All categories

3 years ago

If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?

Physics

1 answer:

Agata [3.3K]3 years ago

7 0

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy $mgh$ , and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature ${\Delta}T$ then will be

${\Delta}T=\frac{{\Delta}Q}{mc} .$

Where m is the mass of water in the pool, c is the specific heat capacity of water, and ${\Delta}Q$ is the added heat which in this case is the energy of the diver.

Since we do not know the mass of the water in the pool, we cannot make this calculation.

You might be interested in

. For transverse waves, compare the direction that the particles move to

Elza [17]

Answer:

The particles in transverse waves move perpendicular to the direction of how the wave moves.

Explanation:

5 0

3 years ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

a) How many parsecs are there in one astronomical unit?

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

(e) How many light-years in a parsec?

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

3 years ago

Q 1: Calculate the pressure.

Oksana_A [137]

Answer:

1: 300pa

2: 10cm2

3: 430pa

4: a: 1.6667pa

b:2.5m2

c:20pa

Explanation:

7 0

3 years ago

When magma cools quickly, what kind of texture or

Ksivusya [100]

Answer:

Explanation:

when magma cools Crystal's form because the solution is super saturated with respect to some minerals if the magma cools quickly the crystals do not have much time to form hence they are small and also the resulting rock is fine grained

6 0

3 years ago

Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is

s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: $\frac{∆v}{t}$

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 $\frac{feet}{second}$ then sped up to 5 $\frac{feet}{second}$ , then the change in my speed would be 2 $\frac{feet}{second}$ , because I started walking 2 $\frac{feet}{second}$ faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 $\frac{meters}{second}$ to 16 $\frac{meters}{second}$ in 4 seconds. What would the acceleration be? Let's set up an equation:

a = $\frac{∆v}{t}$

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 $\frac{meters}{second}$ by 4 $\frac{meters}{second}$ . That makes sense, right? Back to the equation.

a = $\frac{∆v}{t}$
a = $\frac{16-4}{4}$

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = $\frac{12}{4}$

a = 3 ( $\frac{meters}{second^{2}}$ )

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit $\frac{\frac{meters}{second}}{second}$ , which can be simplified down to $\frac{meters}{second^{2} }$ )

Let me know if you need clarification on anything I explained here!
- breezyツ

6 0

2 years ago