Answer:
-7/19
Explanation:
Is Geothermal Energy renewable? Why or why not? Use in your own words.
Answer:
Yes, geothermal energy is a renewable energy resource because the water can be heated by pumping it through the rocks.
An 7.40 kg block drops straight down from a height of 0.83 m, striking a platform spring having a force constant of 9.50 102 N/m. Find the maximum compression of the spring.
Answer:
0.25 m.
Explanation:
mass of the block = 7.40 kg, height = 0.83 m, force constant of the spring = 9.50 x [tex]10^{2}[/tex] N/m.
The maximum compression on the spring can be determined by;
Potential energy stored in the spring = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
But, potential energy = mgh
So that,
mgh = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]
7.4 x 9.8 x 0.83 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
60.1916 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]
[tex]x^{2}[/tex]= [tex]\frac{60.1916}{9.50*10^{2} }[/tex]
= 0.06336
x = 0.2517
x = 0.25 m
The maximum compression of the spring is 0.25 m.
An elastic conducting material is stretched into a circular loop of 13.6 cm radius. It is placed with its plane perpendicular to a uniform 0.871 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 73.9 cm/s. What emf is induced in volts in the loop at that instant?
Answer: The Answer is attached to the image below
It is more difficult to climb a vertical staircasethan a slanted one give reason
Answer:
The reason is similar to the reason why it is difficult to roll an object on a surface with a positive incline than rolling it on the ground
The more the path becomes vertical, the more force we have to apply to oppose the force of gravity
but when we are moving horizontally, we don't have to move against the gravity and hence, it is less difficult than going vertically upwards
If a rock is dropped from the top of a tower at the front of and it takes 3.6 seconds to hit the ground. Calculate the final velocity of the penny in m/s.
Answer:
36 m/s
Explanation:
t = 3.6s
u = 0m/s
a = +g = 10m/s²
v = ?
using,
v = u + at
v = 0 + 10(3.6)
v = 36 m/s
A ball is thrown 24 m/s into the air. How high does it go?
556.4 m
0 m
29.4 m
-556.4 m
Answer:
option c is correct
Explanation:
we know that
2as=vf^2-vi^2
vf=24 m/s
vi= 0 m/s
a=g= 9.8 m/s^2
s=vf^2-vi^2/2a
s=(24)²-(0)²/2*9.8
s=576/19.6
s=29.4 m
therefore option c is correct
. A car going initially with a velocity 15 m/s accelerates at a rate of 2 m/s2 for 10 seconds. It then accelerates at a rate of -1.5 m/s until stop. Find the car’s maximum speed. Calculate the total distance traveled by the car.
Answer:
The maximum speed of the car is 35 m/s
The total distance traveled by the car is 658.33 m
Explanation:
Given;
initial velocity of the car, u = 15 m/s
acceleration of the car, a = 2 m/s²
time of car motion, t = 10 s
(i)
Initial distance traveled by the car is given by;
d₁ = ut + ¹/₂at²
d₁ = (15 x 10) + ¹/₂(2)(10)²
d₁ = 150 + 100
d₁ = 250 m
The maximum speed of the car during this is given by;
v² = u² + 2ad₁
v² = (15)² + (2 x 2 x 250)
v² = 1225
v = √1225
v = 35 m/s
(ii)
The final distance cover by the car during the deceleration of 1.5 m/s².
Note: the final or maximum speed of the car becomes the initial velocity during deceleration.
v² = u² + 2ad₂
where;
v is the final speed of the car when it stops = 0
0 = u² + 2ad₂
0 = (35²) + (2 x - 1.5 x d₂)
0 = 1225 - 3d₂
3d₂ = 1225
d₂ = 1225 / 3
d₂ = 408.33 m
The total distance traveled by the car is given by;
d = d₁ + d₂
d = 250 m + 408.33 m
d = 658.33 m
A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .
Complete Question
A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex] to [tex]t_2 = 6.0s[/tex]
What is its average velocity over this time interval?
