A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used?
Answer:
2.86×10⁻¹⁸ seconds
Explanation:
Applying,
P = VI................ Equation 1
Where P = Power, V = Voltage, I = Current.
make I the subject of the equation
I = P/V................ Equation 2
From the question,
Given: P = 0.414 W, V = 1.50 V
Substitute into equation 2
I = 0.414/1.50
I = 0.276 A
Also,
Q = It............... Equation 3
Where Q = amount of charge, t = time
make t the subject of the equation
t = Q/I.................. Equation 4
From the question,
4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs
Q = 7.899×10⁻¹⁹ C
Substitute these value into equation 4
t = 7.899×10⁻¹⁹/0.276
t = 2.86×10⁻¹⁸ seconds
A 50-kg copper block initially at 140°C is dropped into an insulated tank that contains 90 L of water at 10°C. Determine the final equilibrium tempera
Answer:
16.33°C
Explanation:
Applying,
Heat lost by copper = heat gained by water
cm(t₁-t₃) = c'm'(t₃-t₂).............. Equation 1
Where c = specific heat capacity of copper, m = mass of copper, c' = specific heat capacity of water, m' = mass of water, t₁ = initial temperature of copper, t₂ = initial temperature of water, t₃ = final equilibrium temperature.
From the question,
Given: m = 50 kg, t₁ = 140°C, m' = 90 L = 90 kg, t₂ = 10°C
Constant: c = 385 J/kg°C, c' = 4200J/kg°C
Substitute these values into equation 1
50(385)(140-t₃) = 90(4200)(t₃-10)
(140-t₃) = 378000(t₃-10)/19250
(140-t₃) = 19.64(t₃-10)
140-t₃ = 19.64t₃-196.6
19.64t₃+t₃ = 196.4+140
20.64t₃ = 336,4
t₃ = 336.4/20.6
t₃ = 16.33°C
A certain electric stove has a 16 Ω heating element. The current going through the element is 15 A. Calculate the voltage across the element.
The voltage across the element is = 240 V
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What actually heats up the atmosphere?
Answer:
The heat source for our planet is the sun. Energy from the sun is transferred through space and through the earth's atmosphere to the earth's surface. Since this energy warms the earth's surface and atmosphere, some of it is or becomes heat energy.
12) A negatively-charged balloon touching a wooden wall
A) pulls positive charge on the wall surface toward it.
B) pushes negative charge in the wall away from it.
C) polarizes molecules in the wall.
D) all of the above.
Answer:
D) all of the above.
Explanation:
First polarises it, cahrging and discharging occurs at once.
A negatively charged balloon touching a wooden wall then from the given options option D is correct which is all of the above.
What is a charge?Charged matter experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. Positive or negative charges can exist in an electric field (commonly carried by protons and electrons, respectively).
Contrary charges attract one another, while like charges repel one another. A neutral object is one that carries no net charge. Classical electrodynamics, the name given to an early understanding of how charged particles interact, is still accurate for issues that do not call for taking into account quantum phenomena.
In the first step it polarizes molecules in the wall, then charging and discharging in the wall will take place at once.
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1. An AAMU basketball player is 2.03 meters tall. What is his height given in US customary units of feet and
inches?
Answer:
His height is 6.66 feet or 79.92 inches.
Explanation:
Given that,
An AAMU basketball player is 2.03 meters tall.
Let h is the height.
We know that,
1 m = 3.28 feet
So,
2.03 m = 6.66 feet
Also,
1 m = 39.37 inches
2.03 m = 79.92 inches
Hence, this is the required solution.
It takes a minimum distance of 98.26 m to stop a car moving at 17.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Answer:
x_f = 212.5m
Explanation:
t = (x_f-x_0)/(.5*(v_f-v_0))
t = (98.26m-0m)/(.5(0m/s-17m/s))
t = 11.56s
a = (v_f-v_0)/t
a = (0m/s-17m/s)/11.56s
a = -1.47m/s²
t = (v_f-v_0)/a
t = (0m/s-25m/s)/-1.47m/s²
t = 17s
x_f = x_0+(.5*(v_f-v_0))*t
x_f = 0m+(.5*(0m/s-25m/s))*17s
x_f = 212.5m
A student must use an object attached to a string to graphically determine the gravitational field strength near Earth's surface. The student attaches the free end of the string to the ceiling and pulls the object-string system so that the string makes an angle of 5 degrees from the object's vertical hanging position. The student then releases the object from rest and uses a stopwatch to measure the time it takes for the object to make one complete oscillation. Which of the following is the next step that will allow the student to determine the gravitational field strength?
