Are the coils in an electric heater blackbody radiators? Why?
A. No, because they do not emit different wavelengths of light
B. Yes, because they glow more brightly as they increase in
temperature
O C. No, because they absorb no energy and emit no light
D. Yes, because they reflect most of the energy they receive
Answer:
B. Yes, because they glow more brightly as they increase in
temperature
Explanation:
Did the test !!
The coils in an electric heater blackbody are not radiators. Because, they absorb no energy and emit no light. Thus, the correct option is C.
What is an electric heater?Electric heating is a process in which the electrical energy is converted directly into the heat energy at around about 100% efficiency, using rather the cheap devices. Common applications which are included are space heating, cooking, water heating and industrial processes.
A radiator is the key component of the engine's cooling system. Radiator has the main role to disperse a mix of antifreeze and water molecules throughout the fins, which releases some of the engine's heat energy while taking in the cool air before continuing to pass the rest of the electric engine.
Therefore, the correct option is C.
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Ben runs from a position 3 m west of Main Street to a new position 45 m
west of Main Street in 6 seconds. What is Ben's velocity?
O
A. 7 m/s east
B. 7 m/s west
C. 8 m/s west
O
O
D. 8 m/s east
SUBMIT
Answer:
Option B. 7 m/s West.
Explanation:
From the question given above, the following data were obtained:
Position 1 (P₁) = 3 m West
Position 2 (P₂) = 45 m West
Time (t) = 6 s
Velocity (v) =?
Next, we shall determine the displacement of Ben. This can be obtained as follow:
Position 1 (P₁) = 3 m West
Position 2 (P₂) = 45 m West
Displacement =?
Displacement = P₂ – P₁
Displacement = 45 – 3
Displacement = 42 m West
Finally, we shall determine the velocity of Ben. This can be obtained as follow:
Time (t) = 6 s
Displacement (d) = 42 m West
Velocity (v) =?
v = d/t
v = 42 m West / 6 s
v = 7 m/s West
Thus the velocity of Ben is 7 m/s West
what is happening in terms of matter and energy, while wood burns?
Answer:
The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. For example, when wood burns, the mass of the soot, ashes, and gases, equals the original mass of the charcoal and the oxygen when it first reacted. So the mass of the product equals the mass of the reactant
An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons. Calcuate the acceleration of the object
Answer:
this answer you question
The metal wire is stretched so that its cross-section is still circular but its total length is now 10 meters. What is the resistance of the wire after stretching
What average net force is required to accelerate a 3950 kg bus to a speed of 25m/s in 10.5 s?
Answer:
Explanation:
ACCELERATION = CHANGE IN VELOCITY
TIME TAKEN
∴ a = 25 - 0
10.5
∴ a = 50
21
by using f = ma,
we get , f = 3950 × 50
21
∴ A FORCE OF 9404.8 N IS NEEDED TO ACCELERATE A BUS OF 3950 KG.
Estimate the net force needed to accelerate a 1000-kg car at 3m/s/s from rest.
FILL IN THE BLANKS.
If this force acts on this car for 4 seconds, then the car will increase its speed by ————————- m/s a second each second reaching a final speed of ————————- m/s. The distance traveled during this motion is ——————- meters.
Answer:
First Blank. 30,000
Second Blank. 12,000
Third Blank. 20.0
Explanation:
If a small motor does 520 J of work to move a toy car 260 m, what force does the engine exert on the car?
Answer:
2 NExplanation:
The force engine exert on the car can be found by using the formula
[tex]f = \frac{w}{d} \\ [/tex]
w is the workdone
d is the distance
From the question we have
[tex]f = \frac{520}{260} = 2 \\ [/tex]
We have the final answer as
2 NHope this helps you
What do C-microwaves reveal to scientists?
Answer:
Different wavelengths of microwaves (grouped into "sub-bands") provide different information to scientists. Medium-length (C-band) microwaves penetrate through clouds, dust, smoke, snow, and rain to reveal the Earth's surface.
