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RUDIKE [14]
3 years ago
6

A 250 g beach ball rolls across the sand with a speed of 11.16 km/h. First convert units to kg and m/ then determine the momentu

m of the ball
Physics
1 answer:
yuradex [85]3 years ago
6 0

Answer:

0.775 kg-m/s

Explanation:

Convert the units to the right unit forms necessary

250 g -> 0.25 kg

11.16 km/h -> 3.1 m/s

Now use the formula:

                         velocity

                   mass /  

momentum  /    /

       \           /    /

         \       /    /

          p = mv

p = 0.25 × 3.1 = 0.775 kg-m/s

Hope this helps you!

Bye!

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Determine the magnitude and direction of the resultant force of the following free body diagram.
Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force (\vec F), in newtons, by vectorial sum:

\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j} (1)

\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}

\|\vec F\| \approx 599.923\,N

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)

Where \theta is the direction of the resultant force, in sexagesimal degrees.

\theta \approx 36.405^{\circ}

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0
3 years ago
Although centuries ago, astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we kn
Nutka1998 [239]

Astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we know that it is as a binary star system.

<h3>What is the binary star system about?</h3>

A binary star system is known to be one where one star is known to be called a white dwarf and there is a mass that is said to be transferred to it

A binary star is known to be a kind of a system that is composed of two stars that are known to be gravitationally held together to and in orbit near each other.

Note that Binary stars in the night sky are ones that are often seen as a single object and thus Astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we know that it is as a binary star system.

Learn more about astronomers  from

brainly.com/question/1141458

#SPJ1

5 0
2 years ago
Time period from one noon to the next
matrenka [14]
That's the period of time known as one solar "day". We subdivide it into 24 slices which we call "hours". Using this system of time units, the day is about 4 minutes longer than one complete axial rotation of the Earth.
3 0
3 years ago
Find the resistance of wire of<br>0.65m Radius 0.25<br>and<br>resistivity 3x10-6 OHM<br>​
densk [106]

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

<u>Given the following data;</u>

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

Resistance = P \frac {L}{A}

Where;

  • P is the resistivity of the material.
  • L is the length of the material.
  • A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}}

Resistance = 3 * 10^{-6} * 3316326.531

<em>Resistance = 9.95 Ohms </em>

5 0
3 years ago
The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
3 years ago
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