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Darya [45]
3 years ago
11

A 4.50-kg centrifuge takes 200 s to spin up from rest to its final angular speed with constant angular acceleration. A point P l

ocated 10.0 cm from the axis of rotation of the centrifuge moves with a speed of 150 m/s when the centrifuge is at full(final) speed. a) [4 points] What was its final angular speed
Physics
1 answer:
Irina-Kira [14]3 years ago
3 0

Answer:

Explanation:

ω = v / R

where ω is angular speed of a point situated at distance R from axis and having a linear speed of v .

Given v = 150 m/s ; R =.10 m

ω = 150 / .10

= 1500 radian /s .

Final angular speed is 1500 radian /s .

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Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is
topjm [15]

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

4 0
3 years ago
What does MA mean for you in real life?
andrezito [222]
To me "Ma" mean in real life is either mama or Massachusttes.
8 0
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In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

7 0
3 years ago
A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is
Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ  = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

  = √(T/A×ρ)

  = √[2.96×10^4/(4.49×10^-3×7860)]

  = 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

7 0
3 years ago
You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi
Elenna [48]

Answer:

\frac{g_{2}}{g_{1}} = \frac{1}{4}

Explanation:

The period of the simple pendulum is:

T = 2\pi\cdot \sqrt{\frac{l}{g} }

Where:

l - Cord length, in m.

g - Gravity constant, in \frac{m}{s^{2}}.

Given that the same pendulum is test on each planet, the following relation is formed:

T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}

\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}

\frac{g_{2}}{g_{1}} = \frac{1}{4}

5 0
3 years ago
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