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gogolik [260]
3 years ago
15

Describe an original example of energy changing from one form to two other forms

Physics
1 answer:
irinina [24]3 years ago
8 0
1).  an electric motor running
Electrical energy is changing into kinetic energy and a little bit of heat

2).  light a match
The chemical energy stored in the match head changes into light and heat energy.

3).  a light bulb
Electrical energy is changed into light and heat energy.
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Two bulbs are connected in parallel across a source of EMF = 8.0V with a negligible internal resistance. One bulb has a resistan
Sonja [21]
<h2>Answer:</h2>

(a) 3.18Ω

(b) 3.18Ω

<h2>Explanation:</h2>

Let the two bulbs be A and B

Given;

R_{A} = Resistance in bulb A = 3.0Ω

R_{B} = Resistance in bulb B = 2.5Ω

Since the two bulbs are connected in parallel;

i. their effective resistance (R_{X}) is given by

\frac{1}{R_{X}} = \frac{1}{R_{A} } + \frac{1}{R_{B} }  ---------------(i)

Substitute the values of R_{A} and R_{B} into equation (i)

=> \frac{1}{R_{X}} = \frac{1}{3.0} + \frac{1}{2.5}

Solve for  R_{X}

R_{X} = 1.36Ω

ii. voltage (potential difference), V, across them is the same;

Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.

Use the Ohm's law;

V = I x R    -----------------(ii)

Where;

V = voltage across them = 2.4V

I = total current flowing through them

R = their effective resistance = R_{X} = 1.36Ω

Substitute these values into equation (ii) as follows;

2.4  = I x 1.36

I = 2.4 / 1.36

I = 1.76A

(a) Now get the value of R

Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.

Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.

Therefore, to get the value of R, we use the relation

V = I x R   ------------------------------(iii)

Where;

V = voltage across the resistor = 5.6V

I = current through the resistor = 1.76A

<em>Substitute these values into equation (iii)</em>

=> 5.6 = 1.76 x R

=> R  = 5.6 / 1.76

=> R = 3.18Ω

Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω

(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω

3 0
3 years ago
A 55 kg roller skater is at rest on a flat skating rink, a 198 N horizontal force is needed to set the skater in motion.
Rufina [12.5K]

Answer:

Explanation:

To get the person Moving you have to overcome the static (means not moving) friction coefficient.  U(static)

To get the person going at the same speed you have to overcome the kinetic friction coefficient. U(Kinetic)

Force to get him moving is 198 N.   Force = ma = U(static)Mg

combining the 2 equations you get 198N = U(static)* 55kg *9.8m/s^2   Solve for U(static)

Same equation to keep him moving except with the dynamic force and the dynamic U

 

175N=  U(kinetic)*55kg*9.8m/s^2  Solve (U dynamic)

8 0
3 years ago
Prove that..<br>please help<br>​
GaryK [48]

\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}

Also,

\large{ \longrightarrow{ \rm{ F \propto  \dfrac{1}{ {d}^{2} } }}}

Combining both, We will get

\large{ \longrightarrow{ \rm{F  \propto  \dfrac{ m_1 m_2}{ {d}^{2}}}}}

Or, We can write it as,

\large{ \longrightarrow{ \rm{F  \propto  \:  G \dfrac{ m_1 m_2}{ {d}^{2} }}}}

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0
3 years ago
A 30-kg child sits at the top of a 3-meter slide. After sliding down, the child is traveling 4 m/s. How much PE does he start wi
Jobisdone [24]
At the top:

         Potential Energy = (mass) x (gravity) x (height)

                                       = (30 kg) x (9.8 m/s²) x (3 meters)

                                       =      882 joules

At the bottom:

           Kinetic Energy  =  (1/2) x (mass) x (speed)²

                                       = (1/2) x (30 kg) x (3 m/s)²

                                       =        (15 kg)  x  (9 m²/s²)

                                       =              135 joules .

He had  882 joules of potential energy at the top,
but only  135 joules of kinetic energy at the bottom.

Friction stole  (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.
6 0
3 years ago
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
3 years ago
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