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hjlf
3 years ago
7

a rock thrown horizontally at 20m/s from the top of a building 63m high. How far from the base of the building does the rock lan

d?
Physics
1 answer:
Lorico [155]3 years ago
4 0

1,260

this might be it

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An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole. At the distance of closest approach, her center of
IrinaVladis [17]

Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

r_1 = Distance between center of mass of person and pole = 0.45 m

r_2 = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s

The new angular speed is 1.70344 rad/s

7 0
3 years ago
When a person plucks a guitar string, the number of half wavelengths that fit into the length of the string determines the _____
Vedmedyk [2.9K]

Answer:

Frequency

Explanation:

Each half wavelength has a point of largest amplitude (aka a node). Depending on the wavelength each node oscillates at a certain rate of swings per unit of time. The latter is referred to as frequency and measure in Hertz [Hz].

8 0
3 years ago
What is an example of a wave that is not mechanical and how is it different?
vovikov84 [41]

Answer:

light is an example of a wave that is not mechanical .

it is different as it does not need material medium for its propagation

5 0
3 years ago
Three charges are arranged as shown in the picture above. Find the magnitude and direction of the electrostatic force on the 6 n
bagirrra123 [75]

Answer:

Bot Nm

Explanation:

jdjdmxjd mdjdcj jdsdj jedidj jddj

6 0
3 years ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
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