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I am Lyosha [343]

2 years ago

9

Suppose a mass of 0.500 kg falls from 3050m. What potential energy is associated with the coin when its speed is 30.0 m/s?

1 answer:

mojhsa [17]2 years ago

7 0

Just want your points

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Find your mass if a scale on earth reads 650 N when you stand on it.

dezoksy [38]

Weight = (mass) x (acceleration of gravity) .

On Earth, acceleration of gravity is 9.8 m/s² (rounded) .

650 N = (mass) x (9.8 m/s²)

Divide each side by (9.8 m/s²): 650 N / 9.8 m/s² = mass

Mass = 66.3 kilograms (rounded)

7 0

3 years ago

Over time Pangaea broke apart to form other continents.

Travka [436]

Answer:

jwhgrewhuejqiwmkosjcdihwbfuqjiwdkmojcshidvwuf hiiii againnnn :)) good luck

6 0

2 years ago

In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla

Harlamova29_29 [7]

Answer:

$\lambda'=78.086\ nm$

Explanation:

Given:

wavelength of light in the air, $\lambda=530\times 10^{-9}\ m$

time taken to travel from the source to the photocell via air, $t=16.7\ s$

time taken to reach the photocell via air and glass slab, $t'=21.3\times 10^{-9}\ s$

thickness of the glass slab, $x=0.87\ m$

<u>Now we have the relation for time:</u>

$\rm time=\frac{distance}{speed}$

hence,

$t=\frac{d}{c}$

c= speed of light in air

$16.7\times 10^{-9}=\frac{d}{3\times 10^8}$

$d=16.7\times 10^{-9}\times 3\times 10^8$

$d=5.01\ m$

For the case when glass slab is inserted between the path of light:

$\frac{(d-x)}{c} +\frac{x}{v} =t'$ (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

$\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}$

$v=4.42\times 10^7\ m.s^{-1}$

Using Snell's law:

$\frac{\lambda}{\lambda'} =\frac{c}{v}$

$\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}$

$\lambda'=78.086\ nm$

5 0

2 years ago

A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge

Norma-Jean [14]

Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c) ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π× $10^{-7}$ ×0.024 / 2π(1.20× $10^{-2}$ ) × ln (0.024 + 0.012/0.012) × 120

solve it and we get current that is

current = 4 × $10^{-7}$ × 1.09861 × 120

current = 52.73 × $10^{-6}$ A

so here current in loops is 52.73 μA

8 0

2 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago