To remove 800j of heat the compressor in the fridge does 500j of work. how much heat is released into the room?

Answers

Answer 1

Answer:

Answer:

Heat released into the room = 1300 J

Explanation:

CONCEPT:

According to second law of thermodynamics , heat cannot flow from a lower temperature to a higher temperature.But the refrigerator transfers heat from lower to higher temperature .For this , we have to do work on the refrigerator.

This work is used to transfer heat from lower to higher temperature.

Heat removed by fridge = 800 J

Work done = 500 J

Heat released into the room = Heat removed + work done

Heat released into the room = 800 +500

Heat released into the room = 1300 J

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A baseball leaves the bat with a speed of 40 m/s at an angle of 35 degrees. A 12m tall fence is placed 130 m from the point the ball was struck. Assuming the batter hit the ball 1m above ground level, does the ball go over the fence? If not, how does the ball hit the fence? If yes, how far beyond the fence does the ball land?

Answers

Answer:

The ball land at 3.00 m.

Explanation:

Given that,

Speed = 40 m/s

Angle = 35°

Height h = 1 m

Height of fence h'= 12 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

$V_(x)=V_(i)\cos\theta$

$V_(x)=40*\cos35$

$V_(x)=32.76\ m/s$

We need to calculate the time

Using formula of time

$t = (d)/(v)$

$t=(130)/(32.76)$

$t=3.96\ sec$

We need to calculate the vertical velocity

$v_(y)=v_(y)\sin\theta$

$v_(y)=40*\sin35$

$v_(y)=22.94\ m/s$

We need to calculate the vertical position

Using formula of distance

$y(t)=y_(0)+V_(i)t+(1)/(2)gt^2$

Put the value into the formula

$y(3.96)=1+22.94*3.96+(1)/(2)*(-9.8)*(3.96)^2$

$y(3.96)=15.00\ m$

We need to calculate the distance

$s = y-h'$

$s=15.00-12$

$s=3.00\ m$

Hence, The ball land at 3.00 m.

A toy airplane is flying at a speed of 6 m/s with an acceleration of 0.3 m/s2How fast is it flying after 4 seconds?
A. 5.7 m/s
B. 2.6 m/s
C. 13.9 m/s
D. 7.2 m/s
SUBMIT

Answers

Answer:

Explanation:

0.3=v-6/4

make v the subject

v=7.2ms^2

For a certain optical medium the speed of light varies from a low value of 1.80 × 10 8 m/s for violet light to a high value of 1.92 × 10 8 m/s for red light. Calculate the range of the index of refraction n of the material for visible light.

Answers

Answer:

1.56 - 1.67

Explanation:

Refractive index of any material is given as the ratio of the speed of light in a vacuum to the speed of light in that medium.

Mathematically, it is given as:

n = c/v

Where c is the speed of light in a vacuum and v is the speed of light in the medium.

Given that the speed of light in the optical medium varies from 1.8 * 10^8 m/s to 1.92 * 10^8 m/s, we can find the range of the refractive index.

When the speed is 1.8 * 10^8 m/s, the refractive index is:

n = (3 * 10^8) / (1.8 * 10^8)

n = 1.67

When the speed is 1.92 * 10^8 m/s, the refractive index is:

n = (3 * 10^8) / (1.92 * 10^8)

n = 1.56

Therefore, the range of values of the refractive index of the optical medium is 1.56 - 1.67.

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing?
b. is the electric field in the direction of the current or opposite to the current?

Answers

Answer:

$(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The direction of the electric field is opposite that of the current

Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = (d)/(2)$

$r = (0.03)/(2)$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^(-4 ) \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o( d \phi )/(dt )$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

$\mu_o I + \epsilon_o \mu _o( d \phi )/(dt ) = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^(-12 ) \ m^(-3) \cdot kg^(-1)\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^(-7) N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

$\mu_o I + \epsilon_o \mu _o ( d [EA] )/(dt ) = 0$

=> $(dE)/(dt) =- (I)/(\epsilon_o * A )$

=> $(dE)/(dt) =- (17)/(8.85*10^(-12) * 7.07*10^(-4) )$

=> $(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

A javelin is thrown in the air. Its height is given by ( ) 1 2 8 6 20 h x x x = − + + , where x is the horizontal distance in feet from the point at which the javelin is thrown. a. How high is the javelin when it was thrown? b. What is the maximum height of the javelin? c. How far from the thrower does the javelin strike the ground?'

Answers

The maximum height, the location on the ground and the initial vertical height of the javelin is required.

The initial height of the javelin is 6 feet.

The maximum height of the javelin is 326 feet.

The javelin strikes the ground at 160.75 feet.

The given equation is

$h(x)=-(1)/(20)x^2+8x+6$

where $x$ is the horizontal distance

At $x=$ we will get the initial vertical height.

$h(0)=-(1)/(20)*0+8* 0+6\n\Rightarrow h(0)=6$

Vertex of a parabola is given by

$x=-(b)/(2a)=-(8)/(2* -(1)/(20))\n\Rightarrow x=80$

$h(80)=-(1)/(20)(80)^2+8* 80+6=326$

At $h(x)=0$ the javelin will hit the ground

$0=-(1)/(20)x^2+8x+6\n\Rightarrow -x^2+160x+120=0\n\Rightarrow x=(-160\pm √(160^2-4\left(-1\right)*120))/(2\left(-1\right))\n\Rightarrow x=-0.75,160.75$

Learn more about parabolas from:

brainly.com/question/14477557

This question is incomplete, the complete question is;

A javelin is thrown in the air. Its height is given by h(x) = -1/20x² + 8x + 6

where x is the horizontal distance in feet from the point at which the javelin is thrown.

a. How high is the javelin when it was thrown?

b. What is the maximum height of the javelin?

c. How far from the thrower does the javelin strike the ground?'

Answer:

a. height of the javelin when it was thrown is 6 ft

b. the maximum height of the javelin is 326 ft

c. distance from the thrower is 160.75 ft

Explanation:

Given h(x) = -1/20x² + 8x + 6

we determine the height when x = 0

h(0) = -1/20(0)² + 8(0) + 6 = 6 ft

therefore height of the javelin when it was thrown is 6 ft

to determine the maximum height of the javelin;

we find the vertex of the quadratic

h = - [ 8 / ( 2(-1/20) ) ] = 80

therefore

h(80) = -1/20(80)² + 8(80) + 6

= -320 + 640 + 6 = 326 ft

therefore the maximum height of the javelin is 326 ft

Now the thrower is at the point ( 0,0 ) and the javelin comes down at another point ( x,0 )

this is possible by calculating h(x) = 0

⇒ -1/20x² + 8x + 6 = 0

⇒ x² - 160x - 120 = 0

⇒ x = [ -(-160) ± √( (-160)² - 4(1)(-120) ) ] / [ 2(1) ]

x = [ 160 ± √(25600 + 480) ] / 2

[x = 160.75 ; x = -0.75 ]

distance cannot be Negative

therefore distance from the thrower is 160.75 ft

A ship is traveling at 154 m/s and accelerates at a rate of 1.80 m/s^2 for 1 minute. What will its speed be after that minute? Calculate the answer in both meters per second and kilometers per hour.

Answers

Given:

u(initial velocity): 154 m/s

accelerates (a): 1.8 m/s^2

t= 1 min=60 secs

Now we know that

s= ut + 1/2(at^2)

s= 154 x 60 + (1.8 × 60 ×60) ÷ 2

s= 12,480 m

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