Answer:
Heat transfer is an exchange of thermal energy between two objects. The rate of heat transfer depends upon the temperatures of each entity and the medium through which the thermal energy is being transferred. In cooking, heat transfer refers to heating your food items through a cooking appliance, such as a stove, fryer, microwave, or oven.
Explanation:
I hope it helps
Answer:
Convection
Convection is heat transfer by mass motion of a fluid such as air or water when the heated fluid is caused to move away from the source of heat, carrying energy with it. Convection above a hot surface occurs because hot air expands, becomes less dense, and rises (see Ideal Gas Law).
Explanation:
hope this helps:-)
A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe very large accelerations.
Part A) What is the acceleration at the end of a test tubethat is 10 cm from the axis of rotation?
Part B) For comparison, what is the magnitude of theacceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hardfloor?
Answer:
A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²
Explanation:
Part A) The relation of the test tube is centripetal
a_c = v² / r
the angular and linear variables are related
v = w r
we substitute
a_c = w² r
let's reduce the magnitudes to the SI system
w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s
r = 1 cm (1 m / 100 cm) = 0.10 m
let's calculate
a_c = 418.88² 0.1
a_c = 1.75 10⁴ m / s²
part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor
as part of rest the initial velocity is zero and on the floor the height is zero
v² = v₀² - 2g (y- y₀)
v² = 0 - 2 9.8 (0 + 1)
v =√19.6
v = -4.427 m / s
now let's look for the applied steel to stop the test tube
v_f = v + a t
0 = v + at
a = -v / t
a = 4.427 / 0.001
a = 4.43 10³ m / s²
In an effort to be the star of the half-time show, the majorette twirls a highly unusual baton made up of four mases fastened to the ends of light rods. Each rod is 1.0 m lone. Find the moment of inertia of the system about an axis perpendicular to the page and passing through the point where the rods cross.
Answer:
"0.25 kg-m²" is the appropriate answer.
Explanation:
The diagram of the question is missing. Find the attachment of the diagram below.
According to the diagram, the values are:
m₁ = 0.2
m₂ = 0.3
m₃ = 0.3
m₄ = 0.2
d₁ = d₂ = d₃ = d₄ = 0.5 m
As we know,
The moment of inertia is:
⇒ [tex]I=\Sigma M_id_i^2[/tex]
then,
⇒ [tex]I=m_1d_1^2+m_2d_2^2+m_3d_3^2+m_4d_4^2[/tex]
⇒ [tex]=d^2(m_1+m_2+m_3+m_4)[/tex]
On substituting the values, we get
⇒ [tex]=0.5^2\times (0.2+0.3+0.3+0.2)[/tex]
⇒ [tex]=0.25\times 1[/tex]
⇒ [tex]=0.25 \ Kg-m^2[/tex]
Fossil clues are one of the _____________ clues that support the theory of continental drift.
A. crust B. resource C. climate D. rock
Answer:
a
Explanation:
I think don't get mad if I'm wrong
What is the formula for Ba+2 and F-?
Answer:
BaF2
Explanation:
since you got the valence numbers just do the scissors move where you:
give the F the 2 and give the Ba the 1 so it be like
BaF2 here is the chemical compound
Build a second circuit with a battery and a light bulb but this time add a switch. Your circuit might look something like the one at right. When a switch is open in a circuit, it means the two ends are disconnected and current cannot flow between them. When a switch is closed in a circuit, it means the two ends are connected and current can flow between them. Play with the switch to check how it affects the flow of current. With the switch closed, compare the brightness of the bulb and the flow of current in this circuit, with that of your first circuit. Did increasing the length of wire in the circuit change the brightness of the bulb or the curr
Answer:
the resistance of the wire has no effect on the brightness of the bulb.
Explanation:
Let's apply ohm's law for your light bulb circuit plus wires plus switch
V = I R_{bulb} + I R_ {wire}
the current in a series circuit is constant
V = I (R_{bulb} + R_{wire})
To know the effect of the wires on the brightness of the bulb, we must look for the value of the typical resistance of these elements.
Incandescent bulb
Power 60 W
let's use the power ratio
P = V I = V2 / R
R = V2 / P
the voltage value for this power is V = 120 V
R = 120 2/60
R_bulb = 240 Ω
Resistance of a 14 gauge copper wire (most used), we look for it on the internet
R = 8.45 Ω/ km
in a laboratory circuit approximately 2 m is used, so the resistance of our cable is
R = 8.45 10⁻³ 2
R_wire = 0.0169 Ω
let's buy the two resistors
R_{bulb} = 240
R_{wire} = 0.0169
[tex]\frac{R_{bulb} }{R_{wire} } = \frac{240 }{ 0.0169}[/tex]
[tex]\frac{ R_{bulb} }{ R_{wire} } = 1.4 \ 10^4[/tex]
therefore resistance of the bulb is much greater than that of the wire, therefore almost all the power is dissipated in the bulb.
