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Anna35 [415]
2 years ago
12

Anne is working on a research project that involves the use of a centrifuge. Her samples must first experience an acceleration o

f 100 , but then, the acceleration must increase by a factor of six. By how much will the rotational speed have to increase? Express your answer as a fraction of the initial rotation rate.
Physics
1 answer:
oee [108]2 years ago
6 0

Answer:

The final rotation is 2.449 times the initial rotation

Explanation:

Centripetal acceleration and rotation speed of particle is related as

a=rω²    

where ω  is angular speed and r radius of circle

Rearranging equation    

ω=\sqrt{\frac{a}{r} }

The initial acceleration is given by

            a₁ =100g

The final acceleration of particle is

            a₂ = 6 (100g)

                  =600g

The final and initial rotation speed are related as

                  \frac{w1}{w2} =\sqrt{\frac{a2}{a1 }

                  ω2=\sqrt{\frac{600g}{100g} } *w1

                   ω2=2.449ω1

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An object weighs 980N on the earth’s surface (i) What is its mass? (ii) If the same object weighs 360N on another planet, find t
AnnZ [28]
Weight = mass*gravity. Hence mass = 980/9.8 = 100kg. Gravity of planet 2 = weight/mass = 3.6 m/s^2
4 0
2 years ago
Give two ways of reversing the direction of the forces on the coil in the electric motor?​
nikklg [1K]

Answer:

Interchanging the poles of the magnet

Reversing the direction of the applied current

Explanation:

  1. The working of the electric motor is associated with Fleming's left-hand rule.
  2. It states that if a current-carrying conductor is placed inside a magnetic field, it experiences a force in the direction perpendicular to the direction of the electric current and magnetic field.
  3. These three physical quantities are placed in a mutually perpendicular direction.
  4. So, in order to reverse the direction of force, you have to either change the direction of the current or magnetic field.
4 0
2 years ago
A man runs at a velocity of 4.5 m/s for 15.0 min. When going up an increasingly steep hill, he slows down at a constant rate of
madreJ [45]

The man ran  <u>4252.5 meters.</u>

Why?

To solve the problem, we need to divide the exercise into two movements, the first on while the was running at 4.5 m/s for 15 min, and then, while he was slowing down (going up because of the hill).

First movement: Running at 4.5 m/s for 15 min.

We need convert from minutes to seconds,

1min=60seconds\\\\15min*\frac{60seconds}{1min}=900seconds

Now, calculating the distance covered for the first movement, we have:

x_{1}=0+v_{1}*t_{1}\\\\x_{1}=4.5\frac{m}{s}*900s=4050m

So, we know that the man covered 4050m for the first movement, it will be our initial position for the second movement.

Second movement:  acceleration -0.05m/s^2 (because he's slowing down) for 90 seconds, at 4.5m/s.

x_{2}=x_{1}+v_{1}*t+\frac{1}{2}at^{2}\\\\x_{2}=4050m+4.5m\frac{m}{s}*90seconds-\frac{1}{2}*(0.05\frac{m}{s^{2}})*(90s)^{2}\\\\x_{2}=4050m+405m-(0.5*0.05\frac{m}{s^{2}}*8100s^{2})=4050m+405m-202.5m\\\\x_{2}=4252.5m

Hence, we have that he ran 4252.5 m.

Have a nice day!

4 0
2 years ago
An electron starts from rest 3.00 cm from the center of a uniformly charged sphere of radius 2.00 cm. if the sphere carries a to
Sidana [21]
Answer:
velocity = 7.26 * 10^6 m/sec

Explanation:
The rule that is used to solve this problem is shown in the attached image.
The variables are as follows:
k = 8.99 * 10^9 Nm^2 / C^2
e is the electron charge = -1.6 * 10^-19 C
q is the charge given = 1 * 10^-9 C
m is the mass of the electron = 9.11 * 10^-31
r1 is the radius of starting point = 3 cm = 0.03 m
r2 is the radius of the sphere = 2 cm = 0.02 m
Substitute with the givens in the equation to get the value of the velocity

Hope this helps :)

7 0
2 years ago
There are two forces on the 2 kg box in the overhead view of the following figure, but only one is shown. For F1=20N, a= 12 m/s2
maw [93]

Answer:

second force = 32.784

Magnitude =\sqrt{32.784

θ = -90°

Explanation:

a)

Fnet = ma

F1 + F2 = ma

20N + F2 = 2(12 × cos30° + 12 ×sin30°)

F2 = 2 × 12 ( sin 30° + cos 30°)

    = 24 × ( 1 + √3 )÷ 2

  =12 (1 +√3 )

  = 32.784

b) \sqrt{12(1 +\sqrt{3}}

= \sqrt{12 ( 1+ 1.732)}

= \sqrt{12 (2.732)}

= \sqrt{32.784}  

c)

θ = 30° + 180°

θ = 210°

210° - 300°

θ = -90°

8 0
2 years ago
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