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3 years ago

5

If we put a convex lens made by crown glass in Carbondia sulphide what will happen?

1 answer:

7nadin3 [17]3 years ago

4 0

Answer:

I don't knowI think it will break

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78. A particle moves along the x- axis. The velocity of the particle at time tis given by 4vt()=3 t +1 . If the position of the

RoseWind [281]

Explanation:

Below is an attachment containing the solution

3 0

3 years ago

Assume we’re able to travel to your planet and decide to take some fireworks with us to celebrate our journey.

julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be $\frac{1876.8}{g}$

Explanation:

Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.

writing equation of motion in vertical direction:

$v_{y}=u_{y}+(-g) t$

$u_{y}= u\sin \phi$

and $v_{y}=0$

therefore $\frac{u\sin \phi }{g} =t$

writing equation of motion in horizontal direction:

$s_{x}=u_{x}t$

$u_{x} = u\cos \phi$

therefore the equation becomes $s_{x}=\frac{u^{2} \sin \phi \cos \phi}{g}$

therefore horizontal distance traveled = $\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}$

5 0

3 years ago

How long will it take Matthew to jog 321 m at 5.8m/s?

Xelga [282]

Give the other person brainest I was gonna answer but found no need if you get me

8 0

3 years ago

Please help me out as soon as you can. Really need help.

olga nikolaevna [1]

The answer would be 6 because 2.0x3= 6

(newton’s 2nd law)

mark me brainliest

4 0

3 years ago

In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

3 0

3 years ago