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hammer [34]
3 years ago
5

A wheel of radius 0.5 m rotates with a constant angular speed about an axis perpendicular to its center. A point on the wheel th

at is 0.2 m from the center has a tangential speed of 2 m/s. Determine the tangential acceleration of the point that is 0.2 m from the center.
Physics
1 answer:
Masja [62]3 years ago
8 0

Answer:

The tangential acceleration is 0 m/s².

Explanation:

Given:

Radius of the wheel = 0.5 m

The point of observation for calculating tangential acceleration = 0.2 m from center.

Tangential speed at the point of observation = 2 m/s

The angular speed of the wheel is a constant.

In order to determine the tangential acceleration, we make use of the following formula:

Tangential acceleration at a point = Angular acceleration × Distance of the point from center

Or, a_t=\alpha \times r

Now, angular acceleration is defined as the rate of change of angular speed.

Here, the angular speed of the wheel is a constant. So, the change of angular speed is 0. Therefore, the angular acceleration is also 0 rad/s².

Now, from the above formula, as angular acceleration is 0, the magnitude of tangential acceleration at a point that is 0.2 m from the center of the wheel is also 0 m/s².

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"C) The movement of the gas particles keeps the balloon inflated" best describes the gas trapped in a sealed balloon, although the movement varies with the heat of the gas. 
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3 years ago
At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another b
maks197457 [2]

Answer:

(a) The two balls collide 2\; \rm s after launch.

(b) The height of the collision is 4\; \rm m.

(Assuming that air resistance is negligible.)

Explanation:

Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of g should be negative. The question states that the magnitude of g\! is 10\; \rm m \cdot s^{-2}. Hence, the signed value of \! g should be \left(-10\; \rm m \cdot s^{-2}\right).

Similarly, the initial velocity of the ball thrown downwards should also be negative: \left(-8.0\; \rm m \cdot s^{-1}\right).

On the other hand, the initial velocity of the ball thrown upwards should be positive: \left(12\; \rm m \cdot s^{-1}\right).

Let v_0 and h_0 denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time t:

\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0.

For both balls, g = \left(-10\; \rm m \cdot s^{-2}\right).

For the ball thrown downwards:

  • Initial velocity: v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right).
  • Initial height: h_0 = 40\; \rm m.

\displaystyle  h(t) = -5\, t^{2} + (-8.0)\, t + 40 (where h is in meters and t is in seconds.)

Similarly, for the ball thrown upwards:

  • Initial velocity: v_0 = \left(12\; \rm m \cdot s^{-1}\right).
  • Initial height: h_0 = 0\; \rm m.

\displaystyle  h(t) = -5\, t^{2} + 12\, t (where h is in meters and t is in seconds.)

Equate the two expressions and solve for t:

-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t.

t = 2.

Therefore, the collision takes place 2\, \rm s after launch.

Substitute t = 2 into either of the two original expressions to find the height of collision:

h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m.

In other words, the two balls collide when their height was 4\; \rm m.

3 0
3 years ago
What is the best way to support your hypothesis
Diano4ka-milaya [45]

Answer:

Making a Hypothesis

Explanation:

-Research the subject of your question. Review the literature and find out as much as you can about previous information and discoveries surrounding your question.

-Develop an educated guess that answers your initial question. This is your hypothesis. Make a prediction based on your hypothesis and state it as a cause-effect relationship.

4 0
4 years ago
A roller-coaster car has a mass of 1040 kg when fully loaded with passengers. As the car passes over the top of a circular hill
rusak2 [61]

Answer:

a.6373.5 N

b.-3837.6 N

Explanation:

Mass of roller coaster=m=1040 kg

Radius=r=24 ,

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Normal force=F_N

According to question

mg-F_N=\frac{mv^2}{r}

Where g=9.81 m/s^2

Substitute the values

1040\times 9.81-F_N=\frac{1040\times (9.4)^2}{24}

10202.4-F_N=3828.9

F_N=10202.4-3828.9=6373.5 N

b.v=18m/s

g=9.81 m/s^2

1040\times 9.81-F_N=\frac{1040\times (18)^2}{24}

10202.4-F_N=14040

F_N=10202.4-14040=-3837.6 N

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4 years ago
A 52-kg snow skier is at the top of a 245-m-high hill. After she has gone down a vertical distance of 112 m, what is her total e
Assoli18 [71]

Answer:

  a. 125 kJ

Explanation:

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After skiing down 112 m, some of her initial energy is converted to kinetic energy, and some remains as potential energy. We assume the ski slope is essentially frictionless, and air resistance is negligible.

7 0
3 years ago
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