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coldgirl [10]
3 years ago
7

Explain with reason that when sunlight is focused on a piece of paper with the help of a convex len the paper will burn.​

Physics
1 answer:
Sloan [31]3 years ago
8 0

Answer:

Convex lenses concentrates the light energy to one spot on the paper so that the heat energy accumulates on that one small spot of paper. ... As the heat increases, combustion will occur when the spot becomes too hot and the paper will burn.

Explanation:

Hope it helps!

If you dont mind can you please mark me as brainlest?

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How to calculate this operation? m=10kg and L=2500J/kg? What is the Energy?
lawyer [7]

el espanol tacos fiesta me amigos sombrero el salsa report me right the ⨍⩂Čk now

6 0
3 years ago
g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor
Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1,  and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0
3 years ago
A spaceship starting from a resting position accelerates at a constant rate of 9.8 meters per second per second. How long and ho
Verizon [17]

Accelerating at 9.8 m/s² means that every second, the speed is 9.8 m/s faster than it was a second earlier.  It's not important to the problem, but this number (9.8) happens to be the acceleration of gravity on Earth.

1% of the speed of light = (300,000,000 m/s) / 100 = 3,000,000 m/s .

Starting from zero speed, moving (9.8 m/s) faster every second,
how long does it take to reach  3,000,000 m/s ?

           (3,000,000 m/s) / (9.8 m/s²)  =  306,122 seconds .
                                                   (That's  5,102 minutes.)
                                                        (That's  85 hours.)
                                                     (That's  3.54 days.)

Speed at the beginning . . . zero .
Speed at the end . . . 3,000,000 m/s
Average speed . . . . . 1,500,000 m/s

Distance = (average speed) x (time)

               = (1,500,000 m/s) x (306,122 sec) = 4.592 x 10¹¹ meters

                                                                     =  459 million kilometers

                         That's like from Earth
                                                  to       Sun
                                                             to    Earth
                                                                    to        Sun. 

4 0
3 years ago
If a m = 74.7 kg m=74.7 kg person were traveling at v = 0.800 c v=0.800c , where c c is the speed of light, what would be the ra
igomit [66]

Answer:

\frac{E}{E_c} =3.125

Explanation:

The kinetic energy of a rigid body that travels at a speed v is given by the expression:

E_c=\frac{1}{2} mv^2

The equivalence between mass and energy established by the theory of relativity is given by:

E=mc^2

This formula states that the equivalent energy E can be calculated as the mass m multiplied by the speed of light c squared.

Where c is approximately 3\times 10^{8} m/s

Hence:

E_c=\frac{1}{2} (74.7)*(0.8*3\times 10^{8} )^2=2.15136\times 10^{18} J

E=(74.7)*(3\times 10^{8} )^2 =6.723\times 10^{18} J

Therefore,  the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy is:

\frac{E}{E_c} =\frac{6.723\times 10^{18}}{2.15136\times 10^{18}} =3.125

4 0
3 years ago
A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90
svetoff [14.1K]

Answer:

\tau=22.13Nm

Explanation:

information we have:

mass: m=0.9kg

lenght: L=0.95m

frequency: f=2.6rev/s

time: t=0.2s

and from the information we have we can calculate the angular velocity \omega. which is defined as

\omega=2\pi f

\omega=2\pi (2.6rev/s)\\\omega=16.336 rev/s

----------------------------

Now, to calculate the torque

We use the formula

\tau=I \alpha

where I  is the moment of inertia and \alpha is the angular acceleration

moment of inertia of a uniform rod about the end of it:

I=\frac{1}{3}mL^2

substituting known values:

I=\frac{1}{3} (0.9kg)(0.95m)^2\\I=0.271kg/m^2

for the torque we also need the acceleration \alpha which is defined as:

\alpha=\frac{\omega}{t}

susbtituting known values:

\alpha=\frac{16.336rev/s}{0.2s} \\\alpha=81.68rev/s^2

and finally we substitute I and  \alpha  into the torque equation \tau=I \alpha:\tau=(0.271kg/m^2)(81.68rev(s^2)\\\tau=22.13Nm

5 0
4 years ago
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