Answer:
The velocity is [tex]v = 3.903 \ m/s[/tex]
Explanation:
From the question we are told that
The first position of the ball is [tex]x_1 = 8.0 \ cm[/tex]
The second position of the ball is [tex]x_2 = - 4.1 \ cm[/tex]
Generally the average velocity is mathematically represented as
[tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]
=> [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]
=> [tex]v = 3.903 \ m/s[/tex]
A car’s brakes decelerate it at a rate of -2.40 m/s2. If the car is originally travelling at 13 m/s and comes to a stop, then how far, in meters, will the car travel during that time?
Answer:
Approximately [tex]35.2\; \rm m[/tex].
Explanation:
Given:
Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].
Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)
Implied:
Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)
Required:
Displacement, [tex]x[/tex].
Not required:
Time taken, [tex]t[/tex].
Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:
[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].
In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.
In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.
Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.
The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2
Fd = (1/2)mv^2
F = (1/2)mv^2/d.
Plug in m = 20 kg, v = 3 m/sec, d = 40 m.
83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.
Explanation:Hope I helped :)
If a dog has a mass of 2.5 kg, what is its weight and what is the normal force that it feels.
I
Answer:
Weight = normal force = 24.5 N
Explanation:
Given that,
Mass of a dog, m = 2.5 kg
We need to find its weight and the normal force that it feels.
The weight of an object is given by :
W = mg
Where g is the acceleration due to gravity
[tex]W=2.5\times 9.8\\\\=24.5\ N[/tex]
The normal force is balanced by the weight of an object. So,
Weight = normal force = 24.5 N
a current of 200 mA through a conductor converts 40 joules of electrical energy into heat in 30 seconds determine the p
otential drop across the conductor
Answer:
ou have I=200mA, E=40J, t=30s, and you want to find the voltage drop.
First, you should know that P=V⋅I , so V=PI
Second, you have the amount of energy converted in a certain amount of time, so E=P⋅t
So, find the power and use it to find the voltage drop.
this works , but i thought energy was defined by W = P * t whitch would then be P = W/t
Need help ASAP..please help
Answer:
option 3
Explanation:
can i get brainliest
A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect absorber, calculate the radiation pressure on it.
Answer:
[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]
Explanation:
Given that,
A radio wave transmits 38.5 W/m² of power per unit area.
A flat surface of area A is perpendicular to the direction of propagation of the wave.
We need to find the radiation pressure on it. It is given by the formula as follows :
[tex]P=\dfrac{2I}{c}[/tex]
Where
c is speed of light
Putting all the values, we get :
[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]
So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].
1. A stone of mass 0.8 kg is attached to a 0.9 m long string. The string will break if the tension exceeds 60 N. The stone is whirled in a horizontal circle on a frictionless tabletop while the other end of the string remains fixed. What is the maximum speed the stone can attain without breaking the string?
A. 8.22 m/s
B. 7.30 m/s
C. 9.34 m/s
D. 7.76 m/s
2. A highway curve with radius 900 ft is banked so that a car traveling at 55 mph will not skid sideways even in the absence of friction. At what angle should the curve be banked to prevent skidding?
A. 14.6°
B. 18.9°
C. 10.9°
D. 12.7°
3. A button will remain on a horizontal platform rotating at 40 rev/min as long as it is no more than 0.15 m from the axis. How far from the axis can the button be placed without slipping if the platform rotates at 60 rev/min?
A. 0.365 m
B. 0.338 m
C. 0.225 m
D. 0.294 m
4. What is the tension in a cord 10 m long if a mass of 5 kg is attached to it and is being spun around in a circle at a speed of 8 m/s?
A. 67 N
B. 28 N
C. 32 N
D. 50 N
5. A 0.5 kg mass attached to a string 2 m long is whirled around in a horizontal circle at a speed of 5 m/s. What is the centripetal acceleration of the mass?