The device shows the relative humidity at 22°C. What’s the water vapor density if the maximum water vapor in air at this temperature is 20 grams/cubic meter?
A device showing that at 22 degrees Celsius the relative humidity is 58%.
Answer:
The answer would be 11.6 g/m^3
Explanation:
Knowing that vapor's density at 100% humidity is 20g/m^3 we can calculate the answer using the following math:
0.58 * 20 = 11.6 grams per cubic meter
the units can also be written as g/m^3
The density of water vapor is 11.6 grams/cubic meter.
What is relative humidity?The quantity of atmospheric moisture that is present compared to the amount that would be there if the air were saturated is known as relative humidity, and it is stated as a percentage. Relative humidity depends on both moisture content and temperature because the latter quantity is temperature-dependent. The related temperature and dew point for the specified hour are used to calculate relative humidity. it is expresses as %.
Now given that the maximum water vapor in air at this temperature is 20 grams/cubic meter at 22°C, that is, density of vapor at 100% humidity is 20g/[tex]m^3[/tex] at 22°C.
And, The device showing that at 22 degrees Celsius the relative humidity is 58%
So, water vapor density at that moment is = 20g/[tex]m^3[/tex] × 58%
= 11.6 g/[tex]m^3[/tex] .
= 11.6 grams/cubic meter.
Hence, according to the measurement of device, the water vapor density is 11.6 grams/cubic meter.
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Which of the following is NOT a possible type of law that a government could pass to help fight climate change?
1. Making greenhouses gas production illegal
2. Regulating the amount of greenhouse gases businesses can produce
3. Funding research on technologies that are better for the environment than those we currently use
4. Providing tax incentives for businesses and people to take action
Answer:
4
Explanation:
because people could lie about it
You push your friend, whose mass is 54kg, down a hill so she can go sledding. Her acceleration is 3m/s2. Calculate the amount of force you applied to your friend.
Answer:
18 newtons
Explanation:
Divide weight by speed
Anthony is standing on the top of a building 10 m high holding a 7 kg
bowling ball. Mildred dug a 2-m-deep hole next to the base of the
building. What is the gravitational potential energy of the bowling ball
relative to the bottom of the hole?
I reallly need help!!
Answer:
548.8 J
Explanation:
If the angle between two forces increases, the magnitude of their resultant-
A Decreases
C. Remain unchanged
D. Decrease than decreases
Why do astronomers use spectroscopes to analyze light from distant objects?
fill in the blank:
the more people (greater mass) you load into a car, the ___ (more or less) it will accelerate with the same force applied by the engine.
the more people (greater mass) you load into a car, the more it will accelerate with the same force applied by the engine.
HELP ASAP!
Everything on screenshot!
Approximating Earth's atmosphere as a layer of gas 6.8 km thick, with uniform density 1.3 kg/m3 , calculate the total mass of the atmosphere.
Answer:
The total mass of the atmosphere is 4.508 x 10⁹ kg
Explanation:
Given;
thickness of gas layer, h = 6.8 km
density of Earth's atmosphere, ρ = 1.3 kg/m³
the surface area of Earth is given as 5.10 x 10⁸ km²
Volume of the atmosphere is calculated as;
V = Ah
where;
A is the surface area of Earth
h is the thickness of gas layer
V = (5.10 x 10⁸ km²)(6.8 km)
V = 3.468 x 10⁹ km³
The total mass of the atmosphere is calculated as;
mass = density x volume
m = ρV
m = (1.3 kg/m³)(3.468 x 10⁹ km³)
m = 4.508 x 10⁹ kg
Therefore, the total mass of the atmosphere is 4.508 x 10⁹ kg.
what must be photographed with an impression before it is moved or cast ?
A car stopped at a red light, begins to accelerate with a constant acceleration of 7.4 m/s^2 during a
time of 2.4 seconds. If the speed limit of the road is 35 mph (about 15.64 m/s), is the car
considered to be speeding at any point?
