Answers
A Thermal energy of the air is 17 J of heat to the surroundings.
Thus, Thermal energy is produced by materials whose molecules and atoms vibrate more quickly as a result of a rise in temperature.
The atoms and molecules that make up matter are always in motion. The increase in temperature caused by heating a substance causes these particles to accelerate and collide.
The energy that arises from a heated substance is referred to as thermal energy. The more the substance's thermal energy and the more its particles travel at higher temperatures.
Thus, A Thermal energy of the air is 17 J of heat to the surroundings.
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Related Questions
A liquid x at 25 degree Celsius is poured to a height of 40cm in a capillary tube of length 70cm and the diameter is 1cm .Assume that volume does not change with temperature. A find the initial volume of the liquid in cm^3. B. The temperature is reduced by 10°C causing the liquid to reduce in height to 37cm.Find the volume coefficient of the thermal expansion of the liquid. C. At the initial height of 40cm and temperature of 25°C , what change in temperature is needed for liquid to rise to a height of 49cm.
Answers
The initial volume of the liquid is 31.4 cm³. The volume coefficient of thermal expansion of the liquid is 0.002 per degree Celsius. A temperature increase of 109.5°C is needed for the liquid to rise to a height of 49cm.
The initial volume of the liquid can be found using the formula for the volume of a cylinder:
V = πr²h
where r is the radius (half the diameter), h is the height, and π is approximately 3.14. Plugging in the given values, we get:
V = π(0.5 cm)²(40 cm)
V = 31.4 cm³
The volume coefficient of thermal expansion (β) is defined as the fractional change in volume per degree Celsius change in temperature. It can be calculated using the formula:
β = ΔV/(VΔT)
where ΔV is the change in volume, V is the initial volume, and ΔT is the change in temperature. We can rearrange this formula to solve for ΔV:
ΔV = βVΔT
We know that ΔT = -10°C (a decrease of 10°C) and that the height decreased from 40cm to 37cm, or by 3cm. The change in volume can be found using the formula for the volume of a cylinder again, with the new height of 37cm:
ΔV = π(0.5 cm)²(40 cm - 37 cm)
ΔV = 0.59 cm³
Plugging in all the values, we get:
0.59 cm³ = β(31.4 cm³)(-10°C)
β = 0.002
To find the change in temperature needed for the liquid to rise to a height of 49cm, we can use the same formula as before, but solve for ΔT:
ΔT = ΔV/(βV)
We know that ΔV is the difference between the initial volume and the volume at the new height, which is:
ΔV = π(0.5 cm)²(49 cm - 40 cm)
ΔV = 6.86 cm³
Plugging in all the values, we get:
ΔT = 6.86 cm³/(0.002)(31.4 cm³)
ΔT = 109.5°C
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what causes a solid air fresher to lose mass and volume
Answers
Answer:
A solid air freshener loses mass and volume as a result of sublimation, where solid particles skip the liquid state and change directly from a solid to a gas. Sublimation requires that the particles in the solid state gain enough energy to immediately become a gas.
Explanation:
Hope this helps.
Mr. Daumer wants to create a magnet. He attaches a battery to a metal wire and
wants to wrap the wire around an object. Which object should he wrap the wire around?
a) A battery
b) A pencil
c) A ruler
d) A piece of steel metal.
Answers
big ideas
key questions and terms notes
Answers
Answer:
what i and confused
Explanation:
Answer:
nerd
Explanation:
I H4+3 U
A 1325 kg car and a 2050 kg pickup truck approach a curve on a highway that has a radius of 255 m. At what angle should the highway engineer bank this curve so that vehicles traveling at 75.0 mi/h can safely round it regardless of the condition of their tires
Answers
Answer:
the banking angle of the road is 24.2⁰
Explanation:
Given;
speed of the vehicles considered, v = 75 mi/h
Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s
75 mi/h --------> ?
= 75 x 0.44704 m/s = 33.528 m/s
radius of the curve, r = 255 m
The banking angle of the road is calculated as;
\(\theta = tan^{-1} (\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{33.528^2}{255\times 9.8} )\\\\\theta = tan^{-1}(0.44983)\\\\\theta =24.2^0\)
Therefore, the banking angle of the road is 24.2⁰
The angle of banking is 24 degrees.
