Answer:
10 Sig Figs
Explanation:
Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.
state Ohm`s law as applied in electricity
Answer:
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. When spelled out, it means voltage = current x resistance, or volts = amps x ohms, or V = A x Ω.
A merry-go-round on a playground consists of a horizontal solid disk with a weight of 810 N and a radius of 1.56 m. A child applies a force 49.0 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s
Answer:
Kinetic Energy of the disk = 252 J
Explanation:
weight of disk = 810 N
radius = 1.56 m
applied force = 49 N
time = 2.95 s
kinetic energy of disk = ?
first, we find the mass of the disk
mass of disk = weight/acceleration due to gravity(9.81 m/s^2) = 810/9.81 m/s^2
mass of disk = 82.57 kg
torque on the disk = force x radius = 49 x 1.56 = 76.44 N-m
moment of inertia I = m[tex]r^{2}[/tex] = 82.57 x [tex]1.56^{2}[/tex] = 200.9 kg-[tex]m^{2}[/tex]
recall that
Torque T = Iα
where α = angular acceleration
76.44 = 200.9α
α = 76.44/200.9 = 0.38 m/s^2
from the equation of angular motion,
ω = ω' + αt
where ω = final angular speed
ω' = initial angular speed = 0 rad/s since disk starts from rest
t = time = 2.95 s
imputing values into the equation, we have
ω = 0 + (0.38 x 2.95)
ω = 1.12 rad/s
kinetic energy of the disk = I[tex]w^{2}[/tex]
KE = 200.9 x [tex]1.12^{2}[/tex]
Kinetic Energy of the disk = 252 J
A single loop of wire with an area of 0.0820 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.220 T/s .
Required:
a. What emf is induced in this loop?
b. If the loop has a resistance of 0.600Ω, find the current induced in the loop.
Answer:
a) emf = 0.01804 V
b) I = 0.03 A
Explanation:
a) The emf is calculated by using the following formula:
[tex]|emf|=|\frac{d\Phi_B}{dt}|=|\frac{d(A\cdot B)}{dt}|[/tex] [tex]=A|\frac{dB}{dt}|[/tex]
A: area of the loop = 0.0820m^2
B: magnitude of the magnetic field
dB/dt: change of the magnetic field, in time: 0.220 T/s
Where ФB is the magnetic flux, the surface vector and magnetic vector are perpendicular between them, and the area A is constant.
You replace the values of A and dB/dt in the equation (1):
[tex]|emf|=(0.082m^2)(0.220T/s)=0.01804V[/tex]
b) The current in the loop is:
[tex]I=\frac{emf}{R}[/tex]
R: resistance of the loop = 0.600Ω
[tex]I=\frac{0.01804V}{0.600\Omega}=0.03A=30mA[/tex]
a. The emf induced in this loop is 18.04mV.
b. The current induced in the loop is 30.06mA.
a. We know that,
[tex]flux(\phi)=B*A[/tex]
Where B is magnetic field and A is the area.
[tex]emf=\frac{d\phi}{dt}=A*\frac{dB}{dt}[/tex]
Given that, Area , [tex]A=0.0820m^{2},B=3.80T,\frac{dB}{dt}=0.220T/s[/tex]
Substituting all values in above equation.
[tex]emf=0.0820*0.220=0.01804V=18.04mV[/tex]
b. Resistance, [tex]R=0.600ohm[/tex]
Current induced in the loop is,
[tex]I=\frac{emf}{R}=18.04/0.6=30.06mA[/tex]
Hence, the emf induced in this loop is 18.04mV.
The current induced in the loop is 30.06mA.
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A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outstretched hand catches the keys 1.60 s later. (Take upward as the positive direction. Indicate the direction with the sign of your answer.)With what initial velocity were the keys thrown?
Answer:
[tex]v_{i}=10.10 m/s[/tex]
Explanation:
The equation of the position is:
[tex]y=y_{i}+v_{i}t-0.5gt^{2}[/tex]
Where:
v(i) is the initial velocity
The initial position y(i) will be zero and the final position y = 3.60 m.
So, we just need to solve this equation for v(i).
[tex]v_{i}=\frac{y+0.5gt^{2}}{t}[/tex]
[tex]v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}[/tex]
[tex]v_{i}=10.10 m/s[/tex]
Therefore, the initial velocity is 10.10 m/s upwards.