A. 11.3 m/s2
B. 12.5 m/s2
C. 5.9 m/s2
D. 10.2 m/s2
6. Find the maximum speed with which a car can round a curve that has a radius of 80 m without slipping if the road is unbanked and the coefficient of friction between the road and the tires is 0.81.
A. 44.3 m/s
B. 20.8 m/s
C. 25.2 m/s
D. 30.6 m/s
Answer:
1. A. 8.22. m/s
Explanation:
1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)10N/m
2. What will be the extension of this spring if the load is a) 4N and b) 75 g?
Answer:
1) k = 10 [N/m]
2) a-) x = 0.4 [m]
b) x = 0.075 [m]
Explanation:
To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.
F = k*x
where:
F = force [N] (units of Newtons]
k = spring constant [N/m]
x = distance = 10 [cm] = 0.1 [m]
Now, the weight is equal to the product of the mass by the gravity
W = m*g = F
where:
m = mass = 100 [g] = 0.1 [kg]
g = gravity acceleration = 10 [m/s²]
F = 0.1*10 = 1 [N]
Now clearing k
k = F/x
k = 1/0.1
k = 10 [N/m]
2)
a ) if the force is 4 [N]
clearing x
x = F/k
x = 4/10
x = 0.4 [m]
m = 75 [g] = 0.075 [kg]
W = m*g = F
F = 0.075*10 = 0.75 [N]
x = .75/10
x = 0.075 [m]
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a calculation in a very artificial situation. Suppose a man standing on frictionless ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg, and the mass of the bullet is 10 g. If the bullet leaves the muzzle at a speed of 500 m/s, what is the final speed of the man?
Answer:
Explanation:
m1v1=m2v2
m1=70 kg
m2=10 g=0.01 kg
v2=500 m/s
m1v1=m2v2
v1=m2v2/m1
v1=0.01*500/70
v1=0.07
Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days.
What is the half-life (T1/2) of this isotope?
Answer:half-life (T1/2) of this isotope =
Explanation:
The number of nuclei of any radioactive substance at a given time is expressed by
Nt = N0e⁻kt
Nt=decay of material at a time t, =3110 decays per minute
N=decays at t=0, 8255 decays per minute
k=constant
Nt=N0e−kt
3110= 8255 e⁻k(4.50)
3110/ 8255=e−k(4.50)
0.3767 = e−k(4.50)
In 0.3767 = -k (4.50)
0.976=-4.5k
k=0.976/4.5
=0.2159
Also we know that t 1/2= time that it takes half the original material to decay.it is related to the rate constant by
T₁/₂=ln 2 / k
Therefore half-life (T1/2) of this isotope
T₁/₂=ln 2/0.2159
T₁/₂=3.12 days
Two identical wind-up cars A and B are released. Car B has a 2 kilogram weight strapped to the back of the car. Which will have the greatest average speed towards the end of the motion?
A)Car A
B)They will both have an average speed of zero.
C)They will have the same average speed.
D)Car B
E)There is not enough information to answer.
Answer:
Car A would have a better average speed
Explanation:
added weight to a object that is self propelled will be slower than a identical object with no added weight
David Wetterman drops a 5 kg watermelon from the top of a 30 m building. What is the velocity of the watermelon as it smashes
into the ground (neglecting air resistance)?
-(1)
A)
24.25 m/s
B)
32.45 m/s
C)
60 m/s
D)
588 m/s
Answer:
A. 24.25 m/s
Explanation:
velocity = [tex]\sqrt{2 * g * d}[/tex]
velocity = sqr 2 * 9.8 * 30 = sqr 588 = 24.25 m/s
The velocity of the watermelon as it smashes into the ground will be 24.2 m/s
State the third equation of motion?
The third equation of motion is -
v² - u² = 2aS
Given David Wetterman drops a 5 kg watermelon from the top of a 30 m building.