What is the angle of banking?As a driver approaches a bend two equal and opposite forces act on him which are the centripetal force and the centrifugal force. The driver will have to ben through a certain angle called the angle of banking to avoid falling off.
The angle of banking depends on the speed of the vehicle and the radius of the curve.
θ = v^2/rg
speed = 75.0 mi/h or 33.5 m/s
r = 255 m
g = 9.8 ms-1
θ = tan-1 (33.5 m/s)^2/ 255 m × 9.8 ms-1
θ = tan-1(1122.3/2499)
θ = 24 degrees
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A 2.5 V battery is connected to a small light
bulb with a resistance of 4.8 Ω .
What is the current in the bulb?
Answer in units of A.
Answers
Answer: I=0.52 A
Explanation:
Given
Voltage is \(2.5\ V\)
Resistance \(R=4.8\ \Omega\)
Current is given by
\(I=\dfrac{V}{R}\)
\(\Rightarrow I=\dfrac{2.5}{4.8}=0.52\ A\)
what is this.
best answer gets brainiest
Answers
Answer:
the Thanksgiving turkey
A 5 kg block rests on a rough horizontal table. A rope is
attached to the block and is pulled with a force of 11N to the
left. As a result, the block accelerates at 2 m/s2. The coefficient
of kinetic friction between the block and the table is (round
to the nearest hundredth)
Answers
Answer:
Approximately \(0.02\) (assuming that \(g = 9.81\; \rm N \cdot kg^{-1}\), and that the block was initially moving to the left.)
Explanation:
Apply Newton's Second Law to find the net force on this block.
Mass of this block: \(m = 5\; \rm kg\).
Acceleration of this block: \(a = 2\; \rm m \cdot s^{-2}\).
\(\begin{aligned}\text{Net force} &= m \cdot a \\ &= 5\; \rm kg \times 2\; \rm m \cdot s^{-1} = 10\; \rm N\end{aligned}\).
The block is moving and accelerating towards the left. Hence, friction (which opposes motion) would be towards the right and would be opposite to the direction of the \(11\; \rm N\) pulling force.
Therefore, the expression for the net force on this object would be:
\((\text{Net force} ) = (\text{Pulling force}) - (\text{Friction})\).
\(10\; \rm N = 11\; \rm N - (\text{Friction})\).
Hence:
\(\text{Friction} = 11\; \rm N - 10\; \rm N = 1\; \rm N\).
The question states that this table is horizontal. Weight of the block (downwards) and normal force from the table (upwards) are the only two forces on this block in the vertical direction. Hence, the magnitude of the normal force on this block would equal to that of the weight of the block:
\(\begin{aligned} |\text{Normal force}| &= |\text{Weight}| \\ &= m \cdot g \\ &= 5 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} = 49.05\; \rm N\end{aligned}\).
Calculate the constant of kinetic friction between this block and the table:
\(\begin{aligned}\mu_{\text{k}} &= \frac{\text{Friction}}{\text{Normal force}} \\ &= \frac{1\; \rm N}{49.05\; \rm N} \approx 0.02\end{aligned}\).
A ball is moving up a hill at 20 meters per second, and is decelerating at a uniform rate of 2 meters per second squared. What is its velocity after 5 seconds? Assume the direction up the hill is positive.
Answers
Answer:
-12meters per secondExplanation:
2^5=32-20=-12 meters per sec (going backwards).
If this is correct pls mark me. brainly to let me know, tyy C:
can two objects exert a force on each other without touching? example
Answers
Answer:
where is the example...?
Consider the system of two blocks shown in Fig. P6.81, but with a different friction force on the 8.00 kg block. The blocks are released from rest. While the two blocks are moving, the tension in the light rope that connects them is 37.0 N. (a) During a 0.800 m downward displacement of the 6.00 kg block, how much work has been done on it by gravity? By the tension T in the rope? Use the work–energy theorem to find the speed of the 6.00 kg block after it has descended 0.800 m. (b) During the 0.800 m displacement of the 6.00 kg block, what is the total work done on the 8.00 kg block? During this motion how much work was done on the 8.00 kg block by the tension T in the cord? By the friction force exerted on the 8.00 kg block? (c) If the work–energy theorem is applied to the two blocks con- sidered together as a composite system, use the theorem to find the net work done on the system during the 0.800 m downward displacement of the 6.00 kg block. How much work was done on the system of two blocks by gravity? By friction? By the tension in the rope?