I hope it helps you!
A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring
Answer:
a) The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex], b) The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].
Explanation:
This question is incomplete and complete version will be presented herein:
A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.
a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)
[tex]m\cdot g = k\cdot \Delta x[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]\Delta x[/tex] - Spring linear deformation, measured in meters.
Now, the spring constant is cleared in this equation and outcome is computed: ([tex]m = 0.573\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta x = 0.198\,m[/tex])
[tex]k = \frac{m\cdot g}{\Delta x}[/tex]
[tex]k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}[/tex]
[tex]k = 28.381\,\frac{N}{m}[/tex]
The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex].
b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Given that [tex]k = 28.381\,\frac{N}{m}[/tex] and [tex]m = 0.573\,kg[/tex], the angular frequency, measured in radians per second, of the block is:
[tex]\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }[/tex]
[tex]\omega = 7.038\,\frac{rad}{s}[/tex]
The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].
Tech A says that as engines gain miles, the spark plug gap increases, which raises the ignition system’s available voltage. Tech B says that misfire occurs when required voltage is higher than available voltage. Who is correct? Group of answer choices
Answer: Tech A is correct
Explanation:
Every vehicle has ignition system and without this system,it will not work. The battery of everything vehicle contain energy that start the vehicle and ignore it to start working. Electrical current move from the vehicle's battery and get to the induction coil, the induction coil increases the voltage in it so that the plug will be ignited. The spark plugs produce fire. The spark plug is connected to the ignition system. Once voltage is produced from the induction coil, electrical impulses move from induction coil to insulated plug wires. The spark plug need a very high voltage from the small voltage battery. Once the high voltage exceed the dielectric strength of the gases, spark jump the gap between the plug's fire end.
Nowdothesameproblemwiththepivotatthe toes. A Ballet dancer puts all her weight on the toes of one foot. If her mass is 60 kg, what is the force that has to be exerted by her leg muscle to hold that pose? Assume the pivot is at the toes.
Answer:
The force is [tex]F = 2400 \ N[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The mass of the dancer is [tex]m_d = 60 \ kg[/tex]
From the diagram the
The first distance is [tex]l_1 = 20 \ cm[/tex]
The second distance is [tex]l_2 = 5 \ cm[/tex]
At equilibrium the moment about the center of the dancers feet is mathematically represented as
[tex]F * l_2 - (mg* l_1)[/tex]
Where [tex]g= 10 \ m/s^2[/tex]
substituting values
[tex]F * 5 - (60* 9.8 * 20)[/tex]
=> [tex]F = \frac{60 * 10 * 30}{5}[/tex]
=> [tex]F = 2400 \ N[/tex]
A 2-kilogram toy car is traveling forward at 1 meter per second when it is hit in the rear by a 3-kilogram toy truck that was traveling at 3 meters per second just before impact. If the two toys stick together, their speed immediately after the collision is
Answer:
v = 1.4 m/s
Explanation:
This problem is about an inelastic collision. The total momentum before the collision is equal to total momentum after (because of the conservation of momentum law):
[tex]m_1v_1-m_2v_2=(m_1+m_2)v[/tex] (1)
m1: mass of the toy car = 2 kg
m2: mass of the toy truck = 3 kg
v1: speed of the toy car = 1 m/s
v2: speed of the truck car = 3 m/s
v: speed of both car and truck after the collision = ?
In the equation (1) the negative sign of m2v2 is because of the opposite direction of the toy truck respect to the toy car.
You solve the equation (1) for v, and you replace the values of all variables involved:
[tex]v=\frac{m_1v_1-m_2v_2}{m_1+m_2}\\\\v=\frac{(2kg)(1m/s)-(3kg)(3m/s)}{2kg+3kg}=-1.4\frac{m}{s}[/tex]
this velocity is negative, then, the direction of motion of both car and truck is in the direction of the truck
Hence, the speed of both car and truck toys is 1.4 m/s
The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one of the following frequencies will NOT resonate in the same pipe
a. 1800 Hz
b. 1000 Hz
c. 1400 Hz
d. 600 Hz
e. 400 Hz
Answer:
e. 400 Hz
Explanation:
In closed organ pipe, only odd harmonics of fundamental note is possible .
The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .
200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc
600 Hz , 1000 Hz , 1400 Hz , 1800 Hz .
Frequency not possible is 400 Hz .