А) Repeat the experiment by adding additional mass to the object for multiple trials
B) Repeat the experiment by changing the length of the string for multiple trials
C) Repeat the experiment by changing the angle that the string makes with the object's vertical hanging position
D) Repeat the experiment by measuring the time it takes to make two oscillations, three oscillations, and additional oscillations for multiple trials
Answer:
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
Explanation:
The student is reacting a simple pendulum experiment where he can determine the value of the relationship of gravity with the expression
T = 2π [tex]\sqrt{\frac{L}{g} }[/tex]
let's analyze each statement
A) False. The mass is not a paramer of the period expression
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
C) False. The angle while it is small does not influence the period
D) True. By changing the number of oscillations the period does not change, so you can get the value of the pull of gravity.
We can see that the expressions B and d are true, the most exact value is obtained using procedure B since the graphs allow to reduce the errors
I provided the question above.
Answer:
Explanation:
since it is connected in parallel combination
use this formula
[tex]\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2}[/tex]
[tex]\frac{1}{R} = \frac{1}{2} + \frac{1}{4}[/tex]
[tex]\frac{1}{R} = \frac{4+2}{2}[/tex]
[tex]\frac{1}{R} = \frac{6}{2}[/tex]
[tex]\frac{1}{R} = 3[/tex] ohm
therefore resistence = 3 ohm
then we should find power
P = VI
P = 12*3
P = 24 watt
now to find current use formula power = current * voltage
24 = current * 12
24/12 = current
2 = current
therefore current is 2 ampere (A).
to find potential difference (emf) use formula
V = IR
V = current * resistence
V = 2 * 3
V = 6 volt .
therefore potential difference is 6 volt.
A 500 Kg block is attached with a rope of length 5m, having area 0.4× 10-4 m2. If final length is 6m, Calculate the Stress, Strain and Young's Modulus?
Answer:
stress = 1.225 x 10^8 N/m^2
strain = 1/5
Young's modulus = 6.125 x 10^8 N/m^2
Explanation:
mass, m = 500 kg
length, L = 5 m
Area, A = 0.4 x 10^-4 m^2
Final length, L' = 6 m
extension, x = L'-L= 6 - 5 = 1 m
Stress is defined as force per unit area.
[tex]stress =\frac{Force}{Area}\\\\stress =\frac{500\times 9.8}{0.4\times 10^{-4}}\\\\stress = 1.225\times 10^8 N/m^2[/tex]
Strain is defined as the ratio of change in length to the original length.
[tex]strain =\frac{x}{L}\\\\strain = \frac{1}{5}[/tex]
Young's modulus is given by the ratio of stress to the strain.
[tex]Y = \frac{1.225 \times 10^8}{\frac{1}{5}}\\\\Y = 6.125\times 10^8 N/m^2[/tex]
What is the torque in ( lbs-ft ) of a man pushing on a wrench with 65 lbs of force 8 unches from the nut / bolt he is trying to turn?
Explanation:
The torque [tex]\tau[/tex] is given by
[tex]\tau=Fd = (65\:\text{lbs})(\frac{8}{12}\:\text{ft}) = 43.3\:\text{lbs-ft}[/tex]
What is the weight on Earth of an object with mass 45 kg. Hint gravity = 10 N/kg *
1 point
45 N
450 N
450 kg
10N
Answer:
450N
Explanation:
weight= m*g
weight=45*10
weight=450N
Determine the values of m and n when the following average magnetic field strength of the Earth is written in scientific notation: 0.0000451 T. Enter m and n, separated by commas.
