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Wewaii [24]
3 years ago
14

A boy of 65 kg is skating around a circle with radius 3.8

Physics
1 answer:
fiasKO [112]3 years ago
5 0

Answer:

A 1049.75

Explanation:

notice that a Nms is equal to a kg(m^2)/s which is what you get if multiply all three given values.

the formula is angular momentum = (mass)(velo)(radius)

therefore

65(3.8)(4.25) = the answer

= 1049.75

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What is the net charge (overall or sum of charges) of the nucleus and what number is it equal to?
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Because the nucleus is made up of positively charged protons and neutrally charged neutrons, and no negatively charged particles, the charge of the nucleus will always be equal to the sum of the charges of its protons. A simpler way to say it is because each proton has a +1 charge, the charge of the nucleus will be the same as the number of protons in it.
7 0
3 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
4 years ago
A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The
S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

\sum F_x = 0

f-N_w = 0

N_w = f

The forces in the horizontal direction would be,

\sum F_y = 0

N_f -W =0

N_f = W

The sum of Torques at equilibrium,

\sum \tau = 0

Wdcos\theta - N_wLsin\theta = 0

WdCos\theta = fLSin\theta

f = \frac{Wd}{Ltan\theta}

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

f_{max} = \mu W=\frac{Wd}{Ltan\theta}

\theta = tan^{-1} (\frac{d}{\mu L})

Replacing,

\theta = tan^{-1} (\frac{0.9}{0.42*2})

\theta = 46.975\°

Therefore the minimum angle that the person can reach is 46.9°

8 0
3 years ago
four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?
Kay [80]
There was no net force on the stuffed toy, because the kids might have the same strength,  The same force is on both sides of it.  T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move.  </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving. 
3 0
3 years ago
Read 2 more answers
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

=

7 0
3 years ago
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