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kenny6666 [7]
3 years ago
8

a 2.35 water bucket is swung in a full cirlce of radius 0.824 m just fast enough so that the water doesn't fall out the top mean

ing n equals zero there. What is the speed of the water at the top
Physics
1 answer:
tiny-mole [99]3 years ago
4 0

Answer:

2.84 m/s

Explanation:

At the top position of the circular trajectory, the normal reaction is zero:

N = 0

So it means that the only force that is providing the centripetal force is the gravitational force (the weight of the bucket). Therefore we have:

mg = m \frac{v^2}{r}

where

m is the mass of the water bucket

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed of the bucket

r = 0.824 m is the radius of the circle

Solving for v,

v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(0.824 m)}=2.84 m/s

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A block oscillating on a spring has a maximum speed of 20 cm/s. part a what will the block's maximum speed be if its total energ
yKpoI14uk [10]
It will be 4 times of original thus maximum speed would be 80cm/s
6 0
3 years ago
Combine these three velocity vectors into a resultant: 3.0 m/s north, 4.0 m/s east 1.0 m/s west. Identify the resultant vector
Slav-nsk [51]

Answer:

The resultant vector is 1 m/s

Explanation:

The resultant vector is 1 m/s west based on triangle law of vector addition, when two sides of a triangle is represented by two vectors, the resultant vector is the third side of the triangle.

5 0
3 years ago
Suppose you want to make an interferometer with two slits to show the wave nature of light to your friends. The most convenient
nevsk [136]

Answer:

a) the pattern is bright,  b)  y = 8.25 mm, c) the distance increases with the wavelength increases

Explanation:

This is a team building exercise to perform interference experiments, where the separation of the slits is d = 1 m = 1 10⁻³ m and the distance to the screen L = 1.5 m  

Constructive interference occurs when the path difference is equal to an integer number of wavelengths  

             d sin θ = m λ  

a) The center of the two-ray pattern has the same length so the intensity is MAXIMUM, that is, the pattern is bright  

b) to find the distance let's use trigonometry  

   tan θ= y / L  

   tan θ =sin θ/cos θ= sin θ  

because the angles are small  

we substitute  

     y = \frac{m \lambda L}{d} 

let's calculate  

     y = 10 550 10⁻⁹ 1.5 / 1 10⁻³  

     y = 8.25 10⁻³ m

c) for this part the incident light must be white or polychrome light  

let's find the distance (y) for some colors, in the first order  

Violet λ = 400  

y = 1 400 10⁻⁹ 1.5 / 1 10⁻³  

y = 6 10⁻⁴ m  

green λ = 500 nm  

y = 1 500 10⁻⁹ 1.5 / 1 10⁻³  

y = 7.5 10⁻⁴ m  

yellow λ = 600 nm  

y = 1 600 10⁻⁹ 1.5 / 1.10⁻³  

y = 9 10⁻⁴ m  

lam λ = 700 nm  

y = 1 700 10⁻⁹ 1.5 / 1. 10⁻³  

y = 1.05 10⁻³ m  

violet, green, orange, red  

the distance increases with the wavelength increases

d) To find the wavelength we must measure the distance y from the center of the slit for various interference orders and using the equation we can find the wavelength of rationality

4 0
3 years ago
8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
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