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Eduardwww
3 months ago
15

the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce

ntimeter, how far does the minute hand travel during a 42 minute period
Physics
1 answer:
Ostrovityanka [3.2K]3 months ago
4 0

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

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Density is defined as the mass divided by the volume.

You can alter the density of a substance by adjusting either its mass or volume.

Increasing the volume while maintaining a constant mass will result in a decrease in density (as the denominator of the fraction increases).

Furthermore, reducing the mass while keeping the volume the same will also lower the density (because the numerator is reduced).

Therefore, to achieve a lower density, you should either reduce the mass or increase the volume, keeping the other constant.

I hope this is helpful.




4 0
3 months ago
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You're riding a unicorn at 25 m/s and come to a uniform stop at a red light 20m away. What's your acceleration?
Ostrovityanka [3204]

The result is -15.625 m/s².


Acceleration signifies the alteration of velocity over a specified duration. It can be calculated with this formula:


a = \dfrac{vf-vi}{t}

Where:

vf = final velocity

vi = initial velocity

t = time

Let’s examine the information provided in your query:

Initially, the vehicle was traveling at 25 m/s before coming to a halt. Thus, it was in motion and subsequently ceased moving, indicating that the final velocity is 0 m/s.


However, we notice that the problem does not provide a time value. We need to determine the time taken from when it was in motion to when it reached the traffic light located 20 m away.


The time can be calculated using the kinematics equation:

d = \dfrac{vi+vf}{2} *t


We derive the equation by substituting the known values first.

20m = \dfrac{25m/s+0m/s}{2}(t)

20m = 12.5m/s{2}(t)

\dfrac{20m}{12.5m/s}=t
1.6s=t

The duration from when it was in motion until it stopped is 1.6s. Now we can utilize this in our acceleration calculation.


a = \dfrac{0m/s-25m/s}{1.6s}

a = \dfrac{-25m/s}{1.6s}

a = -15.625m/s^{2}


It is important to note that the acceleration is negative, indicating the vehicle slowed down.

8 0
2 months ago
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A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How fa
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A basketball player maintains a steady pace of 2.5 m/s while throwing a basketball vertically at 6.0 m/s. How far does the player advance before getting the ball back? Air resistance is negligible. I was unsure which formula to apply to this scenario. Is there any relevance to an angle? First, we determine the duration to reach peak height. The total time for the flight will be double the ascent duration. According to Newton's equations of motion: v = u + at. At the highest point, v = 0, where u is 6 m/s. Thus, the equation becomes 0 = 6 - 9.81t, leading us to t = 0.61 seconds. Therefore, the total flight time equals 1.22 seconds as the player runs towards the ball at a horizontal speed of 2.5 m/s. The distance traveled can be calculated using distance = speed × time, resulting in distance = 2.5 m/s * 1.22, yielding a final distance of 6.11m.
3 0
2 months ago
A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a
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Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

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Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

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The car moves at a constant speed in this direction, indicating that its acceleration is zero.

a=0

Thus, according to Newton's second law,

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F=0

Learn more about slopes and friction:

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