Answer:
The electromotive force is the maximum potential difference between the terminals of a battery.
The electromotive force is the maximum potential difference between the terminals of a battery. The correct option is b.
What is electromotive force?The electromotive force also called as EMF, is the force which causes current to flow from the positive to negative terminal of the battery.
The electromotive force is the maximum potential difference between the terminals of a battery.
Thus, the correct option is b.
Learn more about electromotive force
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A heavy truck moving with 20 km/hr hits a car at rest. A physics student argued that
the maximum velocity the car suddenly gains is 40 km/hr. Do you agree with it?
Explain with necessary theory
Answer:
Yes
Explanation:
speed of truck = 20 km/h
Initially the car at rest.
maximum velocity of car = 40 km/h
When the truck and the car collide, the momentum of the truck transferred to car.
So, the car can attain the speed of 40 km/h.
Action and reaction are equal in magnitude and opposite in direction.Then Why do not balance each other
Answer:
Action and reaction are equal in magnitude and opposite in direction but they do not balance each other because they act on different objects so they don't cancel each other out.
hope this will help you more
A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?
Explanation:
Given
Acceleration of the pebble is
At t=0, velocity is
considering horizontal motion
[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]
Velocity acquired during this time
[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]
Consider vertical motion
[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]
Net velocity is
[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]
Angle made is
[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]
I need help with this physics question.
Answer:
5.04 m
Explanation:
You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.
Answer:
5.0384m
Explanation:
% increase = 100 x (Final - Initial / | initial | )
( |~~| Bars indicate absolute value since you can't have a negative height)
Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics.
Answer:
Explanation:
First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.
Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.
For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.
The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.
First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.
Find the volume of cuboid of side 4cm. Convert it in SI form
Answer:
0.000064 cubic meters.
Explanation:
Given the following data;
Length of side = 4 centimeters
Conversion:
100 centimeters = 1 meters
4 cm = 4/100 = 0.04 meters
To find the volume of cuboid;
Mathematically, the volume of a cuboid is given by the formula;
Volume of cuboid = length * width * height
However, when all the sides are equal the formula is;
Volume of cuboid = L³
Volume of cuboid = 0.04³
Volume of cuboid = 0.000064 cubic meters.
Many types of decorative lights are connected in parallel. If a set of lights is connected to a 110 V source and the filament of each bulb has a hot resistance of what is the currentthrough each bulb
Answer:
i₀ = V / R_i
Explanation:
For this exercise we use Ohm's law
V = i R
i = V / R
the equivalent resistance for
[tex]\frac{1}{R_{eq}}[/tex] = ∑ [tex]\frac{1}{R_i}[/tex]
if all the bulbs have the same resistance, there are N bulbs
[tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]
R_{eq} = R_i / N
we substitute
i = N V / Ri
where i is the total current that passes through the parallel, the current in a branch is
i₀ = i / N
i₀ = V / R_i
A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p
Answer:
a. 8p
Explanation:
We are given that
Radius of hollow sphere , R1=R
Density of hollow sphere=[tex]\rho[/tex]
After compress
Radius of hollow sphere, R2=R/2
We have to find density of the compressed sphere.
We know that
[tex]Density=\frac{mass}{volume}[/tex]
[tex]Mass=Density\times volume=Constant[/tex]
Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]
Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]
Using the formula
[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]
[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]
[tex]\rho_2=8\rho[/tex]
Hence, the density of the compressed sphere=[tex]8\rho[/tex]
Option a is correct.
Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.
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.
a vechile having a mass of 500kg is moving with a speed of 10m/s.Sand is dropped into it at the rate of 10kg/min.What force is needed to keep the vechile moving with uniform speed
Answer:
1.67 N
Explanation:
Applying,
F = u(dm/dt)+m(du/dt)................ Equation 1
Where F = force, m = mass of the vehicle, u = speed.
Since u is constant,
Therefore, du/dt = 0
F = u(dm/dt)............... Equation 2
From the question,
Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s
Substitute these values into equation 2
F = 10(10/60)
F = 100/60
F = 1.67 N
A long, straight, vertical wire carries a current upward. Due east of this wire, in what direction does the magnetic field point
The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.
10 A turning pork creates sound cares
with
Frequency of 170Hz: To the
speed of sound in is in 340mls
calculate the wave
wave length
of
in air is
the sound wales.
