression for a wave is given by:

\(D=Asin(kx-\omega t)\)    (2)

By comparing the equation (1) and (2) you have:

A: amplitude of the wave = 0.22m

k: wave number = 5.6 m^-1

w: angular velocity = 34 rad/s

a) The wavelength is given by substitution in the following expression:

\(\lambda=\frac{2\pi}{k}=\frac{2\pi}{5.6m^{-1}}=1.12m\)

b) The frequency is:

\(f=\frac{\omega}{2\pi}=\frac{34s^{-1}}{2\pi}=5.41Hz\)

c) The velocity of the wave is:

\(v=\frac{\omega}{k}=\frac{34s^{-1}}{0.22m^{-1}}=154.54\frac{m}{s}\)

d) The amplitude is 0.22m

e) To calculate the maximum and minimum speed of the particles you obtain the derivative of  the equation of the wave, in time:

\(v_D=\frac{dD}{dt}=(0.22)(34)cos(5.6x+34t)\\\\v_D=7.48cos(5.6x+34t)\)

cos function has a minimum value -1 and maximum +1. Then, you obtain for maximum and minimum velocity:

\(v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\\)

Answer:

B. Km/s/hr

Explanation:

Answer:

third option down

Explanation:

The magnitude of the displacement of the object after it travelled for seven seconds is 28 m (Option A).

The displacement of an object is the change in the position of the object.

Velocity = displacement / time

How to determine the displacement from the velocity - time graph;

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

Time = 7 s

Velocity = 4 m/s

Displacement =?

Velocity = displacement / time

4 = displacement / 7

Cross multiply

Displacement = 4 × 7

Displacement = 28 m

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The form of interval training that typically involves a series of resistance and body-weight exercises with short rest intervals between each activity is called circuit training.

 Circuit training is a type of interval training that consists of a series of resistance and body-weight exercises with short rest periods between each activity. It is a form of exercise that can improve cardiovascular fitness and muscular endurance and is ideal for those who want to incorporate resistance training and aerobic exercises into a single routine.

Therefore, the main answer is option B) Circuit training. The explanation is that circuit training is a type of interval training that is made up of resistance and body-weight exercises with short rest periods between each activity.

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a)5m/s b)5

the 5 is because you add the seconds to get 8 seconds and then do the same with the distance to get 40. 40/8 = 5. speed = 5

Velocity = displacement/change in time

V = 40/8

I just realized how unorganised my math looks but I hope this is helpfull

The resistance of the copper wire when the length of the wire is halved is 5 ohms

The length of the copper wire = 3.14 m

The diameter of the wire = 2 x 10⁻² m

The resistance of  the wire = 10 ohms

The electrical resistance of the wire if the length of the wire is cut in half is

        R = ρ L / A

where R is the resistance of the copper wire

           ρ is the resistivity

           L is the length of the copper wire

           A is the area of the copper wire

We can see that the resistance is directly proportional to the length of the copper wire in the above equation.

Thus, if the length of the wire is cut in half, then the resistance of the wire will also be reduced by half the value.

          Resistance = 10 ohms / 2

                             = 5 ohms

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In a space with hyperbolic geometry, the behavior of parallel lines differs from that of Euclidean geometry.

In hyperbolic space, parallel lines diverge from each other as they extend further.If two initially parallel flashlight beams traverse billions of light-years of space in a hyperbolic geometry, they will gradually diverge from each other. The divergence between the beams will increase as they travel a greater distance.

This phenomenon is a consequence of the non-Euclidean geometry of space. In hyperbolic space, the curvature causes parallel lines to "spread out" or diverge. The extent of the divergence will depend on the specific curvature of the space and the distance traveled.

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