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Alex Ar [27]
3 years ago
8

un futbolista patea una pelota que se encuentra en el pasto con un angulo de 30° (medido desde la horizontal) con la intención d

e hacer un gol en un arco que se encuentra a 30m desde su posición. si la altura del arco es de 2m (medido desde el pasto al travesaño) y el jugador patea directamente en dirección del arco, ¿a que velocidad debe patear la pelota para hacer el gol? ¿cuanto tiempo se demora la pelota en llegar al arco?
Physics
1 answer:
Rus_ich [418]3 years ago
8 0

Answer:

i dont really know what it is

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What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?
ratelena [41]
= 80 N/40 kg
= 2 m/s 2
4 0
2 years ago
A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that
kherson [118]

Answer:

= 7.88 × 10^-12 T

Explanation:

From the above question, we are told that:

Kinetic Energy of the proton is K. E = 10.0 MeV

Step 1

We convert 10.0 MeV to Joules

1 Mev = 1.602 × 10-13 Joules

10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

Find the Speed of the Proton

The formula for Kinectic Energy =

K.E = 1/ 2 mv²

Where

m = mass of the proton

v = speed of the proton

K.E of the proton = 1.602 × 10^-12 J

Mass of the proton = 1.6726219 × 10^-27 kilograms

Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

v = 43772331.227m/s

v = 4.3772331227 × 10^7m/s

Approximately = 4.4 × 10^7 m/s

Step 3

Find the Magnetic Field of that region of space

The formula for Magnetic Field =

B = m v / q r

We are told that the proton executes a circular orbit, hence,

mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

4 0
2 years ago
Dr. Aloysius found four homogeneous mixtures, but he thinks that only three are alloys. He tests each one and determines their e
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Answer:

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Explanation:

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6 0
2 years ago
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In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
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Answer:

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∑F = ma

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W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

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Movement of air from land to sea at night, created when cooler, denser air from the land forces up warmer air over the sea.
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The answer is A. LAND BREEZE
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