Answer:
2m
Explanation:
wavelength=speed/frequency
=340/170
=2m
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
Answer:
[tex]\sigma=0.014\ C/m^2[/tex]
Explanation:
Given that,
The radius of sphere, r = 5 cm = 0.05 m
Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C
We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,
[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]
So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].
Which level of government relies the most on income tax?
OA.
federal
state
OC.
local
Answer:
Its the Federal government
As a roller coaster car crosses the top of a 48.01-m-diameter loop-the-loop, its apparent weight is the same as its true weight. What is the car's speed at the top?
Answer:
The speed of the car, v = 21.69 m/s
Explanation:
The diameter is = 48.01 m
Therefore, the radius of the loop R = 24.005 m
Weight at the top is n = mv^2/R - mg
Since the apparent weight is equal to the real weight.
So, mv^2/R - mg = mg
v = √(2Rg)
v = √[2(24.005 m)(9.8 m/s^2)]
The speed of the car, v = 21.69 m/s
Answer:
The speed is 15.34 m/s.
Explanation:
Diameter, d = 48.01 m
Radius, R = 24.005 m
Let the speed is v and the mass is m.
Here, the weight of the car is balanced by the centripetal force.
According to the question
[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]
Please show steps as to how to solve this problem
Thank you!
Explanation:
Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:
[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]
[tex] -m_1gx + m_2gd_2 = 0[/tex]
[tex]m_1x = m_2d_2[/tex]
Solving for x,
[tex]x = \dfrac{m_2}{m_1}d_2[/tex]
[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x-axis.
Answer:
-1.0778×10⁻¹⁰ N/C
Explanation:
Applying,
E = kq/r²................ equation 1
Where E = elctric field, q = charge, r = distance, k = coulomb's law
From the question,
Given: q = -3.0×10 C, r = 5.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values in equation 1
E = (-3.0×10)(8.98×10⁹)/5²
E = -1.0778×10⁻¹⁰ N/C
Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the spring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.
Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]
So, the new work is more than 130 J.
Characteristics or properties of matter or energy that can be measured
Answer:
Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:
color (intensive)
density (intensive)
volume (extensive)
mass (extensive)
boiling point (intensive): the temperature at which a substance boils
melting point (intensive): the temperature at which a substance melts
Explanation:
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?
Answer:
7.9 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:
[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]
Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:
[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]
Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]
Plug everything in:
7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]
(1590kg the initial weight plus the weight of the added passenger)
When should a line graph be used?
A. When the independent variable is continuous and does not show a relationship to the dependent variable
B. When the independent variable is composed of categories and does not show a relationship
C. When the independent variable is continuous and shows a casual link to the dependent variable
D. When there is no independent variable
a tiger covers a distance of 600 m in 15 minutes what is a speed of a tiger?
Answer: 2.4 km/hr
Explanation:
Distance = 600m
Time= 15 minutes = 15 x 60 second/minute = 900 seconds
Speed = [tex]\frac{Distance}{Time}[/tex] = [tex]\frac{600}{900}[/tex] = [tex]\frac{2}{3}[/tex] m/sec
⇒ [tex]\frac{2}{3}[/tex] x [tex]\frac{18}{5}[/tex] = 2.4 km/hr (1 m/sec = 3.6 km/hr)
NEED HELP ASAP. Please show all work.
A point on a rotating wheel (thin hoop) having a constant angular velocity of 200 rev/min, the wheel has a radius of 1.2 m and a mass of 30 kg. ( I = mr2 ).
(a) (5 points) Determine the linear acceleration.
(b) (4 points) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Look at work
Explanation:
a) I am not sure if you want tangential or centripetal but I will give both
Centripetal acceleration = r*α
Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2
Tangential acceleration = ω^2*r
convert 200rev/min into rev/s
200/60= 10/3 rev/s
a= 100/9*1.2= 120/9= 40/3 m/s^2
b) Rotational Kinetic Energy = 1/2Iω^2
I= mr^2
Plug in givens
I= 43.2kgm^2
K= 1/2*43.2*100/9=2160/9=240J
A small ball of uniform density equal to 1/2 the density of water is dropped into a pool from a height of 5m above the surface. Calculate the maximum depth the ball reaches before it is returned due to its bouyancy. (Omit the air and water drag forces).
