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4 years ago

How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)

Physics

1 answer:

I am Lyosha [343]4 years ago

3 0

<h2>Option C is the correct answer.</h2>

Explanation:

We need to find how many calories is 1 BTU.

Given

1 BTU = 1054 J

1 calorie = 4.186 J

So we have

1 BTU = 4.186 x 251.79 J

1 BTU =251.79 calorie

1 BTU = 252 calorie.

Option C is the correct answer.

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In what do electromagnetic waves cause disturbances when they transfer<br> energy?

Lady_Fox [76]

Answer: B - Both electric and magnetic fields

Explanation: Ap*x verified just did the test and got it correct, answer is right infront of me at the moment

8 0

3 years ago

Read 2 more answers

Four charges are located at the corners of a square. Each side of the square is 37 m. The left two charges have a positive charg

nadezda [96]

Answer:

E = 781.12 N/C

Explanation:

Look at the attached graphic:

The 20µC charges are positive , then, the electric fields leave the charge.

The 22µC charges are negative, then, the electric fields enter the charge.

The electric field due to each of the charges is calculated by Coulomb's law:

E= k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Problem development

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

E₁, E₂, E₃, E₄: Electric field at point P due to charge q₁, q₂, q₃, and q₄ respectively

The electric field in the direction of the y axis at the point P in the center of the square is equal to zero because the sum of the vertical components upwards is equal to the sum of the vertical components downwards.

Calculation of the electric field in the direction of the x-axis in the center of the square (point P)

Eₓ = E₁ₓ + E₂ₓ + E₃ₓ + E₄ₓ

$d = \sqrt{18.5^2+ 18.5^2} =26.16m$

E₁ₓ = E₂ₓ = (9*10⁹)*(20*10⁻⁶)*cos(45°)/(d²) = (9*10⁹)*(20*10⁻⁶)*cos(45°)/(26.16²) = 185.98 N/C

E₃ₓ = E₄ₓ = (9*10⁹)*(22*10⁻⁶)*cos(45°)/(d²) = (9*10⁹)*(22*10⁻⁶)*cos(45°)/(26.16²) = 204.58 N/C

Eₓ = E = 2*185.98 +2*204.58 = 781.12 N/C

5 0

3 years ago

A pail of water is rotated in a vertical circle of radius r = 1.6 m . The acceleration of gravity is 9.8 m/s 2 . r v What is the

Snowcat [4.5K]

Answer:3.95 m/s

Explanation:

Given

radius of circle $r=1.6 m$

Water will not spill if the centripetal acceleration is greater than acceleration due to gravity

i.e. $a_c\geq g$

$a_c=\frac{v^2}{r}$

$\frac{v^2}{r}\geq g$

$v_{min}=\sqrt{gr}$

$v_{min}=\sqrt{9.8\times 1.6}$

$v_{min}=\sqrt{15.68}$

$v_{min}=3.95 m/s$

8 0

4 years ago

The system needs an ordinary friction-based brake to bring the train to a full stop. Explain why the magnetic brake is not very

BabaBlast [244]

Answer:

The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.

Explanation:

A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.

Let us find the drag force which is proportional to the velocity of two conducting plates.

The EMF that is induced in the eddy currents are : $E=v(B \times L)$

The force which is due to the induced magnetic field is, $F=l(L \times B)$

Therefore, $F=\frac{E}{R} \times (L \times B)$

$F=\frac{v(B \times L)}{R} \times (L \times B)$

Here, force is directly proportional to the velocity of the two conducting plates.

Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.

6 0

3 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago