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3 years ago

Two enery converrsation that take place when you warm a cup of cocoa in a microwave Oven

1 answer:

6 0

1). Electrical energy from the wall outlet is converted to microwave (radio) energy, and to a little bit of kinetic energy to spin the turntable. 2). Microwave energy is converted to thermal energy in the cocoa.

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Energy transformations when you cook sausages on a campfire burning wood

Rina8888 [55]

<span>Actually in this case heat energy is being transferred. Heat energy or thermal energy is transferred from the burning of wood to the sausages for it to be cooked. The sausage is being heated by the fire and is absorbing the heat or thermal energy.</span>

8 0

3 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

3 years ago

Which best supports the idea that the surface of the moon has changed very little?

likoan [24]

The correct answer is<span> c) Because the moon has no atmosphere, it is not possible for geologic events to occur on the moon.

Atmosphere is related to geological change because of things such as wind and water erosion that over time change the landscape, and there's neither on the moon. It's just a floating rock.</span>

8 0

3 years ago

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin

faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance $L=0.450H=0.45\times 10^{-3}H$

Current in the inductor $i=1.90mA=1.90\times 10^{-3}A$

Voltage v = 13 volt

Inductive reactance of the circuit $X_l=\frac{v}{i}$

$X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm$

We know that

$X_l=\omega L=2\pi fL$

$2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10$

f = 2421.127 kHz

6 0

3 years ago

In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The

Ratling [72]

Answer:

Electric force, $F=-3.59\times 10^6\ N$

Explanation:

Given that,

Electric charge 1, $q_1=+40\ C$

Electric charge 2, $q_2=-40\ C$

Distance, $d=2\ km=2\times 10^3\ m$

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

$F=k\dfrac{q_1q_2}{d^2}$

k is the electrostatic constant

$F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}$

$F=-3.59\times 10^6\ N$

So, the electric force between these two sets of charges is $-3.59\times 10^6\ N$ .

5 0

3 years ago