Answer:
1.67 m
Explanation:
The potential energy change of the small ball ΔU equals the work done by the buoyant force, W
ΔU = -W
Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5
ΔU = mgΔh
ΔU = ρVgΔh
Now, W = ρ'VgΔh' where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h
So, ΔU = -W
ρVgΔh = -ρ'VgΔh'
ρVg(h - 5) = -ρ'Vgh
ρ(h - 5) = -ρ'h
Since the density of the small ball equals 1/2 the density of water,
ρ = ρ'/2
ρ(h - 5) = -ρ'h
(ρ'/2)(h - 5) = -ρ'h
ρ'(h - 5)/2 = -ρ'h
(h - 5)/2 = -h
h - 5 = -2h
h + 2h = 5
3h = 5
h = 5/3
h = 1.67 m
So, the maximum depth the ball reaches is 1.67 m.
How many loops are in this circuit?
I see six (6) loops.
I attached a drawing to show where I get six loops from.
A commuter backs her car out of her garage with an acceleration of . (a) How long does it take her to reach a speed of 2.00 m/s
Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s
Answer:
1.43 s
Explanation:
Applying,
a = (v-u)/t........... Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time
make t the subject of the equation
t = (v-u)/a........... Equation 2
From the question,
Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²
Substitute into equation 2
t = (2-0)/1.4
t = 1.43 s
You drop two balls of equal diameter from the same height at the same time. Ball 1 is made of metal and has a greater mass than ball 2, which is made of wood. The upward force due to air resistance is the same for both balls. Is the drop time of ball 1 greater than, less than, or equal to the drop time of ball 2? Explain why
Answer:
The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.
Explanation:
The net force acting on the ball will be:
⇒ [tex]F_{net}=mg-F_r[/tex]
Here,
F = Force
m = mass
g = acceleration
Now,
According to the Newton's 2nd law of motion, we get
⇒ [tex]F_{net} = ma[/tex]
To find the value of "a", we have to substitute "[tex]F_{net}=ma[/tex]" in the above equation,
⇒ [tex]ma=mg-F_r[/tex]
⇒ [tex]a=g-\frac{F_r}{m}[/tex]
We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.
Answer:
Both the ball takes equal time to reach to the ground.
Explanation:
Two balls of same diameter
Let the height is h.
Mass of ball 1 is more than the mass of ball 2.
The second equation of motion is
[tex]h = u t +0.5 gt^2[/tex]
Here, the buoyant force due to air is same. So, the time of fall is independent of the mass.
So, both the ball takes equal time to reach to the ground.
High-speed stroboscopic photographs show that the head of a -g golf club is traveling at m/s just before it strikes a -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at m/s. Find the speed of the golf ball just after impact.
The question is incomplete. The complete question is :
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a 50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.
Solution :
We know that momentum = mass x velocity
The momentum of the golf club before impact = 0.200 x 60
= 12 kg m/s
The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.
Now the momentum of the club after the impact is = 0.2 x 40
= 8 kg m/s
Therefore the momentum of the ball is = 12 - 8
= 4 kg m/s
We know momentum of the ball, p = mass x velocity
4 = 0.050 x velocity
∴ Velocity = [tex]$\frac{4}{0.050}$[/tex]
= 80 m/s
Hence the speed of the golf ball after the impact is 80 m/s.
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
You may have been surprised to learn that Olympic gold medals are not made from solid gold, but instead have a coating of • Saved gold on the outside.
To see a possible reason why, determine the value of the medal the size (not mass) of the Olympic gold medal if it were made of solid gold. Hint: As of mid-2018, the cost of gold is about $40 per gram.
Answer:
A gold medal has the (minimum) dimensions of:
diameter = 60mm
thickness = 3mm
So we will work with those dimensions.
The medal is then a cyinder of diameter
D = 60mm = 6cm
and height:
H = 3mm = 0.3cm
Remember that the volume of a cylinder is:
V = pi*(D/2)^2*H
where pi = 3.14
Then the volume of a medal is:
V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3
The density of the gold in g/cm^3 is:
d = 19.3 g/cm^3
And remember that:
density = mass/volume
So, if the volume is 3.768 cm^3
Then the mass will be:
mass = density*volume = 19.3 g/cm^3*3.768 cm^3 = 72.7 g
So, a single gold medal would weight 72.7 grams
And each gram of gold costs $40
Then the total cost of the gold medal would be:
value = $40*72.7 = $2,908
Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.
And each gold medal costs $2,908
So the total cost (only for the gold medals, ignoring the others) would be to high.
This is why the gold medals are made mostly of silver.