The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.
a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
1.been both -ve charged or both +be charged particles
2. 3.52mC
Explanation:
For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.
Now the Force sustain by the extended string is
F = Ke;
Where K is the force constant of the string, 320 N/m
e is the extension,0.033 m
F = 320 × 0.033 =10.56N
2.But according to columns law of charge;
F = kQ1 Q2
But Q1=Q2{ since the charge are of the same magnitude}.
Hence F = KQ^2
Where K is columns constant =9×10^9F/m
Hence Q=√F/K
Q= √10.56/9×10^9
=3.52×10^-3C
= 3.52mC
Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 16 m/s at an angle 32 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3) What is the maximum height the ball goes above the ground?
4) What is the distance between the two girls?
5) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)
Answer:
Explanation:
1. [tex]V_{x}[/tex] = [tex]V_{0}[/tex] * cos[tex]\alpha[/tex] ⇒ 16*cos32 ≈ 13.6 m/s (13.56)
2. [tex]V_{y}[/tex] = [tex]V_{0}[/tex] * sin[tex]\alpha[/tex] ⇒ 16* sin32 ≈ 9.4 m/s
3. [tex]y_{max}[/tex] = [tex]\frac{v_{0}^2*sin^2\alpha}{2g}[/tex]= [tex]\frac{16^2*sin^232}{2*9.8}[/tex] (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)
[tex]y_{max}[/tex] ≈ 3.6677+1.5 ≈ 5.2m
4. [tex]x_{max}[/tex] = [tex]\frac{v_{0}^2*sin(2\alpha)}{g}[/tex]=[tex]\frac{16^2*sin(2*32)}{9.8}[/tex] ≈ 23.5m (23.47)
5. -
answer 4 could be wrong, not certain about that one and i don't know 5
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 28.0 m/s. Then the vehicle moves for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the ride-sharing car in motion (in s)?
(b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s.)
Answer:
Explanation:
Time taken to accelerate to 28 m /s
= 28 / 2 = 14 s
a ) Total length of time in motion
= 14 + 41 + 5
= 60 s .
b )
Distance covered while accelerating
s = ut + 1/2 at²
= 0 + .5 x 2 x 14²
= 196 m .
Distance covered while moving in uniform motion
= 28 x 41
= 1148 m
distance covered while decelerating
v = u - at
0 = 28 - a x 5
a = 5.6 m / s²
v² = u² - 2 a s
0 = 28² - 2 x 5.6 x s
s = 28² / 2 x 5.6
= 70 m .
Total distance covered
= 196 + 1148 + 70
= 1414 m
total time taken = 60 s
average velocity
= 1414 / 60
= 23.56 m /s .
What is the period of a wave if the frequency is? 5 Hz
Answer: If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s.
still really need help with these three questions!!
Explanation:
2. No, not always. Normal force is equal to force of gravity only when there's no acceleration in the vertical direction.
For example, when you stand in an elevator that's not moving, or moving at constant speed, then the normal force equals your weight. But when the elevator accelerates upward, the normal force increases (making you feel heavier). And when the elevator slows down, the normal force decreases (making you feel lighter).
3. Yes, it is possible for an object to be moving eastward and experience a net force westward. An example is a car applying the brakes.
4. Friction force allows you to walk. When you push against the floor, the floor's friction pushes back, as Newton's third law says.
If you try to walk on a slippery surface like ice, you won't be able to push against the ice, and the ice won't push back.
Q) Suppose, you are in a sporting event. You notice that everyone stands up when it’s his turn,
creating a wave that moves through the crowd and they sit back down again after a while. This wave
move around the stadium without moving the people around it. Considering this situation, justify
your answer about nature of wave.
Answer:
The nature of the wave formed is a transverse progressive wave.
Explanation:
A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.
A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.
The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.
What is the momentum of an 8kg bowling ball rolling at 2m/s
Answer:
16kg m/s
Explanation:
P=mv
8 times 2=16kg m/s
Answer:
The momentum of moving body is calculated by
p= mv
In this question m= 8kg
v= 2m/s
so p = 8*2 = 16 kg m/s.
"Mass in motion" can be used to describe momentum. Mass exists in all things. Therefore, if an object is moving, it has momentum—its mass is moving. There are two factors that determine an object's momentum level: how much and how quickly the objects are moving.
