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sveta [45]
3 years ago
11

Which simple machine is described as an inclined plane wrapped around a cylinder?

Physics
2 answers:
tatyana61 [14]3 years ago
7 0

Answer:

A  Screw

Because A wedge is 2 inclined planes. For wheel and Axle think A bike and a pulley An elevator so a screw is the only answer.

Ad libitum [116K]3 years ago
3 0

Answer:

A screw.

Explanation:

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In the realm of scientific inquiry, making an observation typically leads to ___.
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4 years ago
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A baseball is struck home plate and acquires a speed of 60m/s. It rises to a height of 100.0 m above its starting level. Find it
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Answer:

40 m/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 60 m/s

Height (h) = 100 m

Acceleration due to gravity (g) = 10 m/s²

Final velocity (v) =?

The velocity at height 100 m can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

v² = 60² – (2 × 10 × 100)

v² = 3600 – 2000

v² = 1600

Take the square root of both side

v = √1600

v = 40 m/s

Thus, velocity at height 100 m is 40 m/s

7 0
3 years ago
When atoms get more and more energy, they can ionize. What can happen to a gas as more energy is added to its atoms?
Ulleksa [173]

Answer: Option (A) is the correct answer.

Explanation:

When more and more energy is provided to a gas then its atoms move more rapidly.

This rapid and continuous movement converts the gas into hot ionized ions which have positively charged ions and negatively charged electrons.

Therefore, we can conclude that as the atoms move faster, the gas can change into a plasma.

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3 years ago
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Give an example of Newton's 1st law​
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Answer:

A ball moving until gravity pulls it back down to the ground

Explanation:

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3 years ago
The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the
Ivanshal [37]

Answer:

(a) \alpha = - 1.32\ rev/m^{2}

(b) \theta = 13674\ rev

(c) \alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}

(d) a = 22.458\ m/s^{2}

Solution:

As per the question:

Angular velocity, \omega = 190\ rev/min

Time taken by the wheel to stop, t = 2.4 h = 2.4\times 60 = 144\ min

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

\omega' = \omega + \alpha t

omega' = final angular velocity

\omega = initial angular velocity

\alpha = angular acceleration

Now,

0 = 190 + \alpha \times 144

\alpha = - 1.32\ rev/m^{2}

Now,

(b) The no. of revolutions is given by:

\omega'^{2} = \omega^{2} + 2\alpha \theta

0 = 190^{2} + 2\times (- 1.32) \theta

\theta = 13674\ rev

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}

(d) The radial acceleration is given by:

\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s

Linear acceleration is given by:

a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}

a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}

5 0
3 years ago
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