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xxMikexx [17]
3 years ago
13

A kettle transfers 1500 J of energy, 1200 J to a thermal energy store and 300 J to a vibrational energy store (sound). How effic

ient is the kettle?
Physics
1 answer:
il63 [147K]3 years ago
7 0

Answer: 0.8

Explanation:

The efficiency is defined as the ratio of total imput energy that is actually utilized to the end of the device.

So if we have a total transfer of 1500j of energy, and 1200j are used to heat the thing inside the kettle, 300j are not used to the actual function of the kettle.

So the efficiency is n = 1200j/1500j = 0.8

This means that a 80% of the energy imput is actually used to heat the kettle.

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An electron moving with a velocity = 5.0 × 10 7 m/s enters a region of space where perpendicular electric and a magnetic fields
inna [77]

Answer:

The magnetic field is 2 \times 10^{-4} T

Explanation:

Given:

Velocity of electron v = 5 \times 10^{7} \frac{m}{s}

Electric field E = 10^{4} \frac{V}{m}

The force on electron in magnetic field is given by,

 F = qvB \sin \theta                      ......(1)

The force on electron in electric field is given by,

 F = qE                               ......(2)

Compare both equation,

   qE = qvB \sin \theta

Here \sin \theta = 1

  E = vB

  B= \frac{E}{v}

  B = \frac{10^{4} }{5 \times 10^{7} }

  B = 2 \times 10^{-4} T

Therefore, the magnetic field is 2 \times 10^{-4} T

5 0
3 years ago
As time goes on, the ENTROPY in a closed system should increase. This is because of which Law?
olga_2 [115]

Answer:

The answer is D

The second law of thermodynamics

5 0
3 years ago
while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w
Olenka [21]
A: 132.9w because 2525\19 is how much energy transferred per second which is also known as the power
7 0
4 years ago
Read 2 more answers
Explain why the types of technology valued can vary.
Elan Coil [88]
Over time, the types of technology can vary and be improved upon so that more advanced techniques become more valued. This could be the situation with mining whereby back in the 1500's in underground mines the rock was broken by fire setting ie lighting a fire below the rock face to heat up the rock and then throwing cold water on it to crack it, so that it could be dug by hand. With the advent of explosives, this all changed so that the rock could be blasted. The increase in advance rates for an underground heading have thus gone from 5-20 feet per month to up to 300meters (984 ft) per month for a 24/7 mining operation, which is a huge improvement.
4 0
3 years ago
Read 2 more answers
I NEED HELP PLEASE, THANKS! :)
mrs_skeptik [129]

Answer:

1. Largest force: C;  smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

F=K\dfrac{ q_{1}q_{2}}{r^{2}}

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write  

F=\dfrac{k}{r^{2}}

1. Net force on each particle

Let's

  • Call the distance between adjacent charges d.
  • Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

F_{A} : F_{B} : F_{C} : F_{D}  =  \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\

2. Ratio of largest force to smallest

\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}

7 0
3 years ago
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