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astra-53 [7]
3 years ago
10

What is the proper tool for cutting keyways or slots in metal?

Physics
1 answer:
Alika [10]3 years ago
6 0
I think it's a End milling of slots<span> </span>
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In the diagram, R1 = 40.0 ,
Nostrana [21]

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

4 0
3 years ago
A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.05 m/s. If the roof is pitched at
AlekseyPX

(a) The time the baseball spends in the air is 0.92 s.

(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

<h3>Time spent in air by the baseball</h3>

h = vt - ¹/₂gt²

-2.1 = (4.05 x sin 34)t  - ¹/₂(9.8)(t²)

-2.1 = 2.26t - 4.9t²

4.9t² - 2.26t - 2.1 = 0

t = 0.92 s

<h3>Horizontal distance traveled by the baseball</h3>

R = Vx(t)

R = (4.05 x cos 34)(0.92)

R = 3.1 m

Thus, the time the baseball spends in the air is 0.92 s.

The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

Learn more about horizontal distance here: brainly.com/question/24784992

#SPJ1

8 0
2 years ago
Which Model represents a lithium isotope? Use your periodic table for this one! <br> Answer:-D
steposvetlana [31]

Answer:

The model D

Explanation:

6 0
2 years ago
A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot
spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance <em>s</em><em> </em><em> </em>between the spokes along the rim is

s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}

The 20-cm arrow, traveling at 6 m/s, will travel its length in

t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}

The fastest speed that the wheel can spin without clipping the arrow is

v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}

The angular velocity \omega of the wheel is given by

\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}

In terms of rev/s, we can convert the answer above as follows:

23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

3 0
3 years ago
Deena made a chart to summarize features of a velocity vs. time graph for objects with constant acceleration and objects with co
Irina-Kira [14]
The error is:
"<span>Find acceleration by finding the area of the rectangle above the line"

The acceleration of any body is the rate of change of velocity of that body. If the motion of the object being considered is plotted on a graph, with the velocity on the y-axis and the time on the x-axis, this ratio will be equivalent to the slope of the graph, not the area above the line.</span>
5 0
3 years ago
Read 2 more answers
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