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andrezito [222]
3 years ago
9

The blades in a blender rotate at a rate of 6800 rpm . When the motor is turned off during operation, the blades slow to rest in

2.8 s .
Physics
1 answer:
tangare [24]3 years ago
5 0
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
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Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo
kiruha [24]

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

W=F*d*cos(\theta)

here we have the work done by Paul and the friction force, so:

W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J

W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J

Now the change of energy is:

W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s

4 0
3 years ago
Read 2 more answers
What is the weight of an object with a mass of 19 kg?
Natalija [7]

Answer:

Every 2.2 kg is 1 pound. So mulitply 19 * 2.2. It's gonna be equal to 41.8

Explanation:

7 0
2 years ago
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Which sentence best describes a role of gravity in the formation of the
enyata [817]

Answer:

I think it's option D

Explanation:

I think it's option D but not so sure

8 0
3 years ago
Describe free fall and weightlessness ?<br>​
yaroslaw [1]

weightlessness is the complete or near complete absense of the sensation of weight.

when in freefall,the only force acting upon your body is the force of gravity a noun contact force.since the gravity cannot be felt without any other opposing forces,you would have no sensation of it.you would feel weightless when in a state of freefall.

8 0
2 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
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