PHYSICS COLLEGE

A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?

Answers

Answer 1

Answer:

Answer:

The no. of electrons is $7.59* 10^(21)$

Solution:

According to the question:

The rate at which the charge is delivered is given by:

$(dQ)/(dt) = - 0.75$

Now,

$\int_(0)^(Q)dQ = - 0.75\int_(0)^(27 min) dt$

$Q = -0.75t|_(0)^(27 min)$

$Q= -0.75* 27* 60 = - 1215 C$

No. of electrons, n can be calculated from the following relation:

Q = ne

where

e = electronic charge = $1.6* 10^(- 19) C$

Thus

$n = (Q)/(e)$

$n= (1215)/(1.6* 10^(- 19))$

$n = 7.59* 10^(21)$

Related Questions

A 1,200 kg car travels at 20 m/s. what is it’s momentum ?
Which law of physics relates electric fields and current
A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?
A local meteorologist reports the day’s weather. "Currently sunny outside, 34°F. Skies will become overcast later this afternoon, as temperatures drop to 25°F, with windy conditions out of the north at 10–15 miles per hour. Radar indicates 2–3 inches of snow expected to fall later tonight.” Which information is qualitative? These are non-numerical, descriptive data. These are numerical data that have been measured. “sunny” “25°F” “2–3 inches of snow” “10–15 miles per hour”
An athlete can exercise by making mechanical waves in ropes. What is themedium of these waves?A. EnergyB. The ropeC. The athleteD. Air

1. Draw a quantitative motion map for the following description: A bicyclist speeds along a road at 10 m/s for 6 seconds. Then she stops for three seconds to make a 180˚ turn and then travels at 5 m/s for 3 seconds.

Answers

Answer:

Please find the attached file for the figure.

Explanation:

Given that a bicyclist speeds along a road at 10 m/s for 6 seconds.

Its acceleration = 10/6 = 1.667 m/s^2

The distance covered = 1/2 × 10 × 6

Distance covered = 30 m

That is, displacement = 30 m

Then she stops for three seconds to make a 180˚ turn and then travels at 5 m/s for 3 seconds.

The acceleration = 5/3 = 1.667 m/s^2

The displacement = 1/2 × 5 × 3

Displacement = 7.5 m

The resultant acceleration will be equal to zero.

While the resultant displacement will be:

Displacement = 30 - 7.5 = 22.5 m

Please find the attached file for the sketch.

If a person’s weight is W on the surface of the earth, calculate what it would be, in terms of W, at the surface of (a) the moon;
(b) Mars;
(c) Jupiter.

Answers

Answer:b

Explanation:

saad has mass 80kg when resting on the ground at the equator what will be the centripetal acceleration on saad if the radius of earth is 6.4×10^6 meter

Answers

I did try to solve. I hope it is correct, below is the solution:

put everything in s.i units
then the answer what u wrote is acceleration to get is divide by mass(80)G=6.011*10^-11
M=6*10^24
R=6.4*10^6
m=80

Hope it helps.

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4 = 407 N , and MA = 1504 N⋅m . Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.

Answers

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

Final answer:

The Cartesian components of the resultant force and the couple moment are calculated by summing up all the forces and moments acting on the object. The resultant force is 1724 N and the couple moment is 29.764 N*m.

Explanation:

The resultant force and couple moment in the Cartesian coordinate system can be obtained by summing up all the forces and moments acting on the object. In this case, we have the forces F1, F2, F3, F4 and the couple moment MA acting on the object. The resultant force (FR) can be calculated as the sum of all the forces, i.e., FR = F1 + F2 + F3 + F4. Using the values given, FR = 510 N + 306 N + 501 N + 407 N = 1724 N. The resultant moment (MR) can be calculated as the sum of all the moments, i.e., MR = d1*F1 + d2*F2 + d3*F3 + d4*F4 - MA. Using the values given, MR = 0.880 m * 510 N + 1.11 m * 306 N + 0.560 m * 501 N + 2.08 m * 407 N - 1504 N*m = 29.764 N*m. Therefore, the Cartesian components of the resultant force and the couple moment are 1724 N and 29.764 N*m respectively.