Answers
a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.
b) We cannot calculate the work done by the friction force.
c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.
(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:
Work_gravity = force_gravity * displacement * cos(theta),
where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).
The weight of the block is given by:
force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.
Plugging in the values, we get:
Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.
The work done on the 6.00 kg block by the tension in the rope is given by:
Work_tension = tension * displacement * cos(theta).
Plugging in the values, we get:
Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.
The negative sign indicates that the tension is in the opposite direction of the displacement.
Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:
Work_net = change_in_kinetic_energy.
Since the block starts from rest, its initial kinetic energy is zero. Therefore:
Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.
Solving for velocity, we get:
velocity = sqrt(2 * Work_net / mass).
The net work done on the block is the sum of the work done by gravity and the tension:
Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.
Plugging in the values, we get:
velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.
Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.
(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:
Work_net = change_in_kinetic_energy.
Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:
Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.
The work done on the 8.00 kg block by the tension in the rope is given by:
Work_tension = tension * displacement * cos(theta).
Plugging in the values, we get:
Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.
The work done on the 8.00 kg block by the friction force can be calculated using the formula:
Work_friction = force_friction * displacement * cos(theta),
where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.
(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:
Work_net = change_in_kinetic_energy.
Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:
Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.
Simplifying, we get:
Work_net = 1/2 * 14.00 kg * velocity^2.
Using the value of velocity calculated in part (a), we get:
Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.
The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:
Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.
The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:
Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.
Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.
Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.
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6. Displacement vectors of 3 m and 5 m in the same direction combine to make a displacement vector that is
in magnitude.
4 meters
8 meters
15 meters
2 meters
Answers
Answer:
8m is the answer.......
Which of the following is not part of the control system? steering wheel exhaust pipe brake accelerator
Answers
Answer:
Brakes unevenly adjusted: Brakes pulling in one direction or the other can lead to a skid. Tires with worn tread: Tread is necessary for traction in wet weather
Explanation:
Answer: exhaust pipe
Explanation: It doesn't really control anything so I'm assuming this is the answer
True Or False weather conditions in the atmosphere can be recognized through direct observation.
Answers
Answer:
yes, we can tell if there will be rain by the swelling in rain clouds, we can also see if a tornado is forming based on the look of the wall cloud. we can use satellites to predict the amount of precipitation or wind.
Explanation:
thank you for ur generosity
Calculate the mass expressed in kg of a container of 3 liters of olive oil The density of olive oil is 900 g/l
Answers
Mass = density x volume
= 3 x 900
= 2700 g
If 1000 g = 1 kg
Then 2700g = 2700/1000=2.7 kg
Hence the mass is 2.7 kg
Which property of gases allows them to be stored at high concentrations in a bottle of air freshener?
A. attraction
B. compressibility
C. intermolecular force of attraction
D. effusion
E. fluidity
Answers
Answer:
b
Explanation:
The velocity (in meters/sec) of a particle moving along a straight line is given by v(t)=2t^2 + 3t - 1 where t is measured in seconds.
Answer the following questions given that the initial position s (0) =
Answers
Despite the fact that it was screwed up in, the displacement will be S of seven minus.
What is the initial position s (0) ?The goal is to make the position at Times Two equal to 4. This just serves to plug in four for tea. We're going to assess this and see if we can repeat the amount after we just typed in for everywhere for T. That is the current situation. T is equal to 4, therefore that's what it is. The location would be helpful to us. The concept is the same. It is seven times a third. Seven squared plus seven plus six equals that. You may calculate this to obtain 25 0.3 repeating. We want to determine a particle's displacement. It appears to have taken place between the hours of 4 and 7. Despite the fact that it was screwed up in, the initial position will be S of seven minus.To learn more about initial position refer to:
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No work is done by an object at rest because.
1) no force is acting on the object.
2) no distance is moved.
3) heat is not produced.
4) friction is acting on the object
Answers
Answer:
no distance is moved as w=f×d
Question 12 of 20
To produce a magnetic field, what does an electromagnet require?
A. A solenoid with no current flowing through it
B. A current flowing through a coil of wire
C. A core made from a magnetic material
D. A copper core
Answers
Answer:
B a current flowing through a coil of wire :)
Explanation:
A 27.0 kg box is pulled with a 113.6 N force at an angle of 35.1 degrees along a
surface that has a coefficient of friction of 0.4. What is the net force on the box?