Height of building [S] = 30 m
Mass of watermelon [M] = 5 Kg
Initial velocity [v] = 0 m/s
acceleration [g] = 9.8 m/s²
Using the third equation of motion -
v² - u² = 2aS
v² = 2aS
v² = 2 x 9.8 x 30
v² = 588
v = 24.2 m/s
Therefore, the velocity of the watermelon as it smashes into the ground will be 24.2 m/s.
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Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) What is the change in the temperature of the gas?________ ? K(b) Find the change in its internal energy.________ ? J(c) Determine the change in pressure.________ ? Pa
Answer:
A) 140 k
b ) 5.22 *10^3 J
c) 2910 Pa
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J
number of moles (n) = 3
A ) calculate the change in temp of the gas
since the volume of gas is constant no work is said to be done
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R
= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
= 140 K
B) Calculate the change in its internal energy
ΔU = Q this is because no work is done
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure
applying ideal gas equation
P = nRT/V
therefore ; Δ P = ( n*R*ΔT/V )
= ( 3 * 8.3144 * 140 ) / 1.20
= 2910 Pa
A) The change in the temperature of the gas is; ΔT = 139.5 K
B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J
C) The change in pressure of the gas is; ΔP = 2899.5 Pa
We are given;
Volume of Monatomic ideal gas; V = 1.2 m³
Amount of heat added; Q = 5.22 × 10³ J
number of moles; n = 3
A) To calculate the change in temperature of the monatomic idea gas, we will use formula;
Q = ³/₂nRΔT
Where R is a constant = 8.314 J/mol.K
ΔT is the change in temperature
Making ΔT the subject of the formula;
ΔT = ²/₃(Q/(nR))
ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)
ΔT = 139.5 K
B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;
ΔU = Q
Thus;
change in internal energy; ΔU = 5.22 × 10³ J
C) The change in pressure will be calculated from the formula;
ΔP = (n*R*ΔT)/V
ΔP = (3 * 8.314 * 139.5)/1.2
ΔP = 2899.5 Pa
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What is magnet made of
Answer:
metals like iron or nickel
Explanation:
Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?
Answer:
Acceleration = 18g
Explanation:
Given the following data;
Initial velocity, u = 26m/s
Final velocity, v = 0
Time = 0.15 secs
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation, we have;
[tex]a = \frac{0 - 26}{0.15}[/tex]
[tex]a = \frac{26}{0.15}[/tex]
Acceleration = 173.33m/s2
To express it in magnitude of g;
Acceleration = 173.33/9.8
Acceleration = 17.7 ≈ 18g
Acceleration = 18g
When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) what is the effective force constant of the springs? in N/m (b) The four people get out of the car and bounce it up and down. What is the frequency of the car's vibration?
Answer:
Explanation:
F=kx
x=F/k
F=2000 kg
x=100 cm=9*10^-3
effective spring constant=k=F/x
k=2000/9*10^-3=2.2*10^-5
now frequency
f=1/2π√k/m
f=1/2*3.14√2.2*10^-5/310
f=1/6.28√7.097*10^-8
f=1/6.28*2.7*10^-4
f=0.16*2.7*10^-4
f=4.32*10^-5
The effective spring constant of the springs is 33755.55 N/m.
The frequency of the car's vibration is 2.07 Hz.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
Weight of the four people: F = 310 × 9.80 N = 3038 Newton.
The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.
Hence, the effective spring constant of the springs: k= force/compression
= 3038 N/0.90 × 10⁻² m
= 33755.55 N/m.
The frequency of the car's vibration is: f = 1/2π√(k/m)
=1/2π√(33755.55/2000)
= 2.07 Hz.
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someone help please
waves disturb ____, but do not transmit it.
a. energy
b. matter
c. sound
d. none of the above
Answer:
b. matter
Explanation:
Waves disturb matter but do not transmit it.
Waves are disturbances that transmit energy from one point to another. Although they cause disturbances, they do not transfer the matters in the medium.
