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jarptica [38.1K]
2 years ago
5

If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit

hout slipping, how does its linear speed compare to its rotational speed? v = Rω v = ω/R They are unrelated. v = ω
Physics
1 answer:
salantis [7]2 years ago
6 0

Answer:

v = r\omega

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t

The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

So the linear speed compared to angular speed is

\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r

v = r\omega

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The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer
makkiz [27]

Answer:

A \to rad/s

B  \to rad/s^3

Explanation:

\omega_z(t)=A + Bt^2

Required

The units of A and B

From the question, we understand that:

\omega_z(t) \to rad/s

This implies that each of A and Bt^2 will have the same unit as \omega_z(t)

So, we have:

A \to rad/s

Bt^2 \to rad/s

The unit of t is (s); So, the expression becomes

B * s^2 \to rad/s

Divide both sides by s^2

B  \to \frac{rad/s}{s^2}

B  \to rad/s^3

5 0
2 years ago
Because atoms of elements in the same group of the periodic table have the same number of neutrons, they have similar properties
AURORKA [14]
False. What actually determines the properties of elements are the electrons, or aka valence electrons. They are used to bond, which determines its properties. 
4 0
3 years ago
What happens to the convection fluid as it heats up? How might this relate to masses of heated air?
Hitman42 [59]

Answer:

The convection process plays an important role in the liquid. Due to the increasing heat supply or high amount of temperature, the fluid gets heated up, as a result of which it becomes warm, less dense and eventually rises up forming convection cells.

In the interior of the earth, the hot molten rocks get heated up due to the heat supplied by the core of the earth. This makes the magma warm and less dense and rises upward forming convection currents in the mantle.

This convection process is similar to the convection cells that form in the atmosphere, where the hot, less dense air rises up in the atmosphere forming a low-pressure zone. This uprising air forms convection cells, in which the warm air rises and as it rises high in the atmosphere, the temperature becomes low, making the air cold and it eventually sinks.

8 0
3 years ago
When making maps of the large-scale universe, astronomers estimate distances to the vast majority of galaxies by using:
Vesnalui [34]

Answer:

<em>The comoving distance and the proper distance scale</em>

<em></em>

Explanation:

The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster.  Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

4 0
3 years ago
to start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 53.0° above the ho
kap26 [50]
So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.

The x location therefore is 290*cos(53)*35 = 6108.4m

The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.

This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m

So your (x,y) coordinates equals (6108.4, 2097.5)
5 0
3 years ago
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