Mass and velocity are two variables that affect momentum. An object's momentum can be expressed mathematically as the product of its mass multiply by its velocity.
The equation above can be rewritten as p = m • v, where m is the mass and v is the velocity, since momentum is represented by the lower case p in physics. The equation demonstrates that an object's momentum is directly proportional to its mass and velocity.
The quantity momentum is a vector. A vector quantity is a quantity that is fully described by magnitude and direction, as was discussed in a previous unit. Information about the bowling ball's magnitude as well as its direction must be included in order to fully describe the momentum of a 5-kg ball traveling westward at 2 m/s. The ball has a momentum of 10 kg m/s.
Until information about the ball's direction is provided, the ball's momentum cannot be fully described. The direction of the ball's velocity and the direction of the momentum vector are identical. It was mentioned in a previous unit that the velocity vector moves in the same way that an object moves. The bowling ball's momentum can be fully described as 10 kg m/s westward if it is moving westward. The magnitude and direction of an object's momentum can be used to fully describe it as a vector quantity.
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World religions: Shinto
Most Shinto rituals are tied to
A) worshiping the kami.
B) the life-cycle of humans and the seasonal cycles of nature.
C) forgiveness of sins.
D) preparing for the afterlife.
A 2 kg car moving towards the right at 4 m/s collides head on with an 8 kg car moving towards the left at 2 m/s, and they stick together. After the collision, the velocity of the combined bodies is:_____________.
a) 2.4 m/s towards the left.
b) 2.4 m/s towards the right.
c) 0.8 m/s towards the left.
d) 0
e) 0.8 m/s towards the right.
Answer:
correct answer is c
v = -0.8 m / s
Explanation:
This is a problem of quantity of movement, for this we must define a system formed by the two cars, so that the forces during the collision are internal and therefore the quantity of movement is conserved
initial
p₀ = m₁ v₁ - m₂ v₂
final
= (m₁ + m₂) v
We have taken the direction to the right as positive
p₀ =p_{f}
m₁ v₁ - m₂ v₂ = (m₁ + m₂) v
v = (m₁ v₁ - m₂ v₂) / (m₁ + m₂)
we calculate
v = (2 4 - 8 2) / (2 + 8)
v = (8 -16) / 10
v = -0.8 m / s
the negative sign indicates that the set is moving to the left
correct answer is c
Why do some nucleus emit electrons?
Answer:
In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
Explanation:
Hope this helps!
A 2.8 kg block slides with a speed of 2.4 m/s on a frictionless horizontal surface until it encounters a spring. Part A If the block compresses the spring 5.6 cm before coming to rest, what is the force constant of the spring
Answer:
5,142.86Explanation:
The kinetic energy possessed by the block when sliding will be equal to the energy needed to compress the string.
Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²
m = mass of the block (in kg) = 2.8 kg
v = speed of the block (in m/s) = 2.4 m/s
k = force constant of the spring
e = extension (in metres) = 0.056m
Since KE = energy stored in the spring
1/2 mv² = 1/2 ke²
mv² = ke²
2.8(2.4)² = k(0.056)²
16.128 = 0.003136k
k = 16.128/0.003136
k = 5,142.86
The force constant of the spring is 5,236.36
The force that constant of the spring is 5,142.86.
Calculation of the force:The kinetic energy that should be possessed by the block at the time when sliding will be equivalent to the energy required to compress the string.
Here
Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²
m = mass of the block (in kg) = 2.8 kg
v = speed of the block (in m/s) = 2.4 m/s
k = force constant of the spring
e = extension (in metres) = 0.056m
Since KE = energy stored in the spring
So,
1/2 mv² = 1/2 ke²
mv² = ke²
Now
2.8(2.4)² = k(0.056)²
16.128 = 0.003136k
k = 16.128/0.003136
k = 5,142.86
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A student writes down several steps of scientific method. Put the steps in the best order
Answer:
Make a hypothesis, conduct an experiment, Analyze the experimental data..
Which formation is one feature of karst topography?
Sinkholes formation is one feature of karst topography. The top of a cave falls if it develops large enough and its top extends near enough to the surface.
What is karst topography?Karst topography is a type of natural environment formed mostly by chemical weathering by water, resulting in caves, sinkholes, cliffs, and steep-sided hills known as towers.
