Wrap a fur coat around a thermometer. how does the temperature change?

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Answer 1

Heat is not generated by fur coats, it's generated by the person wearing the coat, or the animal the fur is on. That's why when you get into bed on a cold night, the sheets are cold at first, but your own body heat warms them up.

Related Questions

To recharge a 12 v battery, a battery charger must move 3.3 × 105 c of charge from the negative terminal to the positive terminal. how much work is done by the battery charger? answer in units of j

Answers

Work done is calculated by change * potential difference
in equation it means: W = Q * V

since the 2 terminals have now dq charge and (V) potential difference

small work done (dW) = V dq
total work done W = V Q = 12 * 3.3*10^5 = 3.96 *10^6 Joules which is approximately 4 mega joules

To calculate the work done by the battery charger, multiply the charge moved (3.3 × [tex]10^5[/tex] C) by the voltage (12 V). The result is 3.96 × [tex]10^6[/tex] J.

To find out how much work is done by the battery charger, we use the relationship between electrical potential difference (voltage), charge, and work. The work done, W, when moving a charge, Q, through a potential difference, V, is given by the equation:

W = Q × V

Here, the battery charger must move 3.3 × [tex]10^5[/tex] C of charge (Q) through a potential difference of 12 V (V). Substituting these values into the equation gives:

W = 3.3 × [tex]10^5[/tex] C × 12 VW = 3.96 × [tex]10^6[/tex] J

Therefore, the work done by the battery charger is 3.96 × [tex]10^6[/tex] J.

A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system moves with a speed equal to one- third the original speed of the 3.00 kg mud ball. what is the mass of the second mud ball?

Answers

Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

[tex] m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1} + m_{2} * v'_{2}[/tex]

In case of perfectly inelastic collision v'1 and v'2 are same.

We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁

Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg

Mass of second mud ball is 6kg.

A 120-meter-long ski ift carries skiers from a station at the foot of a slope to a second station 40 m above. what is the IMA (ideal machincal advantage) of the lift?

Answers

The ideal mechanical advantage (IMA) of an inclined plane is given by:
[tex]IMA = \frac{L}{h} [/tex]
where L is the length of the inclined plane while h is the height.
In our problem, L=120 m and h=40 m. Therefore, the IMA of the ski lift is
[tex]IMA= \frac{120 m}{40 m}=3 [/tex]

A 7 L sample of gas has a pressure of 1.1 atm at a temperature of 285 K. If the pressure decreases to 0.6 atm, causing the volume to increase to 10 L, what is the new temperature? Round your answer to the nearest tenth.

Answers

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
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Answer: 222.1K

At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 930 m/s2 and the vertical or y component of its acceleration is 900 m/s2. the ball's mass is 0.39 kg. what is the magnitude of the net force acting on the soccer ball at this instant?

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just try your best best friend everyone

Two cars of the same mass have different velocities. Which car has more momentum

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the car with the higher velocity would have a more impactful momentum.

A 2.90 kg mass is pushed against a horizontal spring of force constant 23.0 n/cm on a frictionless air table. the spring is attached to the tabletop, and the mass is not attached to the spring in any way. when the spring has been compressed enough to store 15.0 j of potential energy in it, the mass is suddenly released from rest. part a find the greatest speed the mass reaches. vmax = 3.22 m/s submitmy answersgive up correct part b when does this occur? if not friction acts on it forever. if there is right at the beginning when it is just released. submitmy answersgive up

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Which scientists contributed to discovering the universal law of gravitation? Check all that apply. Tycho Brahe Albert Einstein Johannes Kepler Nicolaus Copernicus Sir Isaac Newton Robert Hooke

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Final answer:

Sir Isaac Newton is the primary contributor to the universal law of gravitation, with his precise mathematical formula that unified terrestrial and celestial phenomena. Johannes Kepler's laws of planetary motion were foundational for Newton's work, and Albert Einstein expanded on these ideas with his theory of general relativity.

Explanation:Contributors to the Universal Law of Gravitation

Several prominent individuals contributed to the discovery and understanding of the universal law of gravitation. Sir Isaac Newton is the most well-known figure associated with the law, as he defined the gravitational force, proposing that it was a universal force that explained both why objects fall to Earth and the motions of celestial bodies. He was the first to provide a precise mathematical formula for the law of gravitation.