Learn more about Resultant force and moment here:

brainly.com/question/34367496

#SPJ3

Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?

Answers

Answer:

a) ω₁ = ω₂ = 3.7 rad/sec

b) Δθ₁ = Δθ₂ = 18.5 rad

c) d₁ = 14.5 m d₂ = 57.5 m

d) Fc1 = 273.9 N Fc2 = 1069.8 N

e) The boy near the outer edge.

Explanation:

Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.
However, linear speeds of points at different distances from the center, are different.
Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:

$v = \omega*r (1)$

Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:

$\omega = (v_(out) )/(r_(out) ) = (11.5m/s)/(3.14m) = 3.7 rad/sec (2)$

As we have already said, ωout = ωin = 3.7 rad/sec

Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.
Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:
$\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)$

⇒ Δθ₁ = Δθ₂ = 18.5 rad.

The linear distance traveled by each child, will be related with the linear speed of them.
Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:

$v_(inn) = \omega * r_(inn) = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)$

vout is a given of the problem ⇒ vout = 11. 5 m/s

Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:

$d_(inn) = v_(inn) * t = 2.9m/s* 5.0 s = 14.5 m (5)$

$d_(out) = v_(out) * t = 11.5 m/s* 5.0 s = 57.5 m (6)$

The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:

$F_(c) = m*(v^(2))/(r) (7)$

Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:

$F_(cin) = m*(v_(in)^(2))/(r_(in)) = 25.4 kg* ((2.9m/s)^(2) )/(0.78m) = 273.9 N (8)$

In the same way, we get Fcout (the force on the boy near the outer edge):

$F_(cout) = m*(v_(out)^(2))/(r_(out)) = 25.4 kg* ((11.5m/s)^(2) )/(3.14m) = 1069.8 N (9)$

The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.
The maximum friction force is given by the product of the coefficient of static friction times the normal force.
Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.
As both boys have the same mass, the normal force is also equal.
This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:
$F_(frs) = \mu_(s) * m* g (10)$

If this force is greater than the centripetal force, the boy will be able to hold on.
So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

Answers

A diverging lens always produces a virtual erect image.

The general lens formula is given as;

$(1)/(F) = (1)/(U) + (1)/(V)$

Where;

U = object distance
V = image distance
F = focal length of the lens

A lens can be converging or diverging.

A converging lens produces a virtual image when the object is placed in front of the focal point. The image can also be real when the object is placed beyond focal point.

The image produced by a diverging lens is always virtual and upright.

Thus, we can conclude that a diverging lens always produces a virtual erect image.

Learn more here:brainly.com/question/11788630

Answer:

E) true. The image is always virtual and erect

Explanation:

In this exercise we are asked to find the correct statements,

for this we can use the constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

In diverging lenses, the focal length is negative and the image is virtual and erect

In convergent lenses, the positive focal length, if the object is farther than the focal length, the image is real and inverted, and if the object is at a shorter distance than the focal length, the image is virtual and straight.

With this analysis let's review each statement

A) False. The image is right

B) False. The type of image depends on where the object is with respect to the focal length

C) False. The real image is always inverted

D) False. The image is always virtual

E) true. The image is always virtual and erect

Other Questions

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?

How do lenses and mirrors compare in their interactions with light? A. Lenses spread apart light; mirrors do not.B. Lenses reflect light; mirrors do not.C. Lenses refract light; mirrors do not.D. Lenses focus light; mirrors do not.

A radioactive nucleus has a half-life of 5*108 years. Assuming that a sample of rock (say, in an asteroid) solidified right after the solar system formed, approximately what fraction of the radioactive element should be left in the rock today? The age of the solar system is 4.5*109 years. (No calculator is necessary, but if it helps use it.)

A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very far from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?