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3 years ago

12. A 1.2-kg bucket of water is held by a string and spun in a horizontal circle with an acceleration of 24.6

Physics

1 answer:

frozen [14]3 years ago

6 0

Answer:

1.1 m

Explanation:

a = v² / r

24.6 m/s² = (5.2 m/s)² / r

r = 1.1 m

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A train moves from rest to a speed of 25 m/s in 50.0 seconds. What is its acceleration?

ELEN [110]

Answer:

Wellll. I am assuming the direction of speed is in the same direction as the direction of displacement of the train. (i.e. Velocity is positive)

Acceleration is defined as the rate of change of velocity with respect to time (m^s-2)

Explanation:

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3 years ago

An unwary football player collides with a padded goalpost while running at a velocity of 9.50 m/s and comes to a full stop after

Amanda [17]

Answer:

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

Explanation:

Given:

Initial velocity of the player (u) = 9.50 m/s

Final velocity of the player (v) = 0 m/s (Comes to a stop)

Displacement of the player (S) = 0.250 m

We know that, using equation of motion relating displacement (S), acceleration (a), initial velocity (u) and final velocity (v), we have:

$v^2=u^2+2aS$

Expressing in terms of 'a', we get:

$a=\frac{v^2-u^2}{2S}$

Plug in the given values and solve for 'a'. This gives,

$a=\frac{0-9.50^2}{2\times 0.250}\\\\a=\frac{-90.25}{0.5}=-180.5\ m/s^2$

Therefore, the acceleration of the player is -180.5 m/s². So, the deceleration is 180.5 m/s².

Now, using the first equation of motion, we have:

$v=u+at\\\\t=\frac{v-u}{a}$

Plug in the given values and solve for 't'. This gives,

$t=\frac{0-9.5}{-180.5}\\\\t=0.053\ s$

Therefore, the the collision will last for 0.053 s.

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

8 0

3 years ago

If a freely suspended vertical spring is pulled in downward direction and then released, which type of wave is produced in the s

larisa [96]

Answer:

longitudinal wave

Explanation:

it is perpendicular to the direction of the wave

3 0

3 years ago

Read 2 more answers

A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

5 0

3 years ago

How are frequency and wave period related?

dedylja [7]

-- The unit of frequency is "per second" (Hz), which is [reciprocal time].

-- The unit of period is "second", which is [time].

Do you see where this is going ?

'Frequency' and 'period' are reciprocals of each other.

For any wave ...

Period = (1) / (frequency) .

Frequency = (1) / (period) .

6 0

3 years ago