Answer:
The car is considered to be speeding
Explanation:
The following data were obtained from the question:
Constant acceleration (a) = 7.4 m/s²
Time (t) = 2.4 s
Speed limit = 15.64 m/s
To know if the car speeding at any point, we shall determine the velocity of the car.
This can be obtained as follow:
Constant acceleration (a) = 7.4 m/s²
Time (t) = 2.4 s
Constant velocity (v) =?
a = v/t
7.4 = v / 2.4
Cross multiply
v = 7.4 × 2.4
v = 17.76 m/s
Thus, the velocity of the car is 17.76 m/s.
Comparing the velocity of the car (i.e 17.76 m/s) with the speed limit (i.e 15.64 m/s), we can see that the velocity of the car is higher than the speed limit. Therefore, the car is considered to be speeding.
Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through flourite. The wavelength of the light from the laser is 632.8 nm in air and the index of refraction of flourite is 1.434.
Answer and Explanation:
The computation is shown below:
As we know that
Frequency = Speed ÷ wavelength
= (3 × 10^8 )÷ (632.8 × 10^-9)
= 4.74 × 10^14 Hz
Speed in flourite is
= Vaccum speed ÷ refraction index
= 3 × 10^8 ÷ 1.434
= 2.092 × 10^8 m/s
wavelength is
= Speed in flourite ÷ frequency
= (3 × 10^8 × 632.8 × 10^-9) ÷ (1.434 × 3 × 10^8)
= 441.28nm
short essay your teacher have provided you
Question 3 (1 point)
here were 2cars racing a quarter mile. The green car had a mass of 1200kg and crossed the finish line with a velocity of 53m/s. The red car had
a mass of 1100Kg and crossed the finish line with a velocity of 55m/s.,Which car had the great momentum?
Black car
Blue car
Green car
Red car
Answer:
The green car had the greatest momentum
Explanation:
Momentum
Momentum can be defined as "mass in motion" and is calculated as the product of the mass of the object by its velocity.
Being v the magnitude of the velocity and m the mass of the object, the momentum is calculated with:
p = mv
The green car had a mass of m1=1200 kg and crossed the finish line at v1=53 m/s. Hence, its momentum was:
p1 = 1200 Kg * 53 m/s = 63600 Kg.m/s
The red car had a mass of m2=1100 kg and crossed the finish line at v2=55 m/s. Hence, its momentum was:
p2 = 1100 Kg * 55 m/s = 60500 Kg.m/s
Since p1 > p2, then the green car had the greatest momentum
Hannah, who has a mass of 65 kg, is riding at 25 m/s in her sports car when she must suddenly swerve off the road to avoid
hitting a dog and slams into the guard rail. Her seatbelt brings her to a stop in 0.4 seconds. What amount of force was
exerted on Hannah?
4000 N
4063 N
650 N
O N
Hannah comes to a rest in t = 0.4 s, so her acceleration is
a = (0 - 25 m/s) / (0.4 s) = -62.5 m/s²
(that is, the acceleration points in the direction opposing her movement forward as she brakes)
Then the force exerted on her by the seatbelt has magnitude
F = (65 kg) (62.5 m/s²) = 4062.5 N ≈ 4063 N
A 420 g soccer ball is kicked into the air with an initial velocity of 30 m/s. How much kinetic energy does the soccer ball have?
Answer:189000J
Explanation:KE=1/2mv^2
1/2(420g)(30m/s)^2
=189000J
the combined focal length of two thin lens is 24 cm and the focal length of one converging lens is 8
cm what is the focal length of the other
Answer: f = -12 cm
Explanation: Combined lenses is an array of simple lenses with a common axis. The combination is useful for correction of optical aberrations which cannot be corrected by simple lenses.
