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4 years ago

8

An astronaut in an inertial reference frame measures a time interval Δt between her heartbeats. What will observers in all other

inertial reference fr

1 answer:

Reika [66]4 years ago

3 0

Answer:

It will be more than deta t

Explanation:

Because

deta t' = န deta t

But န= 1/√ (1 - v²/c²

So the observers in all the initial frames will be more than deta t

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The _______ is a factor that's affected by changes in the independent variable. A. dependent variable B. hypothesis C. experimen

Irina18 [472]

A. dependent variable

As the the independent variable (i.e. number of cats being sold) increases, the dependent variable (i.e. money made) will also increase. You can have 1,000,000 cats for sale and make no money, but you cannot make money without having sold some cats first.
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3 years ago

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The forces F 1, F 2, ..., F n acting at the same point P are said to be in equilibrium if the resultant force is zero, that is,

allochka39001 [22]

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3 years ago

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A man rides up in an elevator at 12 m. He gains 6500 J of gravitational potential energy. what is the man's mass?

iren2701 [21]

Answer:

We know that potential energy of a body;

= mass(m)× gravitational acceleration(g) × height(h)

Lets find out the mass of the body

P.E. = mgh

=> 6500J = mass × 9.8m/s^2 × 12m

=>6500J = mass × ( 9.8 × 12 ) × ( m/s^2 × m)

=> 6500 Nm = m × 117.6 × m^2 / s^2

=> 6500/117.6 Ns^2/m = mass [°.° Ns^2/m = kg]

=> 55.272 Kg = mass

Therefore the mass of the body = 55.272 kg ~ <em>6</em><em>0</em><em> </em><em>k</em><em>g</em><em> </em>(Ans)

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3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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3 years ago