Energy is propagated by a wave. When for example, sound waves are produced, the disturbance is propagated via particle - particle interaction But after the wave train moves, the particles remain.A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.
Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.
The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:
n + (-w) = 0
n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N
The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be
f = 0.67 (51.94 N) ≈ 35 N
so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.
The block starts moving as soon as x = 35 N, at which point f = 35 N.
At any point after the block starts moving, we have
f = 0.48 (51.94 N) ≈ 25 N
so that x = 25 N is the required force to keep the block moving at a constant speed.
As x is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force
Magnitude of normal force ( object at rest ); n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x after object start moving x = 25 NGiven data :
mass of block at rest ( m ) = 5.3 kg
Coefficient of static friction ( μ_s ) =0.67
Coefficient of kinetic friction is ( μ_k ) = 0.48
Horizontal force applied to block = x
First step : magnitude of normal force ( n ) when object is at rest
n = w where w = m*g
n - w = 0
n - ( 5.3 * 9.81 ) = 0 ∴ n = 51.94 N
Second step : Required magnitude of x before the movement of object
F = μ_s * n
F = 0.67 * 51.94 = 34.79 N ≈ 35 N
∴ The object will start moving once F and x = 35 N
Final step : Magnitude of x after object start moving
F = μ_k * n
= 0.48 * 51.94 = 24.93 N ≈ 25 N
∴ object will continue to accelerate at a constant speed once F and x = 25N
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Consider a block sliding down a ramp whose motion is opposed by frictional forces. The total energy of this system is modeled by the equation:
Etotal = 1/2mv^2 + mgh + Ff(f is underscore)d
Which part of the equation represents the amount of energy converted to thermal energy?
A. mg
B. Ffd
C. mgh
D. 1/2 mv^2
Answer:
Energy Flows Quick check answers:
1. Ffd.
2. The kinetic energy decreases, and gravitational potential energy increases.
3. The internal energy of the system increases.
4. KEbox= Etotal-mgh
5. Etotal = 1/2m1(v1)^2+1/2m^2(v2)^2+U
The part of the equation that represents the amount of energy converted to thermal energy is [tex]F_f d[/tex].
The given equation for the total energy of a system;
[tex]E_{total} = \frac{1}{2} mv^2 \ +\ mgh\ + \ \ F_fd[/tex]
The definition of the various terms in the energy equation is given as;
[tex]E_{total}[/tex]: this is the total mechanical energy of the system[tex]\frac{1}{2} mv^2[/tex]: this is the kinetic energy of the system[tex]mgh[/tex]: this is the potential energy of the system[tex]F_f d[/tex]: this is the energy lost due to friction.The energy lost due to friction is equal to the energy converted to thermal energy.
Thus, the part of the equation that represents the amount of energy converted to thermal energy is [tex]F_f d[/tex].
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How many significant figures are in 0.0067?
Answer:
2
Explanation:
there are 2 significant figures in there
A radio station can be heard if the receiver is tuned to a frequency of 6x10^5 Hz. What is the wavelength of the radio waves
5 * 10² m
Explanation:We are Given:
Frequency of the wave = 6 * 10⁵ Hz
Wavelength of the wave:
We know the relation:
c = λν
[where λ (Lambda) is the wavelength ,ν (nu) is the frequency and c is the speed of light ]
3 * 10⁸ m/s= (6 * 10⁵ )* λ [replacing known values]
λ = [tex]\frac{3 * 10^{8}}{6 * 10^{5}}[/tex] [dividing both sides by 6 * 10⁵]
λ = 1/2 * 10³
λ = 1/2 * 10 * 10² [10³ can be rewritten as 10 * 10²]
λ = 5 * 10² m
Therefore, the wavelength of the wave is 5 * 10² m
A household refrigerator consumes electrical energy at the rate of 200 W. lf electricity costs 5 k per kWh, calculate the cost of operating the appliance for 30 days
Answer:
= 720000 [k]
Explanation:
The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.
[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]
= 720000 [k]