which type of graph shows data seperated into intervals? A. stemplot B. Line Graph C. Scatterplot D. Histogram

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

Elenna [48]

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.

8 0

3 years ago

Please help me with physics ks3!! Sound waves and hearding (pic incl)

Nina [5.8K]

Your answers for a), b), and c) are correct. Good work !

d). The high noise levels in Technology and PE are more of
a concern to the teachers than to the students because the
students are only in there for 1 or 2 periods a day, but the
Technology and PE teachers are in there ALL day.

e). Mr. Jones can't hear as high frequency as Jenny can because
he is much older than Jenny is. Sadly, even without damage due to
loud noise, the ability to hear high frequencies does decrease with age.

f). The wooden surfaces in the gym cause the gym to be louder
than it would be if it had carpet on the floor. Carpet ... and soft
walls and ceilings ... absorb a lot of the sound that hits them.
But hard surfaces don't absorb much of the sound that hits them,
so it just keeps bouncing around until it finally fades away.

You can see this easily ... just go into the gym at your school, clap
your hands once, and notice how long you keep hearing the sound
after you clap.

7 0

3 years ago

Read 2 more answers

A 1.15-kg mass oscillates according to the equation where x is in meters and in seconds. Determine (a) the amplitude, (b) the fr

ANEK [815]

The complete question is;

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when x = 0.360 m.

Answer:

A) Amplitude; A = 0.650 m

B) Frequency; f = 1.337 Hz

C) total energy = 17.142 J

D) Kinetic energy = 11.884 J

Potential Energy = 5.258 J

Explanation:

We are given;

Mass;m = 1.15 kg

Equation; x = 0.650 cos (8.40t)

(a) The standard form of a wave function is in the form y(x,t) = Asin(kx−ωt+ϕ)

So, comparing terms in our equation in the question to this, the amplitude is;

A = 0.650 m

(b) we know that formula for frequency is;

f = ω/2π

Again, comparing terms in the standard equation and our question, we can see that ω = 8.4

Thus;

f = 8.4/(2π)

f = 1.337 Hz

(c) Formula for the total energy is given by;

E = m•ω²•A²/2

Plugging in the relevant values, we have;

E = (1.15)(8.40)²(0.650)²/2

E = 17.142 J

(d) we want to find the kinetic energy and potential energy when x = 0.360 m.

The formula for kinetic energy in this case is given by;

K = (1/2)•m•ω²•(A² - x²)

Thus;

K = (1/2) × (1.15) × (8.40)² × ((0.650)² - (0.360)²)

K = 11.884 J

Also, the formula for the potential energy in this case is given by;

U = (1/2)•m•ω²•x²

Thus;

U = (1/2) × (1.15) × (8.40)² × (0.360)²

U = 5.258 J

3 0

3 years ago

A wood block is sliding up a wood ramp. if the ramp is very steep, the block will reverse direction at its highest point and sli

DanielleElmas [232]

<span>Answer: So it gets to the top of the ramp and stops. The parallel force pushing it down the ramp is mg sin Î¸, but for it to move, the frictional force must be overcome. This frictional force is ÎĽmg cos Î¸, where ÎĽ is the coefficient of static friction. For movement, then, mg sin Î¸ > ÎĽmg cos Î¸ ==> tan Î¸ > ÎĽ ==> Î¸ > arctan 0.5 = 26.565Â° ==> Î¸ = 27Â°</span>

5 0

3 years ago

Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

b

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV

Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$