Answer:
B = 4.51×10⁻⁵ T
Explanation:
Given that,
The average magnetic field strength of the Earth is 0.0000451 T.
We need to write the value in the form of scientific notation. Any number in scientific notation is written as follows :
N=a×bⁿ
Where
n is any integer and a is a real no
So,
0.0000451 = 4.51×10⁻⁵ T
So, the required answer is equal to 4.51×10⁻⁵ T.
A speedometer in a car gives the car’s speed at that given moment, or the?
A. General speed
B. Instantaneous speed
C. Average speed
D. Constant speed
It’s not C or D!
Answer:
a because it is at a given moment
Explanation:
did u
how many atoms are in a 4.7 g copper coin?
Answer:
x = 4.45 * 10 ^22 Note. Technically, this should be rounded to 4.5 * 10^22. There are only 2 sig digits.
Explanation:
You have to assume that the coin is pure copper, which I doubt. What a coin is actually made of depends on when it was minted. But for the sake of this question, we'll assume coins are pure copper.
Copper has an atomic mass of 63.546 grams / mol
So 4.7 g of copper = 4.7 / 63.545 mol
We have 0.07396 mol of copper
1 mol of anything = 6.02 * 10^23 atoms (in this case).
0.07396 mol = x
Cross Multiply
1 * x = 0.07396 * 6.02 * 10^23
x = 4.45 * 10 ^22 atoms of copper
New alleles arising from mutations in a population will
A car of mass M traveling with velocity v strikes a car of mass M that is at rest. The two cars’ bodies mesh in the collision. The loss of the kinetic energy the moving car undergo in the collision is
a) a quarter of the initial kinetic energy.
b) half of the initial kinetic energy.
c) all the initial kinetic energy.
d) zero.
Answer:
the correct answer is B
Explanation:
Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved
instantly starts. Before the crash
p₀ = M v +0
final instant. After the crash
m_f = (M + M) v_f
the moment is preserved
p₀ = p_f
M v = 2 M v_f
v_f = v / 2
let's look for kinetic energy
before the crash
K₀ = ½ M v²
after the crash
K_f = ½ 2M (v_f)²
K_f = ½ 2M (v/2)²
K_f = (½ M v²) ½
K_f = K₀ / 2
therefore the correct answer is B
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate
Complete Question
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate(and 2.00 cm from the other) is 450 V?
Answer:
[tex]V'=562.5v[/tex]
Explanation:
From the question we are told that:
Separation distance [tex]d=10cm[/tex]
Voltage at 8cm [tex]V_8=450v[/tex]
Generally the equation for Voltage is mathematically given by
[tex]|V|=|E.d|[/tex]
Where
E=electric field
Therefore
At [tex]d=0.8[/tex]
[tex](450-0)V=E*(0.08m)[/tex]
[tex]E=\frac{450}{0.08}[/tex]
[tex]E=5625v/m[/tex]
Therefore
At [tex]d=10[/tex]
[tex]V'=Ed[/tex]
[tex]V'=5625*0.1m[/tex]
[tex]V'=562.5v[/tex]
Suppose there are 3 molecules in a container. If each molecule has a 1-in-2 chance of being in the left half of the container, what is the probability that there are exactly 2 molecules in the left half of the container?
Answer:
Total probability = 3/8
Explanation:
Below is the calculation:
Number of molecules in the container = 3
The probability of one molecule in the left half, P = 3 / 2 = 1.5 or 1/2
The probability of second molecule in the left half, P1 = (3/4)
Total probability = P x P1
Total probability = (1/2) x (3/4)
Total probability = 3/8
If you warm up the volume of a balloon but keep the pressure the same, you would be using which gas law?
Answer:
Charles law
Explanation:
Charle's law states that the volume (V) of a given gas is directly proportional to the absolute temperature (T) at a constant pressure.
That is;
: V ∝ T
: V/T = K
According to this question, the volume of a balloon is warmed up but the pressure is kept the same. Charles law will be used because it shows the relationship between the volume (V) and the temperature (heat) at a constant pressure (P).