Before Newton, Johannes Kepler discovered three laws of planetary motion, which Newton found crucial for his own work, as they showed gravitation's effects on planetary orbits. Moreover, Albert Einstein expanded upon the concept of gravitation with his theory of general relativity which showed that there is more to the gravity story than Newton's law suggested. Although their contributions were indirect, scientists such as Galileo Galilei and Robert Hooke, also helped set the stage for Newton's discoveries through their work on planetary motions and gravitational investigations respectively.

For a temperature increase of Î´t1, a certain amount of an ideal gas requires 30 j when heated at constant volume and 50 j when heated at constant pressure. how much work is done by the gas in the second situation?

Answers

Work done by the ideal gas in the second situation is 20 Joule.

We know that amount of energy given to a ideal gas is distributed in it to increase its internal energy and to work done by increasing it volume.

Mathematically:

energy given to a ideal gas (dQ) = Increase in internal energy (dU) + work done (dW).

Now in this question: a certain amount of an ideal gas requires 30 j when heated at constant volume. So, this energy is used to increase internal energy (as no volume change occurs).

So, dQ₁ = dU = 30 Joule.

When heated at constant pressure, the certain amount of an ideal gas requires 50 J. So, this energy is used to increase internal energy and work done.

So, dQ₂ = dU + dW

⇒ dW = dQ₂ - dU = 50 Joule - 30 joule = 20 Joule.

Hence, work is done by an amount 20 joule by the ideal gas in the second situation.

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If vx = 7.90 units and vy = -6.50 units, determine (a)the magnitude and (b)direction of v

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a) To find the magnitude of the resultant, we can apply Pythagorean theorem, because vx and vy are the two sides of a rectangle triangle, of which the resultant v is the hypothenuse:
[tex]v= \sqrt{v_x^2+v_y^2}= \sqrt{(7.90)^2+(6.50)^2}=10.2 units [/tex]

b) To find the direction (the angle with respect to the horizontal), let's keep in mind that the tangent of the angle is equal to the ratio between the opposite side and the adiacent side:
[tex]\tan \alpha = \frac{v_y}{v_x}= \frac{-6.50}{7.90}=-0.82 [/tex]
From which we find
[tex]\alpha = arctan (-0.82)=-39.4 ^{\circ}[/tex]

The part of the flower responsible for producing pollen (sperm) is the _______. A. stigma B. anther C. style D. filament

Answers

the answer to ur question is B

Based on their locations on the periodic table, which best compares the properties of the metalloids arsenic (As) and antimony (Sb)?

Answers

Arsenic has a more nonmetallic behavior than antimony because arsenic's atoms have fewer electron shells.

What type of motor operates at a constant steady-state speed regardless of the load?

Answers

The type of motor that allows for constant speed regardless of load are called Geared Speed Control motors. This type of motor has a tachometer feedback device attached at the rear of the motor that gives constant feedback to the speed controller giving the advantage of constant speed regardless of load. The tachometer allows for varied frequency delivery to the motor to maintain pre-set output speed.

If the initial velocity of the system were not zero, how would this affect your results?

Answers

just try your best best friend everyone

A 3.7-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant) is 450 n/m. the block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest. the block then executes simple harmonic motion along the horizontal x-axis. the maximum elastic potential energy of the system is closest to

Answers

Final answer:

In the simple harmonic motion of a block attached to a spring, the maximum elasticity potential energy (U) occurs when displacement from equilibrium is the greatest. Given the provided spring constant (k) and displacement (x), the energy can be calculated using the formula U = (1/2)kx². The maximum elastic potential energy in this scenario is close to 1.44 joules.

Explanation:

In physics, the problem you're dealing with relates to simple harmonic motion associated with a block attached to a spring on a frictionless surface. When the object is displaced from equilibrium and let go, it performs simple harmonic motion. During this motion, there is a constant interconversion between the kinetic and potential energy within the system.

