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3 years ago

A 1.5 kg tether ball is hit so that it circles the pole with an angular speed of 4m/s

1 answer:

5 0

Well, you didn't ask a question, and 4 m/s is not an angular speed.
So all I can offer is a couple of observations:

1). The tension in the rope is

M V² / R = (1.5 kg) x (4 m/s)² / R

   = (24 kg-m²/s²) / (distance of the ball from the pole).

2). Tetherball was the only thing I played at camp,
   more than 60 years ago, and I loved it !
       It was a tough game, because we had to skin
   our own T.Rex and use his hide to make the ball
   and his guts for the rope.

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The even distribution of weight in a compostion is known as

mojhsa [17]

The answer is Balance. hopei helped? ahahahaha

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3 years ago

Let's talk skateboard velocity how much mean when you get my going down a 3 x 3" metal ramp. No posers are allowed to answer th

denpristay [2]

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quarte embri Oslo

Explanation:

8 0

2 years ago

Read 2 more answers

imagine that 501 people are present in a movie theater of volume 8.00 x10^3 that is sealed shut so no air can escape. Each perso

Semenov [28]

Answer:

The temperature of air will increase by $\Delta T=41044.967\ K$

Explanation:

Given:

no. of person in a theater, $n=501$

volume of air in the theater, $V=8\times 10^3\ m^3$

rate of heat given off by each person, $P=110\ J.s^{-1}$

duration of movie, $t=2\ hr=7200\ s$

initial pressure in the theater, $p_i=1.01\times 10^5\ Pa$

initial temperature in the theater, $T_i=20+273=293\ K$

specific heat capacity of air at the given conditions, $c=1.0061\ J.kg^{-1}.K^{-1}$

<u>The total quantity of heat released by the total people in the theater during the movie:</u>

$Q=n.P.t$

$Q=501\times 110\times 7200$

$Q=396792000\ J$

<u>Form the relation of heat capacity:</u>

$Q=m.c.\Delta T$

∵ $p_i.V=m.R.T$

$Q=(\frac{p_i.V}{R.T}) \times c\times (T_f-T_i)$

$396792000=(\frac{1.01\times 10^5\times 8\times 10^3}{287\times 293}) \times 1.0061\times (T_f-293)$

$T_f=41337.967\ K$

Change in temperature of air:

$\Delta T=41044.967\ K$

8 0

3 years ago

A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci

Vlad [161]

Answer:

Approximately $\rm 2.5\; m \cdot s^{-1}$ .

Explanation:

Let the increase in the rocket's velocity be $\Delta v$ . Let $v_0$ represent the initial velocity of the rocket. Note that for this question, the exact value of $v_0$ doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

Mass of the rocket with the 5 kg of fuel: $1000$ .
Initial velocity of the rocket and the fuel: $v_0$ .
Hence the initial momentum of the rocket: $1000\,v_0$ .

Mass of the rocket without that 5 kg of fuel: $1000 - 5 = 995$ .
Final velocity of the rocket: $v_0 + \Delta v$ .
Hence the final momentum of the rocket: $995\,(v_0 + \Delta v)$ .

Mass of the 5 kg of fuel: $5$ .
Final velocity of the fuel: $v_0 - 500$ (assuming that the the 500 m/s in the question takes the rocket as its reference.)
Hence the final momentum of the fuel: $5\,(v_0 - 500)$ .