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valentinak56 [21]
3 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]3 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]3 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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A car with a mass of 2000 kg is moving around a circular curve at a uniform velocity of 25 m/s per second. The curve has a radiu
IrinaK [193]
The centripetal force is:
 F = mv² / R
 Where:
 m: mass of the object
 v: object speed
 R: radius of the curve.
 We have to:
 m = 2000kg
 v = 25 m / s
 R = 80 meters.
 Then the centripetal force acting on the vehicle is:
 F = (2000kg * (25m / s) ²) / 80m
 F = 15625 N
4 0
3 years ago
PLEASE HELP ASAP!!!! What displacement did the object undergo in the time interval between t= 2.0s and t=8.0s?
Katyanochek1 [597]
Displacement s = (u+v)*t/2 (t refers to delta time)
= (0.45 + 2.7)*6/2
= 3.15*3
= 9.45 m
3 0
3 years ago
What is black body radiation? Explain in detail.
tangare [24]

An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called black-body radiation

hope it helps

3 0
2 years ago
A 7 kg ball is moving at a constant speed of 5 m/s. A force of 300 N is applied to the ball for 4 s. The new speed of the ball i
olasank [31]

Answer:

The new speed of the ball is 176.43 m/s

Explanation:

Given;

mass of the ball, m = 7 kg

initial speed of the ball, u = 5 m/s

applied force, F = 300 N

time of force action on the ball, t = 4 s

Apply Newton's second law of motion;

F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v =  \frac{Ft}{m} + u

where;

v is new speed of the ball

v =  \frac{Ft}{m} + u\\\\v =\frac{300*4}{7} + 5\\\\v = 176.43 \ m/s

Therefore, the new speed of the ball is 176.43 m/s

8 0
3 years ago
What is the equivalent resistance of a circuit that contains three 10.0 Ω
yanalaym [24]

Answer:

O D.30.0 Ω

Explanation:

this <em>is </em><em>the </em><em>correct </em><em>answer</em><em>!</em>

8 0
2 years ago
Read 2 more answers
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