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2 years ago

9

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.91 J1.91 J of work to set the ball s

pinning from rest, at what angular speed ????ω will the ball rotate? Model a basketball as a thin-walled hollow sphere. For a men's basketball, the ball has a circumference of 0.749 m0.749 m and a mass of 0.624 kg0.624 kg .

1 answer:

Black_prince [1.1K]2 years ago

7 0

Answer:

ω = 4.07 rad/s

Explanation:

By conservation of the energy:

W = ΔK

$1.91J = I/2*\omega^2$

where $I = 2/3*m*R^2=0.23kg.m^2$

Solving for ω:

$\omega = \sqrt{W*2/I} =4.07rad/s$

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar. In a survey conduc

jarptica [38.1K]

Answer: The expected population that wants flexible working hours in ranging from 58% to 62 %.

Explanation:

Percentage of sample respondents agreed for flexible working hours = 60%

The margin of the error = ± 2%

The expected population that wants flexible working hours in ranging from:

$\text{percentage of agreed}-2\%$ lower range

$\text{percentage of agreed}+\2%$ upper range

$60\%-2\% to 60\%+2\%$ that is 58% to 62 %.

The expected population that wants flexible working hours in ranging from 58% to 62 %.

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2 years ago

From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:

bazaltina [42]

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

$T^2\propto a^3\\\\T^2=ka^3$

It is mentioned that, an asteroid with an orbital period of 8 years. So,

$(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU$

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

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1 year ago

Two identical ladders are 3.0 m long and weigh 600 N each. They are connected by a hinge at the top and are held together by a h

ruslelena [56]

Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

Consider half of the ladder,

let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.

$Y+F=600$ ....(I)

$X=T$ .....(II)

On taking moment about base

$X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0$

Put the value into the formula

$X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0$

$3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30$

$1.598T=450-1.5(600-F)$ ....(III)

We need to calculate the force for ladder

$2F=600\trimes 2$

$F=600\ N$

We need to calculate the tension in the rope

From equation (3)

$1.598T=450-1.5(600-600)$

$1.598T=450$

$T=\dfrac{450}{1.598}$

$T=281.60\ N$

Hence, The tension in the rope is 281.60 N.

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2 years ago

Read 2 more answers

In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis

Rus_ich [418]

M1 descending
−m1g + T = m1a

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a

a = (m2 − m1)g/m1 + m2
= (5.60 − 2)/(2 + 5.60) x 9.81
= = 4.65m/s^2

5 0

1 year ago

A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a

Korvikt [17]

Answer:

$v^2=v_o^2-2\times a\times (v_o.t)$

Explanation:

Given:

Initial velocity of the vehicle, $v_o$

distance between the car and the tree, $x_f$

time taken to respond to the situation, $t$

acceleration of the car after braking, $a$

Using equation of motion:

$v^2=u^2+2a.s$ ..............(1)

where:

$v=$ final velocity of the car when it hits the tree

$u=$ initial velocity of the car when the tree falls

$a=$ acceleration after the brakes are applied

$s=$ distance between the tree and the car after the brakes are applied.

$s=v_o\times t$

Now for this situation the eq. (1) becomes:

$v^2=v_o^2-2\times a\times (v_o.t)$ (negative sign is for the deceleration after the brake is applied to the car.)

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1 year ago