When two lenses are in contact and are thin, focal lengths are related as:
[tex]\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}[/tex]
If there is a distance between the lenses, the focal length will be:
[tex]\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}} -\frac{d}{f_{1}f_{2}}[/tex]
Since the lenses in the question above are thin and in contact, the focal length of one of them will be:
[tex]\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}[/tex]
[tex]\frac{1}{f_{2}} =\frac{1}{f_{1}} -\frac{1}{F}[/tex]
[tex]\frac{1}{f_{2}} =\frac{1}{8} -\frac{1}{24}[/tex]
[tex]\frac{1}{f_{2}} =\frac{-2}{24}[/tex]
[tex]f_{2}=[/tex] -12
The focal length of the other lens is -12 cm, with the negative sign meaning it's a converging lens.
HELP ASAP!!! PLEASE!!
Answer:
Explanation:
The weaker force is pulling to direction East(450 newton)
the net force is (500-450=50 newton) and will move toward west ( in 500 newton direction)
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
the following ATTENTION：using the energy/work formulae only: 3.The kinetic energy of the Rock half way down ？ 4.the speed of the Rock half way down？
5.The speed of the Rock as it hits the ground？
Answer:
See the answers below.
Explanation:
In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].
[tex]E_{A}=E_{B}[/tex]
The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.
So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.
[tex]E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}[/tex]
At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.
[tex]E_{B}=m*g*h+\frac{1}{2}*m*v^{2}[/tex]
Therefore we will have the following equation:
[tex](6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s][/tex]
The kinetic energy can be easily calculated by means of the kinetic energy equation.
[tex]KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J][/tex]
In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.
[tex]E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s][/tex]
A man weighs himself twice in an elevator. When the elevator is at rest, he weighs 824 N; when the elevator starts moving upward, he weighs 932 N. Most nearly how fast is the elevator accelerating, assuming constant acceleration?
a. 0.64 m/s
b. 1.1 m/s^2
c. 1.3 m/s
d. 9.8 m/s^2
Answer: c. 1.3 m/s^2
Explanation:
When he is at rest, is weight can be calculated as:
W = g*m
where:
m = mass of the man
g = gravitational acceleration = 9.8m/s^2
We know that at rest his weight is W = 824N, then we have:
824N = m*9.8m/s^2
824N/(9.8m/s^2) = m = 84.1 kg
Now, when the elevators moves up with an acceleration a, the acceleration that the man inside fells down is g + a.
Then the new weight is calculated as:
W = m*(g + a)
and we know that in this case:
W = 932N
g = 9.8m/s^2
m = 84.1 kg
Then we can find the value of a if we solve:
932N = 84.1kg*(9.8m/s^2 + a)
932N/84.1kg = 11.1 m/s^2 = 9.8m/s^2 + a
11.1 m/s^2 - 9.8m/s^2 = a = 1.3 m/s^2
The correct option is C
A certain planet has a radius of 4990 km. If, on the surface of that planet, a 95.0 kg object has a weight of 591 N, then what is the mass of the planet?

Answer:
3743.489 kg
Explanation:
F_g = 591 N
G = 6.674x10^-11 constant of gravity
m_1 = 95 kg
m_2 = unknown
r = 4990*1000 =
F_g = G[(m_1*m_2)/r^2]
591 N = 6.674x10^-11[(95*m_2)/4990^2]
8.855 = [(95*m_2)/4990^2]
355631.472 = 95*m_2
m_2 = 3743.489 kg
The mass of the planet, which has a radius of 4990 km, is 1.81×10²³kg.
What is Newton's law of universal gravitation ?Newton's law of universal gravitation states that
The force of attraction between any two bodies is inversely proportional to the square of the distance between them and directly proportional to the product of their masses.
Given parameters:
Mass of the object: m = 95 kg
Radius of the planet: r = 4990 km = 4990 × 1000 m.
Weight f the object: F= 591 N
We know that: universal gravitational constant: G = 6.674x10^-11 SI unit.
We have to find: mass of the planet: M = ?
Now, F= GMm/r^2
591 N = 6.674x10^-11[(95×M)/(4990×1000)^2]
⇒ M =1.81×10²³kg
Hence, mass of the planet is 1.81×10²³kg.
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