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3 years ago

Using the rules for significant figures, what do you get when you subtract 15.54 from 508.9538?

1 answer:

4 0

According to the rules for significant figures, simple decimal subtraction is done. But since 15.54 is only accurate to the hundredths place, then the resulting difference should also be rounded off to the nearest hundredths. This is shown below:

508.9538 - 15.54 = 493.4138

Rounding off to the nearest hundredths:

= 493.41

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3 years ago

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

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2 years ago

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puteri [66]

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4 years ago

A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor

Lorico [155]

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

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3 years ago