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Alex_Xolod [135]
3 years ago
14

The moment of inertia of a cylinder is 0.016 kg m^2 with radius 6.0 cm. If the cylinder has a linear speed is 7.7 m/s, what is t

he magnitude of the angular momentum of the cylinder?
Physics
1 answer:
Greeley [361]3 years ago
4 0

Answer:

2.05 kgm²/s

Explanation:

I = moment of inertia of cylinder = 0.016 kgm²

r = radius of the cylinder = 6 cm = 0.06 m

v = linear speed of the cylinder = 7.7 m/s

w = angular speed of cylinder

Using the equation

v = r w

7.7 = (0.06) w

w = 128.33 rad/s

Magnitude of angular momentum of the cylinder is given as

L = I w

L = (0.016) (128.33)

L = 2.05 kgm²/s

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How to find a planet’s gravitational field strength using its radius?
grin007 [14]

The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

g = \frac{F}{m}

Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

And F =\frac{GMm}{r^{2} }

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11}  \times 1}{(6.6 \times 10^{6}) ^{2} }

g = 0.977 \times 10^1= 9.77\ N

So, the gravitational field strength is approximately equal to 10 N.

5 0
3 years ago
A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h
MariettaO [177]

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

coefficient of friction between the box and the surface μ= 0.3

the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

to move the ball horizontal force required is 161.86 N

8 0
3 years ago
6.
S_A_V [24]

The distance in meters she would have moved before she begins to slow down is 11.25 m

<h3>LINEAR MOTION</h3>

A straight line movement is known as linear motion

Given that Ann is driving down a street at 15 m/s. Suddenly a child runs into the street. It takes Ann 0.75 seconds to react and apply the brakes.

To know how many meters will she have moved before she begins to slow down, we need to first list all the given parameters.

  • speed = 15 m/s
  • time t = 0.75 s

From definition of speed,

speed = distance / time

Make distance the subject of the formula

distance = speed x time

distance = 15 x 0.75

distance = 11.25m

Therefore, the distance in meters she would have moved before she begins to slow down is 11.25 m

Learn more about Linear motion here: brainly.com/question/13665920

4 0
2 years ago
A wave traveling in a string has a wavelength of 35 cm, an amplitude of 8. 4 cm, and a period of 1. 2 s. What is the speed of th
AlekseyPX

0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)

35 / 1.2 = 29.16

29.16 ÷ 100 = 0.29

Wave velocity in string:

The properties of the medium affect the wave's velocity in a string. For instance, if a thin guitar string is vibrated while a thick rope is not, the guitar string's waves will move more quickly. As a result, the linear densities of the two strings affect the string's velocity. Linear density is defined as the mass per unit length.

Instead of the sinusoidal wave, a single symmetrical pulse is taken into consideration in order to comprehend how the linear mass density and tension will affect the wave's speed on the string.

Learn more about density here:

brainly.com/question/15164682

#SPJ4

3 0
2 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an
castortr0y [4]

Answer:

  • Fx = -9.15 N
  • Fy = 1.72 N
  • F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

  • Fx = -9.15 N
  • Fy = 1.72 N

Then the magnitude and direction of the resultant are

  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°

4 0
3 years ago
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