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drek231 [11]
3 years ago
5

An observer measures the length (L), width (w), and height (h) of a box while stationary relative to the box. The observer then

travels at near light speeds parallel to the length (L) of the box. If the observer measures the width (w) of the box while moving, the measured value will now be:________
a. equal to w.
b. less than w.
c. greater than w.
d. information is needed. You need to know the density of the box.
Physics
1 answer:
Elenna [48]3 years ago
8 0

Answer:

b. less than w.

Explanation:

In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).

This is supported by the length contraction equation below:

L = L_0\sqrt{1-\frac{v^2}{c^2} }

Here, L is the observed length

L_0 is the original length of the object

v is the relative speed between the object and the observer

and c is the speed of light

Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0

This effectively decreases the length that is observed.

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What is heart sound?
Dafna11 [192]

Answer:

its the sound that a heart produces when beating, this can help doctors detect abnormalities

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3 years ago
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A psychologist is interested in exploring the effect tutorial support on students academic performance and assign students in to
NikAS [45]

Answer:

The dependent variable is academic performance

The independent variable is the presence/absence of tutorial support

The control group are students who did not get the tutorial support.

The experimental group were students that got the tutorial support

Explanation:

In every experiment, there is a dependent and independent variable as well as an experimental and a control group.

The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.

The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.

The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).

5 0
3 years ago
True or False:
bekas [8.4K]
I think it false. Sorry if i'm wrong.

5 0
3 years ago
Read 2 more answers
2.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test
kow [346]

Answer:

0.15625 grams

Explanation:

Half life: It is related to the decay of radioactive material. The duration in which  half of the material will be degraded/decayed. That means after half life 50% of the radioactive material will be left. Here the half life is 28 years.

Initial quantity of the sample: 2.5 grams.

After 28 years, the leftover quantity = 1.25 grams

After 56 years, the leftover quantity = 0.625 grams

After 84 Years, the leftover quantity = 0.3125 grams

After 112 years, the leftover quantity = 0.15625 grams

5 0
3 years ago
Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long
nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

R = \dfrac{\rho L }{A}

where;

A = πr²

R = \dfrac{\rho L }{\pi r ^2}

For the shorter cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{ 4 \rho L }{\pi \ D   ^2}

For the longer cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{32\rho L }{\pi \ (4 D)   ^2}

R = \dfrac{2\rho L }{\pi \ (D)   ^2}

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D   ^2}}{ \dfrac{2\rho L }{\pi \ (D)   ^2}}

\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D   ^2}}* { \dfrac  {\pi \ (D)   ^2} {2\rho L}}

\dfrac{R_s}{R_L} =2

{R_s}=2{R_L}

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0
3 years ago
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