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makkiz [27]
3 years ago
15

Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?

Physics
1 answer:
eduard3 years ago
7 0

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

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I need to find the current resistance and voltage for each in this complicated circuit plz help
konstantin123 [22]

Explanation:

The 11Ω, 22Ω, and 33Ω resistors are in parallel.  That combination is in series with the 4Ω and 10Ω resistors.

The net resistance is:

R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)

R = 20Ω

Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:

V = IR

120 V = I (20Ω)

I = 6 A

So the voltage drops are:

V = (4Ω) (6A) = 24 V

V = (10Ω) (6A) = 60 V

That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:

V = 120 V − 24 V − 60 V

V = 36 V

So the currents are:

I = 36 V / 11 Ω = 3.27 A

I = 36 V / 22 Ω = 1.64 A

I = 36 V / 33 Ω = 1.09 A

If we wanted to, we could also show this using Kirchhoff's laws.

7 0
3 years ago
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v
Mademuasel [1]

Answer:

a = -04978 m / s²

Explanation:

For this exercise we can use the kinematics equations, as the initial speed, distance and time indicate, we can use

        x = x₀ + v₀ t + 1 /2 a  t²

       ½ a t² = x-x₀ - v₀ t

       a = (x-x₀ - v₀ t) 2 / t²

let's reduce the magnitudes to the SI system

     x₀ = 1000 mm = 1 m

     x  = 5000mm = 5m

let's calculate

     a = (5 - 1 - 15 60) 2/60²

     a = -04978 m / s²

negative sign indicates that braking is related

7 0
3 years ago
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kolezko [41]

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Despite their name, standing waves actually do cause the medium to move
Gemiola [76]

Answer:

true

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5 0
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