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4 years ago

10

Please help!! The graph in the figure shows the position of a particle as it travels along the x-axis. What is the magnitude of

the average speed of the particle between t = 1.0 s and t = 4.0 s?

1 answer:

Bogdan [553]4 years ago

6 0

Answer:

1.3 m/s

Explanation:

average speed = total distance/ total time

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The sun dictates the direction of leaves and the direction that the plant will grow towards.

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3 years ago

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A gyroscope flywheel of radius 1.96 cm is accelerated from rest at 13.0 rad/s2 until its angular speed is 2270 rev/min. (a) What

nika2105 [10]

Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².

Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.

Given,

Radius of flywheel (r) = 1.96 cm = 0.0196m

Angular acceleration (α)= 13.0 rad/s²

The tangential acceleration formula is at=rα

where, α is the angular acceleration, and r is the radius of the circle.

using the formula; at=rα = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².

The tangential acceleration is 0.2548 m/s².

Learn more about the Tangential acceleration with the help of the following link:

brainly.com/question/15743294

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1 year ago

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

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3 years ago

Solve the equation 3x^2+8x+5=0

Pachacha [2.7K]

$3x^2+8x+5=0\\
3x^2+3x+5x+5=0\\
3x(x+1)+5(x+1)=0\\
(3x+5)(x+1)=0\\
x=-\dfrac{5}{3} \vee x=-1$

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3 years ago

Andre is playing air hockey with Alexa and shoots his puck across the essentially frictionless surface to score a goal. What fre

DaniilM [7]

Answer:

The answer is C because there is no friction there will be no friction force only applied and since its on ice you have to account for gravity

Explanation:

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3 years ago