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4 years ago

7

A merry go round (mass of 200kg and radius 5m) is rotating so that the outside edge is moving 10m/s. A giant bug of 6kd at a dis

tance of 4m from the center. What is the angular velocity of the merry go round?

1 answer:

romanna [79]4 years ago

8 0

Answer:

2 rad/s

Explanation:

For a rotating object, the linear velocity is given by

$v=\omega r$

where $\omega$ is the angular velocity and $r$ is the radius.

$\omega=\dfrac{v}{r}$

The edge has a linear velocity of 10 m/s and the radius at the edge is 5 m.

$\omega=\dfrac{10 \text{ m/s}}{5 \text{ m}} = 2 \text{ rad/s}$

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Answer:

The mass number will decrease by eight (8).

Explanation:

Given;

successive beta (b), alpha (a), alpha (a) emissions.

Generally, when a radioactive element emits a beta-particle (b), its mass number doesn't increase but its atomic number increases by 1 . $(^{0}_{-1}\beta )$

Also when a radioactive element emits an alpha-particle (a), its mass number decreases by 4, while its atomic number decrease by 2. $(^4_2\alpha)$

For the given question, a successive beta (b), alpha (a), and alpha (a) emissions = (0) + (-4) + (-4) = -8

Thus, when a radioactive element emit a successive beta (b), alpha (a), alpha (a) particles, the mass number will decrease by eight (8).

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3 years ago

Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea

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Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, $t=0.50\ s$

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, $v=0\ m/s$

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, $a=g=-9.8\ m/s^2$

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

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3 years ago

A rigid tank initially contains 1.4 kg saturated liquid water at 200◦C. At this state, 25 percent of the volume is occupied by w

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Answer:

(a) Volume of the tank is $6.47\times 10^{- 3}\ m^{3}$

(b) Temperature is $371^{\circ}C$

Pressure is 21.3 kPa

(c) The change in internal energy is 1373.54 kJ/kg

Solution:

As per the question:

Mass of liquid in the tank, m = 1.4 kg

Temperature, T = $200^{\circ}C$

Volume occupied by water, V = 25% $V_{t}$ = 0.25 $V_{t}$

Volume occupied by air, V' = 75% $V_{t}$

where

$V_{t}$ = Volume of tank

Now,

(a) In order to calculate the volume of the tank, we make use of the steam table for specific volume at as given temperature:

At $200^{\circ}C$ , the specific volume, v = 0.001157 $m^{3}/kg$

At $200^{\circ}C$ , the internal energy, u = 850.46 kJ/kg

Now, Volume of water, V = mv = $1.4\times 0.001157 = 1.62\times 10^{- 3} m^{3}$

Thus

$V = 0.25V_{t}$

$V_{t} = \frac{1.62\times 10^{- 3}}{0.25} = 6.47\times 10^{- 3}\ m^{3}$

(b) For the final temperature and pressure, we calculate the specific volume, v' and then find the corresponding temperature and pressure from the steam table:

v' = $\frac{V_{t}}{m} = \frac{6.47\times 10^{- 3}}{1.4} = 4.63\times 10^{- 3}\ m^{3}/kg$

The corresponding temperature to this specific volume, T' = $371^{\circ}C$

The corresponding pressure to this specific volume, P' = 21.3 kPa

The corresponding internal energy to this specific volume, u' = 2224 kJ/kg

(c) The change in the internal energy of water is given by:

$\Delta U = u' - u = 2224 - 850.46 = 1373.54 kJ/kg$

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