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3 years ago

I have no clue what to do to find the problem.

Physics

1 answer:

oksano4ka [1.4K]3 years ago

7 0

kess dense material comes up. glycerin dencity= 37.8 / 30 syrup density = 80 / 60 so glycerin is less dense and comes up

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Electronic flash unit cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash l

Stolb23 [73]

Answer:

(A) 421 J energy stored in the capacitor for one flash.

(B) The value of capacitance is 0.0537 F

Explanation:

Given :

(A)

Time $t = \frac{1}{675}$

Average power $P = 2.7 \times 10^{5}$ W

From power equation,

$P= \frac{E}{t}$

So energy in one light is given by,

$E = Pt$

$E = 2.7 \times 10^{5} \times \frac{1}{675} = 400$ J

Since efficiency is 95 % so we can write, energy stored in one flash,

$E_{tot} = \frac{400}{0.95} = 421$ J

(B)

From the formula of energy stored in capacitor,

$E = \frac{1}{2}C V^{2}$

Where $E = E_{tot}$ and $V = 125$ V

$C = \frac{2E}{V^{2} }$

$C = \frac{2 \times 421}{15625}$

$C = 0.0537 F$

8 0

3 years ago

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

$EMF = 15.67 \times 10^{-3} V$

5 0

3 years ago

M = 30.3kg<br>M = 40.17kg 9<br>R = 0.5m<br>G = 6. 67x10^11<br>F ?

Lena [83]

Answer:

m¹=30.3kg

m²=40.17kg

R=0.5m

G=6.67*10¹¹

F=Gm¹m²/R²

=160.68

4 0

2 years ago

At what rate does gravity cause objects to accelerate at sea level on earth

VLD [36.1K]

The acceleration due to gravity on Earth is 9.8 m/s per second.

5 0

2 years ago

9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T

andrey2020 [161]

Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

4 0

3 years ago