Answers
Answer: 2,509.42
Explanation: using the equation net force=force x cos(angle) we can find the net force by multiplying the kg and the Newtons to get 3,067.2 then find the cos of 35.1 then multiple those two together to get 2,509.42
Two stars M1 and M2 of equal mass make up a binary star system. They move in a circular orbit that has its center at the midpoint of the line that separates them. If M1 = M2 = 5.45 sm (solar mass), and the orbital period of each star is 3.20 days, find their orbital speed. (The mass of the sun is 1.99E30 kg.)
Answers
Since Two stars M1 and M2 of equal mass make up a binary star system. their orbital speed is 2.69E4 m/s.
What is the orbital speed about?The orbital period of a binary star system is given by the equation:
T = 2 * pi * sqrt(a³ / (G * (M1 + M2)))
where T is the orbital period, a is the semi-major axis of the orbit (half of the distance between the two stars at their closest approach), G is the gravitational constant, and M1 and M2 are the masses of the two stars.
Substituting the given values into this equation, we get:
3.20 days = 2 * pi * sqrt(a³ / (6.67E-11 m³ kg^⁻¹ s^⁻² * (5.45 sm * 1.99E30 kg/sm + 5.45 sm * 1.99E30 kg/sm)))
Solving for the semi-major axis, we find that a = 2.11E11 m.
The speed of each star in the orbit can be found using the equation:
v = sqrt(G * (M1 + M2) / a)
Substituting the values we have calculated, we find that v = 2.69E4 m/s. This is the speed of each star in the orbit.
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what is science ? what qualities do we deal in deal in physic ?
Answers
science is all about the world around us
how you would teach the habit to your students.
Answers
Answer:
Probably just enforcing it little by little. (If you mean a habit. If you mean a certain topic it's different.)
Explanation:
If you are trying to teach your students a habit-
Try to correct them everytime they don't follow it. Expect them to do so and give a reward when they earn it.
If you mean a certain topic-
Make a slideshow or a presentation with an interactive game. Making it not boring and having them more attentive.
I gave a more general answer, so if you would like to go deeper into the subject I would be more happy to.
(If this helped, please let me know. If you have any questions do not be afriad to ask!)
EXPLANATION ⛔
A 20 gram mass is suspended from meter rod at 20cm. The meter rod is balanced on 40cm mark. Weight of meter rod is
A. 0.4N
B. 0.6N
C. 6N
D. 60N
Answers
Answer:b
Explanation:I’m just trynna get more money dude
5. Impulse is...
a. A change in the time it takes an object to come to rest
b. A change in an object's velocity
C. A change in an object's mass
d. A change in an object's momentum
Answers
Answer:
The answer should be C
Explanation:
a What is meant by zero error?
b Give an example of when you would have to allow
for it.
Answers
a) It is the error present in the measuring instrument that causes it to register a value even when there is no input or output being measured.
b) An example of when you would have to allow for zero error is when using a measuring instrument like a vernier caliper or micrometer screw gauge.
a) Zero error refers to the deviation or discrepancy in the measurement instrument, where the indication or reading on the instrument is not zero when the quantity being measured is zero. In other words, it is the error present in the measuring instrument that causes it to register a value even when there is no input or output being measured.
Zero error can occur due to various reasons such as manufacturing defects, wear and tear, misalignment, or improper calibration of the instrument. It can be positive or negative, depending on whether the instrument reads higher or lower than the actual value.
b) An example of when you would have to allow for zero error is when using a measuring instrument like a vernier caliper or micrometer screw gauge. These instruments are commonly used to measure the dimensions of objects with high precision.
In a vernier caliper, for instance, zero error can occur when the jaws do not close perfectly when there is no object being measured. If the caliper shows a reading other than zero when the jaws are closed, it indicates the presence of zero error.
To obtain accurate measurements, the zero error needs to be accounted for and compensated. This can be done by adjusting the position of the zero on the scale or by subtracting the zero error value from the measured readings.
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Check the three scenarios that have balanced forces.
A tow truck tries and fails to pull a car out of a ditch. Neither the car nor the tow truck move.
Jada and Mya run into each other in bumper cars. They bounce off each other and go in new directions
Zeke rests at the top of Stone Mountain and enjoys the view.
