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k0ka [10]
3 years ago
10

The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in

2.5 s. what is the angular acceleration as the blades slow down?
Physics
1 answer:
Tpy6a [65]3 years ago
7 0

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

<em>Angular acceleration = -3,080 rev per minute / sec</em>

That's a perfectly good and true answer to the question, but the units are ugly.  We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

So

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

<em>AngAccel = -322.5 radian/s²</em>

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Nesterboy [21]

Answer:

12.1 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement = 120 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -1.67\times 120-0^2\\\Rightarrow u=\sqrt{2\times 1.67\times 120}\\\Rightarrow u=20.02\ m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20.2}{-1.67}\\\Rightarrow t=12.1\ s

Time taken by the rock to hit the bottom of the crater is 12.1 seconds

5 0
3 years ago
A racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction). It hits the wall of th
zheka24 [161]

The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).

mass m  = 42g = 42/1000 = 0.042kg

initial velocity before collision u = 7 m/s

It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,

velocity after collision v = 7 m/s

To calculate the magnitude of the racquetball's change in momentum, we will use the formula below

Change in momentum = Mv - Mu

Since momentum is a vector quantity, we will consider the direction.

Change in momentum = 0.042 x 7 - ( 0.042 x - 7)

Change in momentum = 0.294 + 0.294

Change in momentum = 0.588 kgm/s

Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Learn more on momentum here: brainly.com/question/402617

5 0
2 years ago
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kakasveta [241]
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8 0
3 years ago
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Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude
aivan3 [116]

Answer:

E=6.8Kv/m

Explanation:

From the question we are told that

Distance b/w plate d=10cm=>0.1m

P_1 Potential at 7.35 V=533v

Generally the equation for electric field at a distance is mathematically given as

E=\frac{v}{d}

E=\frac{533}{7.85*10^-^2}

E=6789.808917

E=6.8*10^3

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7 0
3 years ago
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Please look at the attached image below for the explanation

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