All categories

3 years ago

You might use it to test how cold the pool water is

Physics

1 answer:

ozzi3 years ago

3 0

You<span> should </span>test<span> FC and PH as soon as </span>you<span> take the sample</span>

You might be interested in

The temperature of water in a beaker is 45°C. What does this measurement represent?

Tasya [4]

Answer:

the average kinetic energy of water particles - C.

8 0

2 years ago

A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the

UkoKoshka [18]

Answer:

Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, $KE = \frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

$1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg$

So, mass of bike = 38 kg.

3 0

3 years ago

A dog has a mass of 20 kg. If the dog is pushed across the ice with a force of 40 N, what is its acceleration?

olasank [31]

Answer:

The acceleration is 2 m/s2.

Explanation:

We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:

F = m x a ---> a= F/m

a = 40 N/20 kg <em> 1N= 1 kg x m/s2</em>

a= 40 kgx m/s2/ 20 kg

7 0

3 years ago

A child sits on the edge of a spinning merry go round that has a radius of 1.5 . The child’s speed is a 2m/s. What is the child’

Leto [7]

Answer: 2.67 m/s^2

Explanation:

Centripetal acceleration is defined as v^2/r; in this case, it's 2^2/1.5, which is 2.67.

3 0

2 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago