Answer:
ΔV = - 3274 V
Explanation:
For this exercise we can use conservation of energy
starting point.
Em₀ = U = q ΔV
final point
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q ΔV = ½ m v²
ΔV = [tex]\frac{m \ v^2 }{q}[/tex]
let's calculate
ΔV = [tex]\frac{4.4 \ 10^{-6} \ 80^2 }{ 2 \ 4.3 10^{-6} }[/tex]
ΔV = 3274.4 1 V
since the charge q is negative, the potential at point B must be less than the potential at point A, so the answers
ΔV = - 3274 V
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to volume 1450 cm3. If the expansion is isothermal, what are (a) the final pressure and (b) the work done by the gas
Answer:
a) The final pressure is 1.68 atm.
b) The work done by the gas is 305.3 J.
Explanation:
a) The final pressure of an isothermal expansion is given by:
[tex] T = \frac{PV}{nR} [/tex]
[tex] T_{i} = T_{f} [/tex]
[tex] \frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR} [/tex]
Where:
[tex]P_{i}[/tex]: is the initial pressure = 5.79 atm
[tex]P_{f}[/tex]: is the final pressure =?
[tex]V_{i}[/tex]: is the initial volume = 420 cm³
[tex]V_{f}[/tex]: is the final volume = 1450 cm³
n: is the number of moles of the gas
R: is the gas constant
[tex] P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm [/tex]
Hence, the final pressure is 1.68 atm.
b) The work done by the isothermal expansion is:
[tex] W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J [/tex]
Therefore, the work done by the gas is 305.3 J.
I hope it helps you!
What is the mass of an object if it is moving at a speed of 10 m/s and has 400 J of kinetic energy?
Answers:
8 kg
Explanation:
Kinetic Energy = (mass × velocity × velocity) ÷ 2
We know that Kinetic Energy = 400 J and velocity = 10 m/s.
KE = (m × v × v) ÷ 2
400 J = (m × 10 m/s × 10 m/s) ÷ 2
400 J = m × 50 m^2/s^2
To find the mass you will divide 400 J and 50 m^2/s^2.
m = 8 kg
You can also check it if it gives you 400 J.
KE = (m × v × v) ÷ 2
KE = (8 kg × 10 m/s × 10 m/s) ÷ 2
KE = 400 J
So this means that the mass is 8 kg. I know that it is a bit confusing, but when you do J (joules) ÷ m^2/s^2 = kg (kilograms). Hope this helps, thank you !!
The surface area of a postage
stamp is 0.00600 m^2, and the air
exerts 1.00 atm of pressure on it.
How much force does it exert on
the stamp?
(Hint: The standard unit for
pressure is Pa.)
(Unit = N)
Answer:
Force = 607.95 Newton
Explanation:
Given the following data;
Area = 0.00600 m^2
Pressure = 1 atm to Pascal = 101325 Pa
To find the force;
Pressure = Force/area
Force = pressure * area
Substituting into the equation, we have;
Force = 101325 * 0.00600
Force = 607.95 Newton.
Therefore, the amount of force exerted by the air on the stamp is 607.95 Newton.
REAL ANSWER OR REPORTED, REAL GETS BRAINLIST
2. What is the weight of:
A 32 kg child
An 82 kg linebacker
A 1000 kg car
3. What is the mass of:
A student who weighs 524 Newtons
A table that weighs 441 Newtons
A hammer that weighs 51 Newtons
4. A 15 kg box sits motionless on the floor.
Draw a force diagram for the box.
Write a vertical Fnet equation for the box.
Determine the weight of the box.
What is the magnitude of the normal force?
Answer:
32kg = 313.6N
82kg = 803.6N
1000kg = 9800N
524N = 53.47kg
441N = 45kg
51N = 5.20kg
for the box diagram the weight of the 15kg box is 147N
the net force will be equal to zero because the box is motionless
Fg goes downward arrow of box 147N
Fnorm goes upward arrow of box 147N
Using your knowledge on personal care products, how does sunscreen
lotion protect your skin from the damaging effect of ultraviolet rays?
Explain why of x-rays and gamma rays are commonly used in
radiotherapy.
Answer:
Ultraviolet rays from sun are very harmful from skin and can cause sunburn and skin diseases especially ultraviolet B rays. A sunscreen lotion act as a protection barrier on the skin that restrict the direct contact of UV rays with skin and filter the harmful rays to enter the skin.
Radiotherapy is a medical therapy use to treat cancer. Radiotherapy commonly uses x-rays and gamma rays because they are high-energy particles or waves that kills or destroys the cancer cells.
Sunscreen lotion is able to filter this damaging ultraviolet radiation and prevent it from damaging the skin.
The sun reaches us from outer space brings ultraviolet rays to us. Ultraviolet rays are known to have some damaging effects on the skin. One way to protect our skin from this damaging ultraviolet rays is to use sunscreen lotion which is able to filter this damaging ultraviolet radiation and prevent it from damaging the skin.
X-rays and gamma rays are used in radiotherapy because they are light energy rays which are able to penetrate and destroy malignant cells in the body.
Learn more: https://brainly.com/question/13695751
A ball player catches a ball 3.10 s after throwing it vertically upward. a) with what speed did he throw it? b) what height did it reach?
Answer:
speed=30.41, height=94.27
Explanation:
since the motion is vertical upward means the motion is against gravitational force and the initial velocity of the ball was zero.
1. State the law of conservation of energy and what it means for you as a human considering how energy works.
2. Explain how different forms of energy are related.
PLEASE I NEED HELP!! I NEED IT NOW!! AND PLEASE DO IT IN YOUR OWN WORDS!! THANK YOU!
Answer: 1. The law of consevation of energy sates that energy can neither be created nor destroyed. It can only be transformed or transfered from one form to another. The law of conservation of energy is found everywhere for example, Water falls from the sky, converting potential energy to kinetic energy.
2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.
Explanation:
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
Which two options are homogeneous mixtures?
A. Sand
B. A wet sponge
C. Liquid dishwashing soap
O D. Vinegar
Answer:
Liquid dishwashing soap
Vinegar
According to the law of conservation of energy, how will the sum of the kinetic
and potential energies compare at the four points?
Can you also answer questions 1-5
Answer:
I think that this ans may help you
Carousel conveyors are used for storage and order picking for small parts. The conveyorsrotate clockwise or counterclockwise, as necessary, to position storage bins at the storageand retrieval point. The conveyors are closely spaced, such that the operators travel timebetween conveyors is negligible. The conveyor rotation time for each item equals 1 minute;the time required for the operator to retrieve an item after the conveyor stops rotatingequals 0.25 minute. How many carousel conveyors can one operator tend without creatingidle time on the part of the conveyors
Answer:
the number of carousel conveyors that an operator can operate without any idle time is 5
Explanation:
Given the data in the question;
first we express the equation for number of carousel conveyors that can be operated by an operator;
n' = [tex]\frac{(a + t)}{( a + b)}[/tex]
where a is the concurrent activity time ( 0.25 minute )
b is the independent operator activity time
t is the independent machine activity time( 1 )
Now independent activity time is zero as the operator is not performing any inspection or packaging tasks.
So time taken for the operator to retrieve the finished item at the end of the process is the concurrent activity and independent machine activity time, the conveyor rotation time of each item
so
we substitute
0.25min for a, 1 for t and 0min for b
n' = [tex]\frac{(0.25min + 1min)}{( 0.25min+ 0 min)}[/tex]
n' = 1.25 min / 0.25
n' - 5
Therefore, the number of carousel conveyors that an operator can operate without any idle time is 5