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Kamila [148]
3 years ago
15

3. Looking at the image below label the areas of highest and lowest kinetic energy of both gravitational potential

Physics
1 answer:
MArishka [77]3 years ago
8 0

Explanation:

max potential your welcome i love helping outhers :)

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Evelynn is measuring the pitch of a piano note. What unit of measurement is she most likely recording her value
sweet [91]

Answer:

hertz

Explanation: play piano

5 0
2 years ago
Which was a common goal of Spanish and British explorers in the Southeastern region of North America?
slamgirl [31]

Answer:

B

Explanation:

To discover gold

6 0
3 years ago
A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 4.4 cm from the axis of rotation. (a) Calcul
romanna [79]

Answer:

a) a =0.53 m/s²

b) μ=0.054

c) μ = 0.068

Explanation:

a) If we assume that the turntable is rotating at a constant speed, the only force acting on the seed parallel to the surface, which keeps it  from following a straight line trajectory, is the centripetal force.

So, we can apply Newton's 2nd Law to the seed in this way:

Fnet = m*a = m*ac = m*ω²*r

We have the value of the angular speed, ω, in rev/min, so it is advisable to convert it to rad/sec, as follows:

ω = 33 rev/min*(1 min/60 sec)*(2*π rad/ 1 rev) = 11/10*π rad/sec

So, replacing in (1), we can solve for ac, as follows:

ac = ω²*r = (11/10)²*π²*0.044 m = 0.53 m/s²

b) Now, the centripetal force that we found above, is not a new type of force, it must be a force that explains the behavior of the seed.

As the seed does not slip, the only force acting  on it parallel to the surface, is the static  friction force, which has a maximum value, as follows:

Ff = μ*N

As there is no movement in the vertical direction, this means that the normal force must be equal and opposite to Fg, so we can write the expression for Ff as follows:

Ff = μ*m*g

Now, this force is no other than centripetal force, so we can write this equation:

Ff  = Fc ⇒ μ*m*g = m*ac

⇒ μ*g = ac

Solving for μ:

μ = ac/g = 0.53 m/s² / 9.8 m/s² = 0.054

c) During the acceleration period, added to the centripetal acceleration, as the angular speed is not constant, we will have also an angular acceleration, γ , which we can get as follows:

γ = Δω/Δt = (11/10)*π / 0.37 s = 9.34 rad/sec²

By definition of angular acceleration, there exists a fixed  relationship between the angular acceleration and the tangential acceleration (same as the one between angular and tangential speed), as follows:

at = γ*r = 9.34 rad/sec²*0.044 m = 0.41 m/s

When the turntable has reached to its maximum angular velocity, it will have also the maximum value of the centripetal acceleration, which we have just found out.

So, the magnitude of the total acceleration (at the moment of maximum acceleration) as they are perpendicular each  other) , is given by the following expression:

a = √(ac)²+(at)² = 0.67 m/s²

Now, as friction force opposes to the relative movement between surfaces (the seed and the turntable), it shall be larger than the product of the mass times the total acceleration, acting along  the same action line, so we can say:

Ffmin = μ*m*g = m*a

⇒ μmin = a/g = 0.67 m/s²/9.8 m/s² = 0.068

5 0
3 years ago
A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because
Gnom [1K]

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

\tau = F*d

Where

F= Force

d = Distance

Our values are given as,

m_1 = 0.32Kg

m_2 = 0.4Kg

d = 8.7*10^{-2}m

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

\tau = T_2-T_1

\tau = F_2*d-F_1*d

\tau = m_2g*d-m_1*g*d

\tau = (m_2-m_1)g*d

\tau = (0.4-0.32)(9.8)(8.7*10^{-2})

\tau = 0.068N\cdot m

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is \tau = 0.068N\cdot m

8 0
3 years ago
PLEASE HELP ASAP!!! NO LINKS! Which type of wave does the illustration depict?
Xelga [282]

Answer:

Transverse wave

Explanation:

The wave is moving forwards from the hand to the point of attachment

3 0
3 years ago
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