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zlopas [31]
3 years ago
11

Calculate the rotational inertia of a wheel that has akinetic

Physics
1 answer:
andrew11 [14]3 years ago
4 0

Answer:

J = 14.4 kg*m^2

Explanation:

Assuming that the wheel is not moving anywhere, and the kinetic energy is only due to rotation:

Ek = 1/2 * J * w^2

J = 2 * Ek / (w^2)

We need the angular speed in rad / s

566 rev/min * (1 min/ 60 s) * (2π rad / rev) = 58.22 rad/s

Then:

J = 2 * 24400 / (58.22^2) = 14.4 kg*m^2

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Sugar is considered a crystalline solid because
Nina [5.8K]

Answer:

Im sure the awnser is option

A. arranged in a regular pattern.

8 0
2 years ago
A small lead ball, attached to a 1.75-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
soldier1979 [14.2K]

Answer:

h = 57.6 m

Explanation:

First, we find the linear speed of the ball while in circular motion:

v = rω

where,

v = linear speed of ball = ?

r = radius of circle = length of rope = 1.75 m

ω = angular speed = (3 rev/s)(2π rad/1 rev) = 18.84 rad/s

Therefore,

v = (1.75 m)(18.84 rad/s)

v = 32.98 m/s

Now, we apply the 3rd equation of motion on the ball, when it breaks:

2gh = Vf² - Vi²

where,

g = - 9.8 m/s² (negative sign due to upward motion)

h = height covered = ?

Vf = Final Velocity = 0 m/s (since, the ball finally stops at highest point for a moment)

Vi = Initial Velocity = 32.98 m/s

Therefore,

2(- 9.8 m/s²)h = (0 m/s)² - (32.98 m/s)²

h = ( - 1088.12 m²/s²)/( - 19.6 m/s²)

h = 55.5 m

since, the ball was initially at a height of 2.1 m from ground. So, the total height from ground, will now become:

h = 55.5 m + 2.1 m

<u>h = 57.6 m</u>

7 0
3 years ago
A softball is fouled off with a vertical velocity of 20 m/s and a horizontal velocity of 15 m/s. what is the resultant velocity
raketka [301]
25 m/s is the answer
8 0
3 years ago
I NEED HELP ASAP!!! 100 points if you answer these 7 questions:
netineya [11]

Answer:

#1. B. Alfred Wegener

#2. C. Volcanic activity at the ocean bottom.

#3. B, C. Continental, Sea floor (oceanic).

#4. B. Folding

#5. B. Tectonics

#6. C. Point underground where the earthquake starts

#7. C. Flood waters carrying away soil

Explanation:

#1. Alfred Lothar Wegener was a German polar researcher, geophysicist and meteorologist. During his lifetime he was primarily known for his achievements in meteorology and as a pioneer of polar research, and is also known as the father of plate tectonics.

#2. The most prominent feature of ocean topography discovered in the 1960s was: Volcanic activity at the ocean bottom. The continents have always been in their current positions.

#3. Tectonic plates are composed of oceanic lithosphere and thicker continental lithosphere, each topped by its own kind of crust. The two types of tectonic plates are continental and oceanic tectonic plates.

#4. Tremendous pushing forces exerted by two of Earth's plates moving together squeezed rock layers from opposite sides. This caused the rock layers to buckle and fold, forming folded mountains. Folded mountains are mountains formed by the folding of rock layers caused by compression forces.

#5. Plate tectonics is the scientific theory explaining the movement of the earth's crust. It is widely accepted by scientists today. Recall that both continental landmasses and the ocean floor are part of the earth's crust, and that the crust is broken into individual pieces called tectonic plates.

#6. The hypocenter is the point within the earth where an earthquake rupture starts. The epicenter is the point directly above it at the surface of the Earth. Also commonly termed the focus. See also epicenter.

#7. This process is known as erosion. In earth science, erosion is the action of surface processes that removes soil, rock, or dissolved material from one location on the Earth's crust, and then transports it to another location.

Hope this helps!

Have a nice day!

If you find my answer helpful

<em>Pls consider marking my answer as </em><em>Brainliest</em><em>! It would mean a lot!</em>

3 0
2 years ago
A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.
LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

W = \frac{1}{2}(50)(5.61^2) = 786.8025 J

Now, we can define work:

\large\boxed{W = Fdcos\theta}}

Now, plug in the values:

786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta

Solve for theta:

cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}

4 0
2 years ago
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