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never [62]
3 years ago
5

EASY BRAINLIEST PLEASE HELP!!

Physics
2 answers:
Soloha48 [4]3 years ago
7 0

Answer:

Solution given:

frequency[f]=x

velocity[V]=15000m/s

wave length=59m

we have

wave length=\frac{V}{f}

59m=\frac{15000}{x}

x=\frac{15000}{59}=254.Hz

frequency=254Hz

seropon [69]3 years ago
6 0

Answer:

look at the picture i have sent

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ANTONII [103]
The correct answer is d. deterioration of brain tissue. A brain ventricle is a fluid filled structure located in the parenchyma. The fluid in the ventricles is called cerebro-spinal fluid. An increase in the volume of this fluid can lead to a condition known as hydrocephalus.
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Chemical energy stored in food cannot be transformed into mechanical energy true or false
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3 years ago
A heavy ball with a weight of 100 NN is hung from the ceiling of a lecture hall on a 4.4-mm-long rope. The ball is pulled to one
allochka39001 [22]

Answer:

175.3 N

Explanation:

The motion of the ball is a uniform circular motion, therefore the net force on it must be equal to the centripetal force.

There are two forces acting on the ball at the lowest point of motion:

- The tension in the string, T , upward

- The weight of the ball, mg, downward

The net force (centripetal force) has the same direction as the tension (upward, towards the centre of the circular path), so we can write:

T-mg=m\frac{v^2}{r}

where the term on the right is the expression for the centripetal force, and where:

T is the tension in the string

mg=100 N is the weight of the ball

m=\frac{mg}{g}=\frac{100}{9.8}=10.2 kg is the mass of the ball

v = 5.7 m/s is the speed of the ball at the lowest point

r = 4.4 m is the length of the rope, so the radius of the circle

Solving for T, we find the tension in the string:

T=mg+m\frac{v^2}{r}=(100)+(10.2)\frac{5.7^2}{4.4}=175.3 N

7 0
3 years ago
A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average
Maksim231197 [3]

Answer: 72.66 AU=1.089(10)^{10} km

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a of its orbit”:

T^{2}\propto a^{3} (1)  

Now, if T is measured in years (Earth years), and a is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:  

T^{2}=a^{3} (2)  

So, knowing T=619.36 years and isolating a from (2) we have:  

a=\sqrt[3]{T^{2}} (3)  

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Finally:

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4 0
3 years ago
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Answer:

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Hence To = 23N, attractive. C ans.

Thanks.

5 0
3 years ago
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