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svetoff [14.1K]

3 years ago

12

In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text

describes the typically accuracy of atomic clocks in terms of two atomic clocks differing by only one second in 6000 years. Assume one of the clocks gains one second in six thousand years, approximately what is the change in the number of oscillations of the light each second?

1 answer:

Aleks [24]3 years ago

5 0

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second , change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

= 4.859 x 10⁻² oscillations .

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If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?

astraxan [27]

Mass of the object is given as

$m = 1417 g = 1.417 kg$

now the speed of object is given as

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

here we know that

$d = 47 m$

$t = 90 s$

now we will have

$v = \frac{47}{90} = 0.52 m/s$

now we will have kinetic energy of the object as

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(1.417)(0.52)^2$

$KE = 0.19 J$

now the power is defined as rate of energy

so here we can find power as

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

$P = \frac{0.19}{90} = 2.14\times 10^{-3} W$

so above is the power used for the object

3 0

3 years ago

A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m

konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, $\gamma _0$ = 0.875

mass of the object in oil, $M_o$ = 0.013 kg

mass of the object in water, $M_w$ = 0.012 kg

let the mass of the object in air = $M_a$

weight of the oil, $W_0 = M_a - 0.013$

weight of the water, $W_w = M_a - 0.012$

The relative density of the oil is given as;

$\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg$

Therefore, the mass of the body is 0.02 kg.

6 0

3 years ago

Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi

Vedmedyk [2.9K]

Answer:

$v = 1.98\ m/s$

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m u₄ = 4.5 m/s

m₅ = m u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

$v = \dfrac{9.9}{5}$

$v = 1.98\ m/s$

5 0

3 years ago

ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of

sesenic [268]

Answer:

a) $V_{2/1}=0.8m/s$

b) The second runner will win

c) d = 10.54m

Explanation:

For part (a):

$V_{2/1} = V_{2} - V_{1} = 0.8m/s$

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

$t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s$

$t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s$

Since it takes less time to the second runner to cross the finish line, we can say the she won the race.

For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

X1 = V1*t2 = 239.46m Since the finish line was 250m away:

d = 250m - 239.46m = 10.54m

6 0

3 years ago

Read 2 more answers

Setler [38]

I think the answer is B.

Hope this helps.

5 0

3 years ago

Read 2 more answers