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zepelin [54]
3 years ago
7

A person standing at the top of mountain aconcagua would be approximately 4.3 mi high. the radius of earth is 3959 mi. what is t

he distance to the horizon from this point?

Physics
2 answers:
Fed [463]3 years ago
6 0

Explanation :

It is given that,

The radius of the earth is, r = 3959 mi

From the attached figure, we have to find the distance to the horizon from this point.

So, using Pythagoras theorem:

AP^2=OA^2-OP^2

x^2=(R+h)^2-R^2

x^2=(3959+4.3)^2-(3959)^2

x^2=34065.89\ mi

x=184.57\ mi

The distance to the horizon from this point is 184.57 mi.

Hence, this is the required solution.

PilotLPTM [1.2K]3 years ago
5 0
Besides ambitious, it would be 184.6 mi
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A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r
Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s²  ,  v = 42 m/s

Explanation:

The uniformly accelerated circular movement,  is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (1)

θ=  ω₀*t + (1/2)*α*t²  Formula (2)

at = α*R  Formula (3)

v= ω*R  Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s  : wheel’s initial angular velocity

R = 1.0 m  : wheel’s radium

(a)  Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ=  ω₀*t + (1/2)*α*t²

θ=  (2)*(10) + (1/2)*(4)*(10)²

θ=  220 rad  

θ=  220 rad  

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

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3 years ago
You have a series circuit powered by a 9V battery. If you double the amount of
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It gets 2 times weaker
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2 years ago
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A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

  • Mass of the first puck = m_1\ =\ 5\ kg
  • Mass of the second puck = m_2\ =\ 3\ kg
  • initial velocity of the first puck = u_1\ =\ 3\ m/s.
  • Initial velocity of the second puck = u_2\ =\ -1.5\ m/s.

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

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3 years ago
Javier wants to improve his flexibility and increase his personal discipline. Which community resource should he select?
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HELP ASAP!!!! WILL GIVE BRAINLIEST!
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