The 'spring constant' (k) of an ideal spring is given as 450 n/m and the displacement from the equilibrium position (x) is 0.080 m. The maximum elastic potential energy is at extremes of the motion, when the displacement is the greatest (+/- x). It is given by the formula: U = (1/2)kx², where U is the elastic potential energy.

Substituting the given spring constant and displacement values into the formula: U = (1/2) * 450 * (0.080)² = 1.44 joules. Hence, the maximum elastic potential energy of the system is closest to 1.44 joules.

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__________ helps us adapt to our environment. It also generally __________ with age. A. Plasticity . . . decreases B. Development . . . increases C. Plasticity . . . increases D. Development . . . decreases

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Not sure this is a physics question (probably biology).
Anyway, the correct answer is A):
"Plasticity helps us to adapt to our environment. It also generally decreases with age".
Plasticity is the ability to adapt to the environment. Since this ability is linked with brain functions, and brain functions worsen with age, then plasticity decreases with age.

Answer: A, Plasticity ... decreases

Explanation:

Which statements accurately describe elements? Check all that apply.
Elements are made up of two or more types of atoms.
Elements are made up of only one type of atom.
Each element has a unique chemical symbol.
Elements can be identified by their atomic number.
One element cannot be combined with another element.

Answers

The correct statements are:
- Elements are made up of only one type of atom.
- Each element has a unique chemical symbol.
- Elements can be identified by their atomic number.

In fact, elements are made up of only one tipe of atom, and each of them is identified by a different symbol. Moreover, each element has a different atomic number (number of protons), therefore elements can be identified by their atomic number.

Answer: The correct statements are:

- Elements are made up of only one type of atom.

- Each element has a unique chemical symbol.

- Elements can be identified by their atomic number.

Explanation: and just for a little bit more information about elements.

An element is a pure substance that is made from a single type of atom. Elements are the building blocks for all the rest of the matter in the world. Examples of elements include iron, oxygen, hydrogen, gold, and helium. An important number in an element is the atomic number.

A block of mass 2.5 kg is attached to a spring with spring constant 150 n/m and slides on a rough, horizontal surface (coefficient of kinetic friction 0.25). the spring is initially extended by a distance of 45 cm from its equilibrium position and released from rest. (a) what is the speed of the block when it is at the equilibrium position of the spring? (b) where does the block stop?

Answers

if a vector a - b = 2c, a + b = 4c and c = 3i + 4j,find a and b

Answers

Ans:
a = 9i + 12j
b = 3i + 4j

Explanation:
We have two equations:

a - b = 2c --- (A)
a + b = 4c --- (B)

and two unknowns:
a = ?
b = ?

Whereas,
c = 3i + 4j

Add equations (A) and (B):
=> a - b = 2c
a + b = 4c
----------------------
2a = 6c

=> a = 3c --- (C)
Put the value of vector c in equation (C):

(C) => a = 3(3i + 4j)
=> a = 9i + 12j

Now that we have vector a, put that vector and vector c in equation (A):
(A)=> (9i + 12j) - b = 2(3i + 4j)

=> b = (9i + 12) - (6i + 8j)
=> b = 3i + 4j

Ans:
a = 9i + 12j
b = 3i + 4j

How much energy is required to move a 1350 kg object from the earth's surface to an altitude twice the earth's radius?

Answers

Final answer:

The calculation of energy required to move an object to an altitude twice the Earth's radius involves understanding and applying principles of gravitational potential energy and the universal law of gravitation.

Explanation:

The question involves calculating the energy required to move a 1350 kg object from the Earth's surface to an altitude twice the Earth's radius. To solve this, we use the formula for gravitational potential energy (GPE), which is GPE = mgh at close distances to Earth's surface, and the universal law of gravitation for larger distances.

However, at distances far from the surface, the formula shifts to GPE = -G * (m1*m2)/r, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers. Since the altitude is twice the Earth's radius, the effective distance r would be 3 times the Earth's radius. This question requires integration of both concepts and understanding of physics to solve comprehensively.

