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Kazeer [188]
2 years ago
14

A sharpening wheel is traveling at 5 rad/s, it slows down to rest in 30 seconds while sharpening an axe. What is its angular acc

eleration?
Physics
1 answer:
Ratling [72]2 years ago
3 0

Answer:

Angular acceleration = 0.167 rad/s^2

Explanation:

Given

Initial Angular velocity (w1) = 5 rad/s

Final Angular velocity (w2) = 0 rad/s

Time taken to change velocity from w1 to w2 = 30 seconds

Angular acceleration is equal to the change in angular velocity to the time taken for making thing change

Hence, Angular acceleration

\frac{w_2 -w_1}{t} \\\frac{5-0}{30}\\0.167rad/s^2

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What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0
2 years ago
A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at â7.0 meters per second2 to a
Harrizon [31]
<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a. Thus, d = [(0^2)-(15^2)]/(2*-7) d = [0-(225)]/(-14) d = 225/14 d = 16.0714 m With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>
8 0
2 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
2 years ago
When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion. Which o
Setler [38]

<u>Answer:</u>

Option: D. Gravity is pulling the crash test dummy in the direction the car is moving.

<u>Explanation: </u>

When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion because the gravity is pulling the crash test dummy in the direction the car is moving.  

Basically when the car is starting, the person inside is in static position and the car is going to move. So it is putting a force on the person to move on the same speed. But as the person is sitting static hence gravity is pulling him behind from moving. Hence, The dummy appears to be pressed backward.

7 0
2 years ago
Assume that segment r exerts a force of magnitude t on segment l. what is the magnitude flr of the force exerted on segment r by
mrs_skeptik [129]
If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R  as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick T_R=T whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that T_L=-T

8 0
2 years ago
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