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3 years ago

A graduated cylinder.measures 15.3 mL. Convert this measurement to DaL

Physics

1 answer:

ololo11 [35]3 years ago

5 0

Answer:

0.000153DaL

Explanation:

We have been given:

15.3mL to convert to DaL

DaL is a unit of volume which indicates a decaliter.

This implies that;

1 Da L = 1 x 10²L

So:

1 mL = 1 x 10⁻³L

So 15.3mL will give 15.3 x 10⁻³L

So;

1 x 10²L = 1 DaL

15.3 x 10⁻³L will give $\frac{15.3 x 10^{-3} }{1 x 10^{2} }$ = 15.3 x 10⁻⁵DaL

Therefore, this is 0.000153DaL

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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

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

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

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

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

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3 years ago

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ryzh [129]

Answer:

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1 year ago