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SVEN [57.7K]
2 years ago
14

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.72 x 10^4 rad/s to an angular speed of 5

.42 x 10^4 rad/s. In the process, the bit turns through 1.72 x 10^4 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 8.42 x 10^4 rad/s, starting from rest?
Physics
1 answer:
Leto [7]2 years ago
4 0

Answer:

The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.

Explanation:

ω1= 1.72x10^4 rad/sec

ω2= 5.42x10^4 rad/sec

ωmax= 8.42x10^4 rad/sec

θ= 1.72x10^4 rad

\alpha = \frac{w2^{2}-w1^{2}  }{2*(\theta2 - \theta1)}

α=7.67 x10^4 rad/sec²

t= ωmax / α

t= 8.42 x10^4 rad/sec  /  7.67 x10^4 rad/sec²

t=1.097 sec

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The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a
mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

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An oscilloscope shows a steady sinusoidal signal of 5 Volt peak to peak, which spans 5 cm in vertical direction on the screen. B
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You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
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If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

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Euglena are _______.<br><br> A heterotrophs<br><br> B autotrophs
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The Euglena is unique in that it is both heterotrophic (must consume food) and autotrophic (can make its own food).
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