In summary, the resistance of the wire has no effect on the brightness of the bulb.
Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader
Answer:
7.55 g
Explanation:
Using the relation :
Δt = temperature change = (6° - 0°) = 6°
Q = quantity of heat
C = specific heat capacity = 4190 j/kg/k
1000 J = 1kJ
333 KJ = 333000 j
The quantity of ice that will melt ;
= 0.419 * 6 * 100 / 333000
= 2514000 / 333000
= 7.549 g
The mass of ice that will melt :
2.514 / 0.333
= 7.549 g
A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding
Answer:
0.6
Explanation:
Given that :
Radius, R = 7m
Period, T = 6.9s
The Coefficient of static friction, μs can be obtained using the relation :
μs = v² / 2gR
Recall, v = 2πR/T
μs becomes ;
μs = (2πR/T)² / 2gR
μs = (4π²R² / T²) ÷ 2gR
μs = (4π²R² / T²) * 1/ 2gR
μs = 4π²R / T²g
μs = 4π²*7 / 6.9^2 * 9.8
μs = 28π² / 466.578
μs = 276.34892 / 466.578
μs = 0.5922887
μs = 0.6
Which one is it? Help ASAP
Answer:
extreme heat, because no physical damage can demagnetize a magnet
Explanation:
Answer:
the 3rd one
Explanation:
Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed
Answer:
The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg
Explanation:
The given percentage by weight of protein solids in raspberries = 10 weight%
The ratio of sugar to raspberries in ja-m = 45:55
The mass of the mixture after boiling = 0.4 weight fraction water
Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry
The mass of raspberry, r = 1 kg
The percentage by weight of water in raspberry = 90 weight %
The mass of water in 1 kg of raspberry = 90/100 × 1 kg = 0.9 kg
The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55
∴ s = 1 kg × 55/45 = 11/9 kg
The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg
The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction
Let 'w' represent the mass of water boiled off, we have;
(0.9 - w)/(20/9 - w) = 0.4
(0.9 - w) = 0.4 × (20/9 - w)
0.9 - w = 8/9 - 0.4·w
9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w
(81 - 80)/(90) = (6/10)·w
1/90 = (6/10)·w
w = ((10/6) × 1/90) = 1/54
w = 1/54
The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg
Which graph represents the relationship between the magnitude of the gravitational force exerted by earth on a spacecraft the distance between the center of the spacecraft the center of earth
Answer:
B as distance increase force decrease, but it is not a linear relationship.
A hot air balloon is rising at a speed of 10 km/hr. One hour later, the balloon
is still rising at 10 km/hr. What is its acceleration?
The Sun is divided into three regions.
True оr False?
Answer:
false I think
Explanation:
hope that help
so it's not divided in 3 regions
A spring in a toy gun has a spring constant of 10 N/m and can be compressed 4 cm.
It is then used to shoot a 1 g ball out of the gun. Find the velocity of the ball as it
leaves the gun
why do the stars rotate
Answer:
Angular momentum
Explanation:
Stars are formed as a result of a collapse of a low-temperature cloud of gas and dust. During the colapse conservation of angular momentum causes any small net rotation of the cloud to increase thus forcing the material into rotating
PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE
Which best explains why species living in Australia are found nowhere else on Earth? This is an example of Geologic Evolution.
A.
Australia has an ecosystem different from any other area on Earth.
B.
Humans have genetically altered many Australian species in laboratories.
C.
Australian species were genetically altered after a comet hit the landmass.
D.
Australia separated from other continents and species there evolved independently.
Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 lb and was traveling eastward. Car B weighs 11251125 lb and was traveling westward at 42.042.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.517.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.7500.750 . How fast (in miles per hour) was car A traveling just before the collision
Answer:
v = 28.98 ft / s
Explanation:
For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision
In the exercise they indicate the weight of each car
Wₐ = 1500 lb
W_b = 1125 lb
Car B's velocity from v_b = 42.0 mph westward, car A travels east
let's find the mass of the vehicles
W = mg
m = W / g
mₐ = Wₐ / g
m_b = W_b / g
mₐ = 1500/32 = 46.875 slug
m_b = 125/32 = 35,156 slug
Let's reduce to the english system
v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s
We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved
we assume the direction to the east (right) positive
initial instant. Before the crash
p₀ = mₐ v₀ₐ - m_b v_{ob}
final instant. Right after the crash
p_f = (mₐ + m_b) v
the moment is preserved
p₀ = p_f
mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v
v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]
we substitute the values
v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]
v = 0.559 v₀ₐ - 26.40 (1)
Now as the two vehicles united we can use the relationship between work and kinetic energy
the total mass is
M = mₐ + m_b
M = 46,875 + 35,156 = 82,031 slug
starting point. Jsto after the crash
K₀ = ½ M v²
final point. When they stop
K_f = 0
The work is
W = - fr x
the negative sign is because the friction forces are always opposite to the displacement
Let's write Newton's second law
Axis y
N-W = 0
N = W
the friction force has the expression
fr = μ N
we substitute
-μ W x = Kf - Ko
-μ W x = 0 - ½ (W / g) v²
v² = 2 μ g x
v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5
v = 28.98 ft / s
What is your hypothesis (or hypotheses) for this experiment?
(about Thermal Energy Transfer)
Answer:
I hypothesis that the motion involving the balls in the experiment were moving to create data.
Explanation:
I hope this helps!
Activities:
1. Name the instrument that is used to measure Air Pressure.
2.Explain what is Cyclone and Anticyclone
Answer: barometer.
A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.
Jim and Sally both do identical jobs. Jim works quickly while Sally works slowly. Which of the following is true?
A) Sally uses more energy.
B) Jim uses more energy.
C) Jim uses more power.
D) Sally uses more power.
Match the following:
machinery part :nickel or chromium
ornamentation and decoration pieces :silver and gold
processed food :tin coated iron can
bridges and automobiles :zinc metal
distilled water:bad conductor
Answer:
iron metal :chromium
machinery part :nickel or chromium
ornamentation and decoration pieces :silver and gold
processed food :tin coated iron can
bridges and automobiles :zinc metal
distilled water:bad conductor
Explanation:
A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answer:
speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
Explanation:
Given:
mass of truck M = 1370 kg
speed of truck = 12.0 m/s
mass of car m = 593 kg
collision is elastic therefore,
Applying law of momentum conservation we have
momentum before collision = momentum after collision
1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)
Also for a collision to be elastic,
velocity of approach = velocity of separation
12 -0 = v2-v1 ....(ii)
using (i) and (ii) we have
So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
A ball is dropped off the side of a bridge,
After 1.55 S, how far has it fallen?
(Unit=m)
Answer:
Distance S = 11.77 m (Approx.)
Explanation:
Given:
Time t = 1.55 Second
Gravity acceleration = 9.8 m/s²
Find:
Distance S
Computation:
S = ut + (1/2)(g)(t)²
S = (0)(1.55) + (1/2)(9.8)(1.55)²
S = (0)(1.55) + (1/2)(9.8)(1.55)²
Distance S = 11.77 m (Approx.)
Part D Here is one last question as a final check on your understanding of your work for this problem, looking at this problem as an example of the Conservation of Energy. The action in this problem begins at location A , with the block resting against the uncompressed spring. The action ends at location B, with the block moving up the ramp at a measured speed of 7.35 m/s . From A to B, what has been the work done by non-conservative forces, and what has been the change in the mechanical energy of the block-Earth system (the ramp is a part of the Earth)
Answer:
The answer is "39.95 J".
Explanation:
Please find the complete question in the attached file.
[tex]\to W_{AC}=(\mu \ m \ g \ \cos \theta ) d[/tex]
[tex]=(0.45 \times 1.60 \times 9.8 \times \cos 26^{\circ}) 6.30 \\\\=(7.056 \times \cos 26^{\circ}) 6.30 \\\\=6.34189079\times 6.30\\\\=39.95 \ J\\\\[/tex]
[tex]\therefore \\\\\bold{\Delta E =39.95 \ J}[/tex]
he inductance of a tuning circuit of an AM radio is 4 mH. Find the capacitance of the circuit required for reception at 1200 kHz.
Answer:
4.4pF
Explanation:
the capacitance of the circuit required for reception is given:
wL = [tex]\frac{1}{wC}[/tex]
w = 2π [tex]f[/tex]
Using both equation
Capacitance is given
C = 1 - 4π2 f2 L
1- 4×9.8969×144×10 10 ×0.004
=4.4pF
Suppose you are watching an ice-hockey game with some classmates. As the puck is sliding across the ice at a relatively constant speed, it passes close to a player, who hits it (with his stick) in the same direction as it is already moving. After the hit, you notice that the puck is moving much faster than it was before the hit. Your classmates propose the following three explanations for why the puck is moving faster after the hit than before. Whose reasoning do you agree with?
