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Dominik [7]
3 years ago
8

A centrifuge accelerates uniformly from rest to 15000 rpm in 330 s . Through how many revolutions did it turn in this time?

Physics
1 answer:
eduard3 years ago
4 0

Answer:

The number of revolutions turned by the centrifuge is 8250 revolutions.

Explanation:

Given;

number of revolution per minutes, ω = 15000 rpm

time of  motion, t = 330 s = 5.5 minutes

The number of revolutions turned by the centrifuge is given by;

N = \frac{1500 \ Rev}{minutes} *5.5 \ minutes\\\\N = 8250 \ revolutions

Therefore, the number of revolutions turned by the centrifuge is 8250 revolutions.

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a baseball is pitched with a speed of 35.0m/s. how long does it take the ball to travel 18.4 m from the pitchers mound to home p
Vlad [161]
Given the speed and the distance, to find time you can use the formula speed is equal to distance over time. From there you can manipulate the equation for time to equal the distance divided by speed. Time is equal to 18.4 meters divided by 35m/s which equals 0.526 seconds.

7 0
3 years ago
A parachutist of mass 20.1 kg jumps out of an airplane at a height of 662 m and lands on the ground with a speed of 7.12 m/s. Th
Klio2033 [76]

Answer:

1.30\cdot 10^5 J

Explanation:

The energy lost due to air friction is equal to the mechanical energy lost by the parachutist during the fall.

The initial mechanical energy of the parachutist (at the top) is equal to his gravitational potential energy:

E_i=mgh

where

m = 20.1 kg is his mass

g=9.8 m/s^2 is the acceleration due to gravity

h = 662 m is the initial heigth

The final mechanical energy (at the bottom) is equal to his kinetic energy:

E_f=\frac{1}{2}mv^2

where

v = 7.12 m/s is the final speed of the parachutist

Therefore, the energy lost due to air friction is:

\Delta E=E_i-E_f=mgh-\frac{1}{2}mv^2=(20.1)(9.8)(662)-\frac{1}{2}(20.1)(7.12)^2=1.30\cdot 10^5 J

4 0
4 years ago
DESCRIBE THE FORMATION OF THE SOLAR SYSTEM ACCORDING TO THE NEBULAR THEORY
marishachu [46]

When a cloud of gas and dust in space was disturbed, maybe by the explosion of a nearby star.This explosion made waves in space which squeezed the cloud of gas & dust.

8 0
3 years ago
Select all the dangers that commonly follow a major earthquake. A. fire B. tsunamis C. tornadoes D. volcanoes E. aftershocks F.
ArbitrLikvidat [17]
F. i hope that helps im sorry if im wrong but that sounds right to me


6 0
3 years ago
A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o
marshall27 [118]

Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, m=5.0 kg

Force acting towards east, F=27 N

Frictional force acting towards west, f=17 N

Let the acceleration be a m/s².

Now, net force acting on the box towards east is given as:

F_{net}=F-f=27-17=10\textrm{ N}

From Newton's second law of motion,

F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2

Therefore, the magnitude of the acceleration of the box is 2 m/s².

6 0
3 years ago
Read 2 more answers
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