Answer:
The description of that same situation has been listed throughout the explanation segment below.
Explanation:
When another huge box or container containing your new machine or device sits on someone's pick-up truck's bed, the third low portion of the operation response force. This same friction force of the box mostly on the truck bed as well as the friction force including its truck bed on either the box from either the immune response pair.So that the above seems to be the right answer.
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.
Convert from scientific notation to standard form
9.512 x 10-8
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.
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 airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal
Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is [tex]H = 318.5 \ m[/tex]
Explanation:
From the question we are told that
The speed at which the plane takes off is [tex]u = 49 \ m/s[/tex]
The angle at which it takes off is [tex]\theta = 30 ^o[/tex]
The time taken is [tex]t = 13 s[/tex]
The vertical distance traveled is mathematically represented as
[tex]H = u sin \theta t[/tex]
Substituting values
[tex]H = (49) * sin (30) *13[/tex]
[tex]H = 318.5 \ m[/tex]
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
Learn more about force here: https://brainly.com/question/3398162
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
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].
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:
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.
For more details about momentum check these:
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Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine their mass by measuring the period of oscillation when sitting in a chair connected to a spring. Suppose a chair is connected to a spring with a spring constant of 600 N/m. If the empty chair oscillates with a period of 0.9s, what is the mass of an astronaut who oscillates with a period of 2.0 s while sitting in the chair
Answer:
ma = 48.48kg
Explanation:
To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:
[tex]T=2\pi\sqrt{\frac{m_c}{k}}[/tex] (1)
mc: mass of the chair
k: spring constant = 600N/m
T: period of oscillation of the chair = 0.9s
You solve the equation (1) for mc, and then you replace the values of the other parameters:
[tex]m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg[/tex] (2)
Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:
T': period of chair when the astronaut is sitting = 2.0s
M: mass of the astronaut plus mass of the chair = ?
[tex]T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg[/tex] (3)
Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :
[tex]m_a=M-m_c=60.79kg-12.31kg=48.48kg[/tex]
The mass of the astronaut is 48.48 kg
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.
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.
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 .
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
At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.
A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.
Answer:
Explanation:
For a convex mirror, the value of its image distance and its focal length are negative.
using the mirror formula 1/f = 1/u+1/v
f is the focal length = Radius of curvature/2 = 0.560/2
f= 0.28m
u is the object distance = 10.6m
v is the position of the image = ?
On substitution;
1/0.28 = 1/10.6 + 1/-v
3.57 = 0.094 - 1/v
3.57 - 0.094 = -1/v
3.476 = -1/v
v = -1/3.476
v = -0.2877m
B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.
C) Magnification = Image distance/object distance
Magnification = 0.2877/10.6
Magnification = 0.0271
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section), and the power fit of your Trendline equation,calculate the drag coefficient. Solve for it first (see video) and then plug in the values.
Answer:
The drag coefficient is [tex]D_z = 1.30512[/tex]
Explanation:
From the question we are told that
The density of air is [tex]\rho_a = 1.21 \ kg/m^3[/tex]
The diameter of bottom part is [tex]d = 0.15 \ m[/tex]
The power trend-line equation is mathematically represented as
[tex]F_{\alpha } = 0.9226 * v^{0.5737}[/tex]
let assume that the velocity is 20 m/s
Then
[tex]F_{\alpha } = 0.9226 * 20^{0.5737}[/tex]
[tex]F_{\alpha } = 5.1453 \ N[/tex]
The drag coefficient is mathematically represented as
[tex]D_z = \frac{2 F_{\alpha } }{A \rho v^2 }[/tex]
Where
[tex]F_{\alpha }[/tex] is the drag force
[tex]\rho[/tex] is the density of the fluid
[tex]v[/tex] is the flow velocity
A is the area which mathematically evaluated as
[tex]A = \pi r^2 = \pi \frac{d^2}{4}[/tex]
substituting values
[tex]A = 3.142 * \frac{(0.15)^2}{4}[/tex]
[tex]A = 0.0176 \ m^2[/tex]
Then
[tex]D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }[/tex]
[tex]D_z = 1.30512[/tex]
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 .
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 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 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
Carbon is added to iron to make steel. Steel is stronger than either carbon or iron by itself.
What does this example show?
Answer:
This example shows that alloys are stronger than either of it's parent materials by themselves.
Explanation:
Since carbon is added to iron to make steel, it means steel is an alloy of iron and carbon.
This is because an alloy is a mixture of two or more elements, where at least one element is a metal.
Now, steel is stronger than either carbon or or iron by itself because Steel contains atoms of other elements including carbon and iron. These atoms have different sizes to iron carbon atoms, so they distort the layers of atoms in the pure iron and carbon. This means that a greater force is required for the layers to slide over each other in steel, so steel is harder than pure iron.
(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]
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.
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.
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]
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
To learn more about the karst topography, refer to the link;
https://brainly.com/question/1167881
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Answer: A) Caves
Explanation:
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