Student A: "The force of the stick was transferred to the puck during the hit. After the hit, the puck has more force, so it moves faster."
Student B: "The puck still has the force that started it moving, which is what keeps it moving at a constant speed. When the player hit it, he adds to this force, which makes the speed of the puck increase. After the hit, this extra force is gone, so the puck stops increasing its speed, but the original force is still there so now it moves at a faster constant speed."
Student C: "While the stick is in contact, it applies a force to the puck that makes its speed increase. As soon is contact is lost, this force disappears, so the speed of the puck stops increasing."
Answer:
the answer, the correct one is C
Explanation:
Let's propose the solution of this problem to know which explanation is correct, when the concha stick with the disc is an impulse exercise
I = ΔP
∫ F dt = pf-po
∫ F dt = m v_f - m v₀
Therefore, during the time that the contact lasts, a force is applied to the disk, which causes that if the amount of movement increases and therefore its speed increases, when the constant ceases the forces are reduced to zero and the disk no longer changes its momentum following with constant velocity.
When reviewing the answer, the correct one is C
A scientist analyzes the light from a distant galaxy and finds that it is shifted to the longer wavelength of the electromagnetic spectrum. What does this data help to study?
1) the color of the galaxy
2) the distance of the galaxy from Earth
3) the existence of life on any planet in the galaxy
4) the study of the amount of light scattered by dust in space
Answer:
Option 2
Explanation:
As per the relation between the distance of the galaxy and shifting of the light of the galaxy towards any specific wavelength of the electromagnetic spectrum, a galaxy at great distance shifts more towards the red spectra that has the highest wavelength.
Thus, this observation give details about the distance of the galaxy from earth.
Answer:
b
Explanation:
Car X is travelling at 30m/s north. Its driver looks at car Y approaching on another road and he estimates it is moving at 15m/s south-west relative to his car. Calculate the velocity of car Y relative to the ground.
Answer: 22.1 m/s
Explanation:
The velocity of Car traveling 30 m/s towards the north
In vector form it is
[tex]v_x=30\hat{j}[/tex]
The velocity of car Y w.r.t X is
[tex]\Rightarrow v_{yx}=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}][/tex]
Solving this
[tex]\Rightarrow v_{yx}=v_y-v_x\\\Rightarrow v_y=v_{yx}+v_x[/tex]
putting values
[tex]\Rightarrow v_y=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}]+30\hat{j}[/tex]
[tex]\Rightarrow v_y=-10.606\hat{i}+19.39\hat{j}[/tex]
absolute velocity relative to ground is
[tex]\left | v_y\right |=\sqrt{(-10.606)^2+(19.39)^2}\\\left | v_y\right |=22.101\ m/s[/tex]
2.Test the age of your eyes. a.Hold a pencil or ballpoint pen vertically at arm's length. b.Close your left eye and focus on the tip. c.Quickly bring the pencil closer to your eye until it is out of focus. d.Have your partner measure the distance between your eye and the pencil. e.Repeat for both eyes. f.Try it with and without glasses (if you wear glasses). Age of your Eyes Cm91013185083
Answer:
See Explanation
Explanation:
Given
Steps: a - f
Table
[tex]\begin{array}{ccccccc}{cm} & {9} & {10} & {13} & {18} & {50} & {83} \ \\ {Age} & {10} & {20} & {30} & {40} & {50} & {60} \ \end{array}[/tex]
Note that: The question is a practical question and the result may differ base on individuals and environment.
So, I will pick up the question from how to determine the age of the eye after the distance between the eyes and the pencil has been established
In my case, the measurement is:
[tex]Length= 10.4[/tex]
Approximate
[tex]Length= 10[/tex]
From the above table, the corresponding age to 10cm is:
[tex]Age = 20cm[/tex]
If in your measurement, the length is approximately (for example):
[tex]Length = 9cm[/tex]
The age will be:
[tex]Age = 10[/tex]
A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 140 m above the launch point. (a) Calculate the horizontal component of its velocity. (b) Calculate the vertical component of its velocity just after launch. (c) At one instant during its flight the vertical component of its velocity is found to be 65 m/s. At that time, how far is it above or below the launch point
Answer:
a). 53.78 m/s
b) 52.38 m/s
c) -75.58 m
Explanation:
See attachment for calculation
In the c part, The negative distance is telling us that the project went below the lunch point.
plz help me with my career!!!
part one...
Answer:
#1 Yes
Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.
Question 1: Crops.
Question 2: Diagnostic Services.
Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.
Question 4: A bachelor's degree in energy research.
Question 5: Environmental Resources.
If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.