Which equations can be used to calculate the electric potential energy stored in a capacitor? Check all that apply. U = QV U = CV U = U = CV2 U = U = QV2

Answers

There are different ways to calculate the electric potential energy stored in a capacitor.
Among these forms we have:
Form 1:
U = (CV ^ 2) / 2
Form 2:
U = (Q ^ 2) / 2C
Form 3:
U = (QV) / 2
Where,
Q: load
C: capacitance
V: voltage.
Answer:
U = (CV ^ 2) / 2
U = (Q ^ 2) / 2C
U = (QV) / 2

U=1/2QV

U=1/2CV^2

U=Q^2/2C

Which graphic design tools help you draw circles and rectangles?

Answers

Geometric Shape Tools
Hope this helps :)

Answer: geometric shape tools

Explanation: plato/edmentum answer.

hope this helps! :)

what causes the end of a star?

Answers

The start likely burned all of its hydrogen fuel and expands in size

Which has a higher acceleration: a 10 kg object acted upon with a net force of 20 N or an 18 kg object acted on by a net force of 30 N

Answers

When a force F is acting on an object of mass m, the acceleration of the object is (Newton's second law)
[tex]a= \frac{F}{m} [/tex]
The first object has a mass of 10 kg and a force of 20 N is acting on it, so its acceleration is
[tex]a_1= \frac{20 N}{10 kg}=2 m/s^2 [/tex]
The second object has a mass of 18 kg and a force of 30 N is acting on it, so its acceleration is
[tex]a_2= \frac{30 N}{18 kg}=1.67 m/s^2 [/tex]
So, the first object has higher acceleration.

Answer:

the 10-kg object has higher acceleration

Explanation:

Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x. find the x component of the velocity of the center of mass (vcm)x at that moment. keep in mind that, in general: vx=dx/dt. express your answer in terms of m1, m2, v1x, and v2x.

Answers

The center of mass is given with this formula:
[tex]x_c=\frac{\sum_{n=1}^{n=i}m_ix_i}{M}[/tex]
Velocity is:
[tex]v=\frac{dv}{dt}[/tex]
So, for the velocity of the center of mass we have:
[tex]\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\ v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\[/tex]
In our case it is:
[tex]v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}[/tex]

Final answer:

The x component of the center of mass velocity is calculated as the momentum-weighted average of the individual blocks' velocities, using the formula (m1*v1x + m2*v2x) / (m1 + m2).

Explanation:

To find the x component of the velocity of the center of mass (vcm)x, we use the formula for the center of mass velocity in one dimension, which is given by:

(vcm)x = (m1*v1x + m2*v2x) / (m1 + m2)

This equation reveals that the center of mass velocity is the momentum-weighted average of the velocities of the individual blocks. Since momentum is mass times velocity, the product m1*v1x is the momentum of block 1 in the x direction, and m2*v2x is the momentum of block 2 in the x direction. The sum of these momenta gives the total momentum in the x direction. By dividing this sum by the total mass (m1 + m2), we obtain the velocity of the center of mass in the x direction.

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A particle moves at a speed such that its kinetic energy just equals its mass energy. what is the speed of the particle?

a. 3 x 108 m/s;

b. 2.6 x 108 m/s;

c. 2.82 x 108 m/s;

d. 2.3 x 108 m/s

Answers

I believe it is B. I could be wrong though.

A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?

Answers

The initial kinetic energy of the car is
[tex]E_1 = \frac{1}{2}mv_1^2 = 1.9 \cdot 10^3 J[/tex]

Then, the velocity of the car is decreased by half: [tex]v_2 = \frac{v_1}{2} [/tex]
so, the new kinetic energy is
[tex]E_2 = \frac{1}{2}mv_2 ^2 = \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4} [/tex]
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
[tex]E_2 = \frac{1.9 \cdot 10^3 J}{4}=475 J [/tex]

A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops of ore vertically into the cart. how fast does the cart move just after being loaded with the ore?

Answers

The value of the final speed depends on the mass of the ore.

Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.