If the car falls down the side of the cliff, what is happening to the gravitational potential energy of the falling car? (Assume the bottom of the cliff is zero)
Answer:
Sentences with many clauses and phrases are difficult to understand because the clauses and phrases typically _____.
modify other clauses and phrases in the sentence
refer to other sentences in the passage
make it hard to determine where the sentence ends
change the intended meaning of the sentence
Explanation:
A homeowner has a new oil furnace which has an efficiency of 60%. For every 100 barrels of oil used to heat his house, how much (in barrels of oil) goes up the chimney as waste heat?
Answer:
below
Explanation:
The red light from a helium-neon laser has a wavelength of 644.6 nm in air. Find the speed, wavelength, and frequency of helium-neon laser light in air, water, and glass. (The glass has an index of refraction equal to 1.50.)speed (m/s)wavelength (nm)frequency (Hz)airwaterglass
Answer:
air f = 4.6527 10¹⁴ Hz
water f = 3.4914 10¹⁴ Hz
glass f = 3.1027 10¹⁴ Hz
Explanation:
The refractive index of a material is given by
n = c / v
where c is the speed of light in a vacuum c = 3 108 m / s and v is the speed of light in the material medium.
the speed of the wave is
v = λ f
we substitute
c / n = λ f
f = [tex]\frac{c}{n \ \lambda}[/tex]
The refractive indices are
air 1,00029
water 1.3330
glass 1.5
let's calculate the frequencies
vaccum
f = 3 10⁸ / 1 644.6 10⁻⁹
f = 4.6540 10¹⁴ Hz
air
f = 3 10⁸ / 1,00029 644.6 10⁻⁹
f = 4.6527 10¹⁴ Hz
Water
f = 3 10⁸ / 1.333 644.6 10⁻⁹
f = 3.4914 10¹⁴ Hz
glass
f = 3 10 ^ 8 / 1.5 644.6 10⁻⁹
f = 3.1027 10¹⁴ Hz
A heat engine with 0.100 mol of a monatomic ideal gas initially fills a 3000 cm3 cylinder at 800 K. The gas goes through the following closed cycle Isothermal expansion to 5000 cm3 ?
Part A How much work does this engine do per cycle? Express your answer with the appropriate units. sochoric cooling to 200 K -Isothermal compression to 3000 cm3. - Isochoric heating to 800 K Value Units
Part B What is its thermal efficiency? Express your answer with the appropriate units.
Answer:
below
Explanation:
Part A) This engine works per cycle is 254.9 J.
Part B) The thermal efficiency is 23.42%
What is the thermal efficiency?The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.
For isothermal expansion, work done is
W₁ =nRT₁ x ln(V₂/V₁)
W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)
W₁ = 339.8 J =Q₁
For isochoric cooling ,
W₂ =0
Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)
Q₂ = -748.3 J
For isothermal compression,
W₃ =nRT₂ ln (V₄/V₃)
W₃ = 0.1 x 8.314 x 200 x ln(3000/5000)
W₃ = -84.9J
For isochoric heating
W₄ =0
Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)
Q₄ = -748.3 J
Total work done in all the process W = W₁ +W₂ +W₃ +W₄
W =254.9 J
Thus, the work done is 254.9 J
Thermal efficiency = Work done/Heat taken
η = W/ Q₁ +Q₄
η = [254.9 / 339.8 +748.3 ] x 100 %
η = 0.2342 x 100 %
η = 23.42%
Thus, the thermal efficiency is 23.42%
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'
A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficients of static and kinetic friction are 0.697 and 0.371, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? If the block does not move, give 0 m/s2 as the acceleration?
Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
The lumberjack pulls on the sled with 40 N at an angle of 30 degrees, pulling so the sled moves at a constant velocity. 1) What is the x component of the applied force? 2) What is the y component of the applied force? 3) If the loaded sled has a mass of 65 kg, what is the magnitude of the force of gravity? 4) What is the magnitude of the normal force acting on the sled? 5) What is the coefficient of friction between the snow and the sled?
