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Nataly_w [17]
3 years ago
7

A projectile is launched vertically at 100 m/s. If air resistance can be ignored, at what speed will it return to its initial le

vel?
Physics
2 answers:
eduard3 years ago
7 0
<h2>Speed with which it return to its initial level is 100 m/s</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 100 m/s  

Acceleration, a = -9.81 m/s²  

Final velocity, v = ?

Displacement, s = 0 m  

Substituting  

v² = u² + 2as

v² = 100² + 2 x -9.81 x 0

v² = 100²

v = ±100 m/s

+100 m/s is initial velocity and -100 m/s is final velocity.

Speed with which it return to its initial level is 100 m/s

madreJ [45]3 years ago
6 0

Answer:

100 m/s

Explanation:

given,

vertical velocity of the projectile = 100 m/s

When the Projectile is launched vertically it will have two-component one is the vertical component and another is a horizontal component.

The horizontal component of velocity will remain constant.

The vertical component of the velocity will be zero at the highest point is zero and the second half motion of the projectile is the mirror of the first half movement of the projectile.

When the object reaches the initial level the velocity will be the same i.e. 100 m/s.

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32 °C.

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En este caso, debemos entender que la relación entre el calor y la temperatura viene dada por:

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3 years ago
A cyclotron is designed to accelerate protons (mass 1.67 x 10^-27 kg) up to a kinetic energy of 2.5 x 10^-13 J. If the magnetic
Dafna11 [192]

Answer:

24 cm

Explanation:

Given:

Mass of proton = 1.67 × 10⁻²⁷ Kg

kinetic energy = 2.5 × 10⁻¹³ J

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Now,

Kinetic energy = \frac{1}{2}mv^2  = 2.5 × 10⁻¹³ J

where, v is the velocity of the electron

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\frac{1}{2}\times1.67\times10^{-27}\times v^2  = 2.5 × 10⁻¹³ J

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on substituting the respective values, we get

\frac{1.67\times10^{-27}\times1.73\times10^7}{r}  = 1.6 × 10⁻¹⁹ × 0.75

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Hence, the correct answer is 24 cm

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