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3 years ago

9

Consider the average speed of a runner who jogs around a track four times. The distance (400m) remains constant for each lap. Ho

wever, each lap is run 5 seconds slower than the first. The time for each lap increases. The average speed for each lap ______________. This is an example of a(n) _____________ relationship.
A) increases, direct
B) decreases, direct
C) increases, inverse
D) decreases, inverse

1 answer:

Yuki888 [10]3 years ago

7 0

D) decreases, inverse

Explanation:

The average speed of each lap decreases. This is an example of an inverse relationship.

Speed is the rate of change of distance with time. In calculating the average speed, we should understand that the total length of path is taken into consideration and so is the total time taken.

Average speed = $\frac{total distance}{total time}$

It was explained that after every 5 seconds the speed of the runner becomes slower.

There is definitely an inverse relationship between average speed and the total time taken.

An inverse relationship is such that as one variable increases, the other diminishes.

As average speed reduces the time taken increases.

learn more:

Speed brainly.com/question/1386181

#learnwithBrainly

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If a photon has frequency = 2.00 x 1014s-1 and the speed of light = 3.00 x 108ms-1, then what is its wavelength?

butalik [34]

Answer:

The photon has a wavelength of $1.5x10^{-6}m$

Explanation:

The speed of a wave can be defined as:

$v = \nu \cdot \lambda$ (1)

Where v is the speed, $\nu$ is the frequency and $\lambda$ is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

$c = \nu \cdot \lambda$ (2)

Where c is the speed of light.

Therefore, $\lamba$ $\lambda$ can be isolated from equation 2 to get the wavelength of the photon.

$\lambda = \frac{c}{\nu}$ (3)

$\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}$

$\lambda = 1.5x10^{-6}m$

Hence, the photon has a wavelength of $1.5x10^{-6}m$

<em>Summary: </em>

Photons are the particles that constitutes light.

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3 years ago

A pendulum on earth swings with angular frequency ω. On an unknown planet, it swings with angular frequency ω/ 4. The accelerati

Afina-wow [57]

Answer:

g / 16

Explanation:

T = 2π $\sqrt{\frac{l}{g} }$

angular frequency ω = 2π /T

= $\sqrt{\frac{g}{l} }$

ω₁ /ω₂ = $\sqrt{\frac{g_1}{g_2} }$

Putting the values

ω₁ = ω , ω₂ = ω / 4

ω₁ /ω₂ = 4

4 = $\sqrt{\frac{g}{g_2} }$

g₂ = g / 16

option d is correct.

6 0

3 years ago

A bullet is fired horizontally at a height of 1.3 meters and a velocity of 950 m/s. How long was the bullet in the air?

seropon [69]

Answer:

<em>The bullet was 0.52 seconds in the air.</em>

Explanation:

<u>Horizontal Motion </u>

It occurs when an object is thrown horizontally with a speed v from a height h.

The object describes a curved path ruled exclusively by gravity until it hits the ground.

To calculate the time the object takes to hit the ground, we use the following equation:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Note it doesn't depend on the initial velocity but on the height.

The bullet is fired horizontally at h=1.3 m, thus:

$\displaystyle t=\sqrt{\frac{2\cdot 1.3}{9.8}}$

$\displaystyle t=\sqrt{\frac{2.6}{9.8}}$

t = 0.52 s

The bullet was 0.52 seconds in the air.

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3 years ago

When a refrigerant enters the compressor, it is a ____ and when it leaves the compressor, it is a ____. A. low pressure low temp

saveliy_v [14]

Answer:

A

Explanation:

The compressor receive hot refrigerant and raises the pressure and temperature even further as it is send to the condenser.

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3 years ago

A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A

nadya68 [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The point charge is $Q_z = -0.0912 \ \mu C$

The inner shell is $Q_t = 0.4168 \ \mu C$

The outer shell is $Q_w = -0.6514 \ \mu C$

Explanation:

From the question we are told that

The inner radius of thin first spherical conducting shell is $r_1$

The outer radius of thin first spherical conducting shell is $r_2$

The inner radius of second thin spherical conducting shell is $R_1$

The outer radius of second thin spherical conducting shell is $R_2$

The magnetic flux for different region is $\phi = -10.3 *10^3 N\cdot m^2 /C \ for \ r < r_1$

The magnetic flux for first shell is $\phi = 36 * 10^3 N \cdot m^2 /C \ for \ r_2 < r$

The magnetic flux for second shell is $\phi = -36 * 10^3 N \cdot m^2 /C \ for \ r$

The magnitude of the point charge is mathematically represented as

$Q_z = \ \phi_z * \epsilon _o$

$Q_z = -10.3*10^{3} * 8.85 *10^{-12}$

$Q_z = -9.115*10^{-8} \ C$

$Q_z = -0.0912 \ \mu C$

Considering the inner shell

$Q_a = \phi_a * \epsilon _o$

=> $Q_a = 36 .8 * 10^3 * 8.85*10^{-12}$

$Q_a = 32.56*10^{-8} \ C$

$Q_a =0.326} \ \mu C$

Charge on the inner shell is

$Q_t = Q_a - Q_z$

$Q_t = 0.326} \ \mu - ( -0.0912 \ \mu)$

$Q_t = 0.4168 \ \mu C$

Considering the outer shell

$Q_y = \phi_y * \epsilon_o$

=> $Q_y = -36.8 *10^{3} * 8.85*10^{-12}$

$Q_y = -32.56*10^{-8} \ C$

$Q_y = - 0.326} \ \mu C$

Charge on the outer shell is

$Q_w = Q_y - Q_z$

$Q_w =- 0.326} \ \mu - ( -0.0912 \ \mu)$

$Q_w = -0.6514 \ \mu C$

5 0

3 years ago