The top of a cave falls if it develops large enough and its top extends near enough to the surface. Sinkholes are formed as a result of this, and they are one of the most distinguishing aspects of karst terrain.
When water absorbs carbon dioxide from the atmosphere and ground, it becomes carbonic acid.
Hence, sinkholes formation is one feature of karst topography
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Answer: A) Caves
Explanation:
(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is 9.00¢/(kW · h) and the specific heat for water is 4184 J/(kg · °C). $ 67 Incorrect: Your answer is incorrect. How much heat is needed to raise the temperature of m kg of a substance? How many joules are in 1 kWh? (b) What current was used by the 220 V AC electric heater, if this took 3.45 h? 88.2 Correct: Your answer is correct. A
Answer:
a) [tex]E = 6.024\,USD[/tex], For m kilograms, it is 4184m J., 3600000 joules, b) [tex]i = 88.200\,A[/tex]
Explanation:
a) The amount of heat needed to warm water is given by the following expression:
[tex]Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})[/tex]
Where:
[tex]m_{w}[/tex] - Mass of water, measured in kilograms.
[tex]c_{w}[/tex] - Specific heat of water, measured in [tex]\frac{J}{kg\cdot ^{\circ}C}[/tex].
[tex]T_{f}[/tex], [tex]T_{i}[/tex] - Initial and final temperatures, measured in [tex]^{\circ}C[/tex].
Then,
[tex]Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)[/tex]
[tex]Q_{needed} = 180748800\,J[/tex]
The energy needed in kilowatt-hours is:
[tex]Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)[/tex]
[tex]Q_{needed} = 50.208\,kWh[/tex]
The electric energy required to heat up the water is:
[tex]E = \frac{50.208\,kWh}{0.75}[/tex]
[tex]E = 66.944\,kWh[/tex]
Lastly, the cost of heating a hot tub is: (USD - US dollars)
[tex]E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)[/tex]
[tex]E = 6.024\,USD[/tex]
The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.
b) The current required for the electric heater is:
[tex]i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}[/tex]
[tex]i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]
[tex]i = 88.200\,A[/tex]
Convert from scientific notation to standard form
9.512 x 10-8
why is India called peninsula?
Answer:
India is a peninsula.
Explanation:
India is called as Indian Peninsula because it is surrounded by the Indian ocean on the south, the Arabian sea on the west and the Bay of Bengal on the east.
02
Blue light has a frequency of about 7.5 x 1014 Hz. Calculate the energy, in Joules, of a single photon associated with this frequency
Answer:
49.725× 10^-24J
Explanation:
The Energy associated with a Photon us defined as;
E = hf
Where h is Planck's constant = 6.63× 10^-34m2kg/s
f is the frequency= 7.5 x 10^14 Hz
Hence
E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J
A sulfur dioxide molecule has one sulfur
atom and two oxygen atoms. Which is its
correct chemical formula?
A. SO2
C. S2O2
B. (SO)
D. S20
Answer:
a. SO2
Explanation:
A 148 g ball is dropped from a tree 11.0 m above the ground. With what speed would it hit the ground
Answer:
14.68m/s
Explanation:
As per the question, the data provided is as follows
Mass = M = 0.148 kg
Height = h = 11 m
Initial velocity = U = 0 m/s
Final velocity = V
Gravitational force = F
Mass = M
Based on the above information, the speed that hit to the ground is
As we know that
Work to be done = Change in kinetic energy
[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]
[tex]V^2 - U^2 = 2gh[/tex]
[tex]V^2 - 0 = 2gh[/tex]
[tex]V = \sqrt{2 g h}[/tex]
[tex]= \sqrt{2\times9.8\times11}[/tex]
= 14.68m/s
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 240 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Answer:
The value of F= - 830 N
Since the force is negative, it implies direction of the force applied was due south.
Explanation:
Given data:
Mass = 1000-kg
Distance, d = 240 m
Initial velocity, v1 = 20.0 m/s
Final velocity, v2 = 0 (since the car came to rest after brake was applied)
v2²= v1² + 2ad (using one of the equation of motion)
0= 20² + (2 x a x 240)
0= 400 + 480 a
a = - 400/480
a = - 0.83 m/s²
Then, imputing the value of a into
F = ma
F = 1000 kg x ( - 0.83 m/s²)
F= - 830 N
The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.