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Maslowich
2 years ago
14

Assume that the turntable deccelerated during time Δt before reaching the final angular velocity ( Δt is the time interval betwe

en the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, ⟨τ⟩, acting on the bottom disk due to friction with the record?
Physics
1 answer:
Zina [86]2 years ago
6 0

Answer:

See below...

Explanation:

Let’s express ⟨α⟩ in terms of ωi , ωf , and Δt. and torque in terms of It , ωi , ωf , and Δt.

STEP 1.  

The rate of change of angular velocity is Angular acceleration.  

The net change in angular velocity is Average angular acceleration divided by the elapsed time.

⟨α⟩ = ω f −ω i/Δt

STEP 2.

Torque is assumed this way

          dω

   τ = I ----

           dt

.

⟨τ ⟩ = I t (ω f −ω i )/Δt

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Sandra's target heart rate zone is 135bpm—172bpm. Marissa's target heart rate zone is 143bpm—176bpm. They stop playing basketbal
Feliz [49]

Answer: Neither Sandra nor Marissa will be in her THR zone.


Explanation:


1) Actual pulse of both Sandra and Marissa : 144 bpm


2) Decrease of 20 bpm ⇒ 144 bpm - 20 bpm = 124 bpm


3) Sandra's TRH is in the range 135 - 172 bpm.


Since 124 < 135, she will be below the range.


4) Marissa's TRH range is 143 - 176 bpm.


Since, 124 < 143, she is below the range


In conlusion, neither Sandra nor Marissa will be in her THR zone.

6 0
2 years ago
Read 2 more answers
A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N
Mnenie [13.5K]

Answer:

8N and 32N

Explanation:

Given that a  light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center.

To calculate the forces that are exerted by the sawhorses on the board, we must consider the equilibrium of forces acting on the board.

Let the two upward forces produce by the saw horses be P1 and P2

Assuming that the weight is negligible

Sum of the upward forces = sum of the downward forces.

P1 + P2 = 40 ....... (1)

Also, the sum of the clockwise moment = sum of the anticlockwise moments.

Let's assume that the board is uniform. The weight will act at the centre.

Taking moment at the centre:

P1 × 5 + 40 × 2 = 0

P1 = 40 / 5

P1 = 8N

Substitute P1 into equation 1

8 + P2 = 40

P2 = 40 - 8

P2 = 32N

3 0
2 years ago
Read 2 more answers
Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l
alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

m_1=M_p\\m_2 = M_s\\r=R

So the force is

F_A=G\frac{M_p M_s}{R^2}=F_0

For the system planet B - Star B, we have:

m_1 = 4 M_p\\m_2 = M_s\\r=R

So the force is

F=G\frac{4M_p M_s}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

m_1 = M_p\\m_2 = 4M_s\\r=R

So the force is

F=G\frac{M_p (4M_s)}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

G\frac{mM}{r^2}=m\frac{v^2}{r}

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

v=\sqrt{\frac{GM}{r}}

For System A,

M=M_s\\r=R

So the tangential speed is

v_A=\sqrt{\frac{GM_s}{R}}

For system B,

M=M_s\\r=R

So the tangential speed is

v_B=\sqrt{\frac{GM_s}{R}}=v_A

So, the speed of planet B is the same as planet A.

For system C,

M=4M_s\\r=R

So the tangential speed is

v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A

So, the speed of planet C is twice the speed of planet A.

3 0
2 years ago
A bullet is fired in a horizontal direction with a muzzle velocity of 300m/s.In the absence of air resistance,how far will it ha
Kruka [31]

Answer:

304.9m

Explanation:

Given

Velocity = 300m/s

Tim = 1sec

Required

Horizontal distance S

Using the formula

S= ut+1/2gt²

S = 300(1)+1/2(9.8)(1)²

S = 300+4.9

S = 304.9m

Hence the distance travelled is 304.9m

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2 years ago
A 1,160 kg satellite orbits Earth with a tangential speed of 7,446 m/s. If the satellite experiences a centripetal force of 8,95
Juli2301 [7.4K]

Answer:

8.02×10⁵ m

Explanation:

Equation for centripetal force:

F = mv²/r

Solving for r:

r = mv²/F

Given:

F = 8955 N

m = 1160 kg

v = 7446 m/s

r = (1160 kg) (7446 m/s)² / 8955 N

r = 7.182×10⁶ m

The height above the surface is:

h = 7.182×10⁶ m − 6.38×10⁶ m

h = 0.802×10⁶ m

h = 8.02×10⁵ m

4 0
2 years ago
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