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valentinak56 [21]
2 years ago
15

A record turntable is rotating at 33 1 3 rev/min. a watermelon seed is on the turntable 5.0 cm from the axis of rotation. (a) ca

lculate the acceleration of the seed, assuming that it does not slip. (enter the magnitude.)
Physics
2 answers:
kvv77 [185]2 years ago
8 0
Since the motion is rotational, we can associate this with centrifugal/centripetal acceleration. The formula for this is:

a = v²/r = (rω)²/r = rω²
where ω is the angular velocity and r is the radius

Let's convert v in terms of rad/min. The conversion from rev to rad is, 1 rev = 360° = 2π. So,
ω = 33 1/3 rev/min * 2π rad/1 rev = 209.44 rad/min

a = 5 cm(209.44 rad/min)²
<em>a = 219,325.568 cm/min²</em>
LenaWriter [7]2 years ago
4 0
Distance D = 5 cm = 0.05m 
Revolutions = 33.33 per min => t = 60 sec 
Acceleration is v^2 / r, so first we need to find velocity 
Velocity = (D x 3.14 x r) / t => (0.05 x 3.14 x 33.33) / 60 
Velocity = 0.0872 m/s 
Acceleration = v^2 / r = 0.0872^2 / 0.05 = 0.152.
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Vladimir [108]

Answer: 65.9^{\circ}

Explanation:

Given

Intensity must be reduced by a factor of 6

Intensity is given by I=I_o\cos ^2\theta\\

Substitute I by \frac{I_o}{6}

\Rightarrow \dfrac{I_o}{6}=I_o\cos ^2\theta\\\\\Rightarrow \cos^2\theta =\dfrac{1}{6}\\\\\Rightarrow \cos \theta=\dfrac{1}{\sqrt{6}}\\\\\Rightarrow \theta=65.9^{\circ}

So, the disk must be rotated by an angle of 65.9^{\circ} .

7 0
2 years ago
The acceleration due to gravity is 9.81 m/s2, towards the Earth. Rain falling from an altitude of 9,000 m would fall for about 1
Anna [14]

Answer:

the final speed of the rain is 541 m/s.

Explanation:

Given;

acceleration due to gravity, g = 9.81 m/s²

height of fall of the rain, h = 9,000 m

time of the rain fall, t = 1.5 minutes = 90 s

Determine the initial velocity of the rain, as follows;

h = ut + \frac{1}{2} gt^2\\\\9000 = 90u +  \frac{1}{2} (9.8)(90)^2\\\\9000 = 90u + 39690\\\\90u = -30690\\\\u = \frac{-30690}{90} \\\\u = -341 \ m/s

The final speed of the rain is calculated as;

v^2 = u^2 + 2gh\\\\v^2 = (-341)^2 + 2(9.8\times 9000)\\\\v^2 = 292681\\\\v = \sqrt{292681} \\\\v = 541 \ m/s

Therefore, the final speed of the rain is 541 m/s.

3 0
3 years ago
Read 2 more answers
Two planets having equal masses are in circular orbit around a star. Planet A has a smaller orbital radius than planet B. Which
vova2212 [387]

Answer:

Explanation:

To solve this, we must know two things.

First, the force of gravity acting on an orbiting object is equal to its mass times centripetal acceleration.

Second, the force of gravity between two objects is defined by Newton's law of universal gravitation: Fg = mMG/r², where Fg is the force of gravity, m and M are the masses of the objects, G is the universal constant of gravitation, and r is the distance between the objects.

Therefore:

Fg = m v²/r

mMG/r² = m v²/r

v² = MG/r

The potential energy of each planet is:

PE = mgr = m (MG/r²) r = mMG/r

The kinetic energy of each planet is:

KE = 1/2 mv² = 1/2 m (MG/r) = 1/2 mMG/r

The total mechanical energy is:

ME = PE + KE = 3/2 mMG/r

Since both planets have the same mass, the only difference is the orbital radius.  Since planet A has a smaller orbital radius, it has more potential energy, more kinetic energy, and more mechanical energy.

6 0
3 years ago
What are some interesting facts about electrons?
Norma-Jean [14]
-- all electrons are identical
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8 0
3 years ago
The diagram shows the parabolic path of a projectile that leaves the foot of a kicker with a horizontal velocity of 15 m/s and a
rusak2 [61]

Answer:

I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.

Explanation:

In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:

R = V₀² Sin 2θ/g

where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.

The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:

V₀ = √(V₀ₓ² + V₀y²)

where,

V₀ₓ = Horizontal Velocity

V₀y = Vertical Velocity

Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.

<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>

4 0
3 years ago
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