Mike took less time to run the 100 meters than mitchell what does this tell us about his speed

The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the poo

DedPeter [7]

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

$n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m$

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

4 0

3 years ago

The single-pole, single-throw switch is normally wired ? between the source and the load to turn devices on and off.

zzz [600]

In series.

Single-pole and single-throw switch:

A switch with only one input and one output is referred to as a Single Pole Single Throw (SPST) switch. This indicates that it has a single output terminal and a single input terminal.

A single pole, one throw switch functions as an on/off switch in circuits. The circuit is turned on when the switch is closed. The circuit is shut off when the switch is open.

Thus, SPST switches are relatively basic in design.

Circuit for a single-pole, single-throw (SPST) switch

Types:

According to the application, it can be divided into three categories, including:

Simple SPST

(ON)-OFF, Push-to-close, SPST Momentary

ON-(OFF), Push-to-Open, SPST Momentary

Inches Switch SPST

Learn more about terminal here:

brainly.com/question/14236970

#SPJ4

7 0

2 years ago

A balloonist happens to drop his metal ballpoint pen from his balloon as he is taking notes on his flight. Because his pen has a

Lerok [7]

So this is dealing with the conservation of energy. So you set kinetic energy equal to potential energy, so it looks like this:

1/2mv^2=mgh. The m's cancel out, so it is 1/2v^2=gh.

To find out what the height h is, divide g on both sides, so...

h=0.5v^2/g. v=22m/s, g=9.81m/s^2, so h=(0.5)(22^2)/(9.81)=24.67m

5 0

3 years ago

At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. thr

svp [43]

The acceleration of the particle at time t is:
$a(t)=24t^2 ft/s^2$
The velocity of the particle at time t is given by the integral of the acceleration a(t):
$v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s$
and the position of the particle at time t is given by the integral of the velocity v(t):
$x(t)=\int v(t) = \int (8t^3)=8 \frac{t^4}{4}=2t^4 ft$

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
$x(2 s)=2 t^4 = 2 (2s)^4=32 ft$

5 0

3 years ago

When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre wi

mixer [17]

Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

5 0

3 years ago