Haleigh throws a baseball from first base to second.
Ryan is playing ice hockey and glides across the ice in a straight line at a constant speed.
Answers
The examples of balanced forces include;
A tow truck tries and fails to pull a car out of a ditch. Neither the car nor the tow truck move.Jada and Mya run into each other in bumper cars. They bounce off each other and go in new directions.Zeke rests at the top of Stone Mountain and enjoys the view.What are balanced forces?Balanced forces are those that are opposite in direction and equal in size. Balanced forces are considered to be in a state of equilibrium.
A tow truck tries and fails to pull a car out of a ditch. Neither the car nor the tow truck move. (forces are balanced that's why they don't move).
Jada and Mya run into each other in bumper cars. They bounce off each other and go in new directions. (forces are balanced that's why they bounce off in opposite direction).
Zeke rests at the top of Stone Mountain and enjoys the view. (forces are balanced that's why he is at rest).
Haleigh throws a baseball from first base to second. (forces are not balanced that's why the baseball is in motion).
Ryan is playing ice hockey and glides across the ice in a straight line at a constant speed. (forces are not balanced that's why he is in motion).
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Question 7
Identify the correct Earth System in order from #1-4.
Reorder answers
1.Atmosphere
Reorder answers
2.Geosphere
Reorder answers
3.Hydrosphere
Reorder answers
4.Biosphere
Answers
Answer:
2, 3, 4, 1
Explanation:
geosphere, hydrosphere, biosphere, and atmosphere
Answer:
Lithosphere hydrosphere biosphere atmosphere
Explanation:
So yea mark me brainliest
Before they were decommissioned, the NASA space shuttles required two solid rocket boosters (SRBs) to launch the shuttle from Earth’s surface. Both SRBs produced 1.7 x10^7 N at liftoff. The combined mass of a shuttle and rocket boosters was about 1.5 x 10^ 6 kga) Calculate the net acceleration of a space shuttle and rockets at the time of liftoff. (b) Calculate the speed of the shuttle and rockets after 10.0 s.
Answers
Given:
The force on the booster is
\(F=\text{ 1.7}\times10^7\text{ N}\)The mass is
\(m=\text{ 1.5}\times10^6\text{ kg}\)Required:
(a) The net acceleration
(b) Speed of the shuttle and rockets after time t = 10 s
Explanation:
(a) The net acceleration can be calculated as
\(\begin{gathered} a=\frac{F}{m} \\ =\frac{1.7\times10^7}{1.5\times10^6} \\ =11.33\text{ m/s}^2 \end{gathered}\)(b)
The initial speed of the rocket and shuttle will be zero.
The speed of the rocket and shuttle after time t = 10 s will be
\(\begin{gathered} v=0\text{ m/s + 11.33}\times10 \\ =\text{ 113.3 m/s} \end{gathered}\)Final Answer:
(a) The net acceleration is 11.33 m/s^2.
(b) The speed of the rocket and shuttle is 113.3 m/s
Locate the virtual F Prescribe a suitable bi-focal spectacles for a person who cannot see clearly the object nearer than 50 cm and 2 m/away or further. 75
Answers
The virtual focus of a person is -2m and the power of bi focal spectacles for a person who can see between 50cm and 2m is -0.5D
What distinguishes actual focus from virtual focus?The true focus is the location of the actual convergence, while the virtual focus is the location from which the diverging rays appear to arise as they are stretched backward. In the case of a convex lens, real focus is attained, and in the case of a concave lens, virtual focus is attained.
Visible object distance (u)= 2m
Image distance (v) = infinity
Focal length (f) formula => 1/v - 1/u = 1/f
=> 1/infinity - 1/2 = 1/f
=> f = -2m
The relation between power (P) and focal length (f) => P = 1/f
P = 1/-(2) => -0.5D
What is presbyopia identified as?The ciliary muscles gradually become less flexible and the eye lens becomes less flexible, which leads to presbyopia. A bifocal lens is used to repair the problem in eyes that have both myopia and hypermetropia.
Visible object distance (u)= 2m
Image distance (v) = infinity
Focal length (f) formula => 1/v - 1/u = 1/f
=> 1/infinity - 1/2 = 1/f
=> f = -2m
The relation between power (P) and focal length (f) => P = 1/f
P = 1/-(2) => -0.5D
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