Initially, there is only the cart, so the momentum is
[tex]p=Mv=(1200 kg)(10.8 m/s)=12960 kg m/s[/tex]
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is [tex]v_f[/tex]. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
[tex]p=12960 kg m/s =(1200 kg+m)v_f[/tex]
and so the final speed is
[tex]v_f = \frac{12960 kg m/s}{1200 kg +m} [/tex]

An automobile engine slows down from 4500 rpm to 1600 rpm in 6.0 s. (a) calculate its angular acceleration, assumed constant. 3036.87 incorrect: your answer is incorrect. rad/s2 (b) calculate the total number of revolutions the engine makes in this time. 8700 incorrect: your answer is incorrect. rev

Answers

1) The angular acceleration of the engine is given by:
[tex]\alpha = \frac{\omega _f - \omega _i}{\Delta t} [/tex]
where [tex]\omega _f [/tex] and [tex]\omega _i[/tex] are the final and initial angular velocity of the engine, while [tex]\Delta t[/tex] is the time interval considered.
We need first to convert the velocities from rpm (revolutions per minute) into rad/s. Keeping in mind that
[tex]1 rev=2\pi rad[/tex]
[tex]1min = 60 s[/tex]
The factor of conversion is
[tex]1 \frac{rev}{min} = \frac{2 \pi rad}{60 s} [/tex]
So, the two velocities become
[tex]\omega _i = 4500 rpm \cdot ( \frac{2 \pi rad}{60 s} )=471 rad/s[/tex]
[tex]\omega _f = 1600 rpm \cdot ( \frac{2 \pi rad}{60 s} )=167 rad/s[/tex]
And using [tex]\Delta t=6.0 s[/tex] we can find the angular acceleration of the engine:
[tex]\alpha = \frac{167 rad/s - 471 rad/s}{6.0 s}=-50.7 rad/s^2 [/tex]
where the negative sign means the engine is decelerating.

2) The total number of revolutions is given by the law of angular motion:
[tex]S(\Delta t)= \omega _i \Delta t + \frac{1}{2} \alpha (\Delta t)^2 = (471 rad/s)(6.0 s) + \frac{1}{2}(-50.7 rad/s^2)(6.0 s)=[/tex]
[tex]=1915 rad [/tex]
And keeping in mind that [tex]1 rev=2 \pi rad[/tex], the number of revolutions is
[tex]1915 rad \cdot \frac{1}{2 \pi}=305 rev [/tex]

The angular acceleration of the automobile is 50.62 rad/s².

The total number of revolutions within the given time is 290 revolutions.

Angular acceleration of automobile

The angular acceleration of the automobile is calculated as follows;

[tex]\alpha = \frac{\omega _f - \omega _i}{t} \\\\[/tex]

ωf = 1600 rpm = 167.57 rad/s

ωi = 4500 rpm = 471.3 rad/s

[tex]\alpha = \frac{167.57 - 471.3}{6} \\\\\alpha = -50.62 \ rad/s^2[/tex]

Total number of revolutions in 6 s

N = (4500 rpm - 1600 rpm)

N = 2,900 rpm

N = 2,900 x (6/60)

N = 290 rev

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) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force on each charg

Answers

Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others.
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
[tex]F_x=F_3+F_2cos(45^{\circ})[/tex]
[tex]F_y=F_1+F_2sin(45^{\circ})[/tex]
Let's figure out each component:
[tex]F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\ F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\[/tex]
[tex]F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}[/tex]
Total force acting on the charge would be:
[tex]F=\sqrt{F_x^2+F_y^2}[/tex]
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
[tex]F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1}
{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N[/tex]
[tex]F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N[/tex]
Now we can find the total force acting on a single charge:
[tex]F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N[/tex]
As said before, intensity of the force acting on charges is the same for all of them.

Final answer:

The magnitude and direction of the force between each pair of charges can be determined using Coulomb's Law.

Explanation:

The magnitude of the force between two charges can be determined using Coulomb's Law. In this case, we have four charges placed at the corners of a square. Since the charges are the same (6.15 mc), the force between each pair of charges will have the same magnitude. Let's label the charges as q1, q2, q3, and q4. The force between q1 and q2 is given by:

F12 = (k * q1 * q2) / r^2

where k is the Coulomb's constant (8.988 × 10^9 N m^2/C^2) and r is the distance between the charges (0.100 m). The direction of the force will be repulsive, since the charges have the same sign.

Using this formula, we can calculate the magnitude and direction of the force for all pairs of charges.

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