1) (40 N) cos(30°) ≈ 34.6 N
2) (40 N) sin(30°) = 20 N
3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N
4) The net force on the sled acting in the vertical direction is made up of
• the sled's weight, 585 N, pointing downward
• the vertical component of the applied force, 20 N, pointing upward
• the normal force, with magnitude n, also pointing upward
The sled does not move up or down, so by Newton's second law,
∑ F = n + 20 N - 585 N = 0 ==> n = 565 N
5) The net force in the horizontal direction consists of
• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)
• kinetic friction, with magnitude f, pointing in the opposite and negative direction
By Newton's second law,
∑ F = 34.6 N - f = 0 ==> f ≈ 34.6 N
Now if µ is the coefficient of kinetic friction, then
f = µn ==> µ = f/n = (34.6 N) / (565 N) ≈ 0.0613
The component of the force is the effective part of that force in that direction.
What is the component of a force?The component of the force is the effective part of that force in that direction.
1) The horizontal component of a force = 40 N cos 30 degrees = 34.6 N
2) The vertical component of the force = 40 N sin 30 degrees = 20 N
3) The magnitude of the gravitational force = mg cos 30 degrees = 65 Kg * 9.8 m/s^2 * cos 30 degrees = 551.7 N
4) The normal force = 551.7 N
5) The coefficient of friction = F/R = 40 N /551.7 N = 0.07
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A 3-kg projectile is launched at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.
1517.4 m
Step-by-step explanation:
Since the projectile broke up at the peak of its flight, it already traveled half its initial range so we can find its initial launch velocity [tex]v_0[/tex] from the equation
[tex]\frac{1}{2}R= \dfrac{1}{2} \left(\dfrac{v_0^2}{g}\sin 2\theta_0 \right)[/tex]
where [tex]\theta_0 = 45°[/tex] and [tex]\frac{1}{2}R = 50\:\text{m}[/tex] so we will get [tex]v_0=31.3\:\text{m/s}[/tex]. Next, we can use the equation
[tex]v_y = v_0y - gt = v_0 \sin 45 - gt[/tex]
and since [tex]v_y=0[/tex] at its peak, we get t = 22.1 s. Let's set this aside for a moment and we'll use it later.
At the top of its peak, we can use the conservation law of linear momentum. Let M be the mass if of the original projectile, [tex]m_1[/tex] be the mass of the larger fragment (2 kg) and [tex]m_2[/tex] be the mass of the smaller fragment (1 kg). We can write the conservation law as
[tex]Mv_0x = m_1V_1 + m_2V_2[/tex]
where [tex]V_1\:\text{and}\:V_2[/tex] are the velocities of the fragments immediately after the break up. But we also know that [tex]V_1=0[/tex] so the velocity of [tex]m_2[/tex] can be calculated from the conservation law as
[tex]Mv_0 \cos 45° = m_2V_2[/tex]
or
[tex]V_2 = \dfrac{M}{m_2}v_0 \cos 45° = 66.4\:\text{m/s}[/tex]
Now we can calculate the horizontal distance the smaller fragment traveled after the break up. Recall that the amount of time for it to go up is also the amount of time to get down so the horizontal distance x is
[tex]x = V_2 t = (66.4\:\text{m/s})(22.1\:\text{s})= 1467.4\:\text{m}[/tex]
Therefore, the total distance traveled from the launch point is
[tex]D = 50\:\text{m} + 1467.4\:\text{m}=1517.4\:\text{m}[/tex]
40 ohms
1.2 A
40 ohms
12 V
Calculate the total energy developed in 5
minutes by the system above.
Answer:
17280 J and 1080 J
Explanation:
Given :
R= 40 ohm
I=1.2A
t= 5 min=60×5=300 sec
Now,
Total energy can be calculated as:
[tex]E=I^{2} Rt\\E=(1.2)^{2} *40*300\\E=17280 J[/tex]
Now,
V=12V
R=40 Ohm
[tex]E=\frac{V^{2} }{R} *t\\E=\frac{(12)^{2} }{40} *300\\E=1080 J[/tex]
Total energy is 17280 J and 1080 J
The tendency for objects to resist acceleration is called *
A) motion
B) inertia
C) reaction force
D) sluggishness
Answer:
B) Inertia is the answer
Record the lengths of the sides of ABC and ADE.
