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Vladimir79 [104]
3 years ago
11

Mario golpea el balón con el pie para lanzárselo a Tamara que está situada a 18 m de distancia. El ángulo de salida del balón es

de 30° sobre la horizontal y la velocidad a la que sale el balón de la bota de Mario es de 15 mis. ¿A qué altura deberá poner el pie Tamara para hacer el control de la pelota que le envía Mario?
Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer:

y = 0.99 m

Explanation:

This is a projectile launching exercise, let's start by finding the components of the initial velocity, using trigonometry

         cos θ = v₀ₓ / v₀

         sin θ = v_{oy} / v₀

         v₀ₓ = vo cos θ

         v_{oy} = I go sin θ

         v₀ₓ = 15 cos 30 = 12.99 m / s

         v_{oy} = 15 sin 30 = 7.5 m / s

Let's find the time it takes to travel x = 18 m

         x = v₀ₓ t

         t = x / v₀ₓ

         t = 18 / 12.99

         t = 1,385 s

at this point it is at a height of

         y = v_{oy} - ½ g t²

         y = 7.5 1.385 - ½ 9.8 1.385²

         y = 0.99 m

therefore the camera must place the foot 99 cm from the ground

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Sholpan [36]

Given the potential energy of 100 J, the height of the hill is 0.128 m (12.8 cm)

Explanation:

The gravitational potential energy of an object is given by:

U=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object, relative to the ground

In this problem, we have

U = 100 J is the potential energy of the person

g=9.8 m/s^2 is the acceleration of gravity

m = 80 kg is the mass of the person

Solving for h, we find at what height the person is:

h=\frac{U}{mg}=\frac{100}{(80)(9.8)}=0.128 m = 12.8 cm

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5 0
3 years ago
A quasar is a distant celestial body in space. Investigators use a special telescope and determine that a certain quasar was giv
iris [78.8K]

Answer:

equipment for the measurement of microwave L bands with a range between 1 GHz and 2 GHz.

Explanation:

The electromagnetic spectrum can be calculated with the relationship between the speed of light, its wavelength and its frequency.

         c = λ f

For reasons of analysis and equipment used, it is artificially divided into ranges, with poorly defined limits and in some cases with overlaps between some.

For the case of analysis, f = 1.41 10⁹ Hz, we have the range called

* Microwave for f> 3 108 Hz to approximately f <3 1011 Hz

For the lower part of the frequency 3 10⁸ <f <3 10⁹ Hz we have UHF television channels and cell phones and military communications.

As the frequency observed by the researchers is in the UHF range, it is possible that they are using microwave equipment for communications, specifically equipment for the measurement of microwave L bands with a range between 1 GHz and 2 GHz.

6 0
4 years ago
Can a constant magnetic field set into motion an electron initially at rest? Explain your answer.
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No,  a constant magnetic field cannot set an electron initially at rest into motion

A force that accelerates a particle is necessary to change its velocity. The magnetic force is inversely proportional to the particle's speed. There cannot be a magnetic force acting on a moving particle, according to Einstein. A flux is a precise description of the greater-than-unity magnetic determine involving energy currents and magnet resources. The magnetic flux in a stage is actually selected apart from each some sort of route and also a degree (or durability); therefore, it is just a vector industry. The magnetic flux is usually defined as the Lorentz force that acts on moving galvanic costs.

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5 0
2 years ago
A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla
kotegsom [21]

Answer:

The power output of the oscillator is 0.350 watt.

Explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

P=\dfrac{1}{2}\times\mu\times\omega^2\times a^2\times v

Put the value into the formula

P=\dfrac{1}{2}\times A\times\rho\times\omega^2\times a^2\times\dfrac{\sqrt{T}}{\mu}

P=\dfrac{1}{2}\times3.14\times(0.0005)^2\times7850\times(2\times\pi\times57.0)^2\times(0.54\times10^{-2})^2\times\sqrt{\dfrac{5.7}{3.14\times(0.0005)^2\times7850}}

P=0.350\ Watt

Hence, The power output of the oscillator is 0.350 watt.

3 0
3 years ago
Which of the waves has the smallest amplitude?
balandron [24]

Answer:

I think its the Blue wave, im not sure so dont take my word for it.

Explanation:

3 0
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