Which tools would be use to find an irregularly shaped object’s mass and volume?

A child is pushing a merry-go-round. the angle through which the merry-go-round has turned varies with time according to θ(t)=γt

Andreyy89

I attached the missing part of the question.
Part A
Angular velocity is simply the rate at which angle changes, in other words, it is the first derivative of the angle function with respect to time:
$\theta (t)=\gamma t+\beta t^3\\ w(t)=\frac{d\theta(t)}{dt}\\ w(t)=\gamma +3\beta t^2$
Part B
Initial value is the value at t=0. To find initial value we simply plug t=0 into the equation.
$w(0)=\gamma +3\beta\cdot 0^2\\
w(0)=\gamma\\
w_0=\gamma= 0.422\frac{rad}{s}$
Part C
To find these values we simply need to plug in t=5 in the equation.
$w(5)=\gamma +3\beta\cdot 5^2\\
w(5)=\gamma +75\cdot \beta=0.422+75 \cdot 1.35\cdot10^{-2}=1.43\frac{rad}{s}$
Part D
Average angular velocity is total angular displacement divided by time:
$w_{av}=\frac{\Delta \theta}{\Delta t}$
$\Delta \theta=\theta(5)-\theta(0)=\gamma\cdot5+\beta 5^3\\
\Delta \theta=0.422\cdot5+125\cdot1.35\cdot10^{-2}=3.7975 $rad$
The average angular velocity is:
$w_{av}=\frac{\Delta \theta}{\Delta t}=\frac{3.7975 }{5}=0.7595\frac{rad}{s}$

3 0

3 years ago

Heavier elements like gold are created in the most massive stars and spread through the universe when these stars explode. This

myrzilka [38]

Answer:

Big bang

Explanation:

Trust me I know :)

If I helped, mark brainliest

Your best regards,

Lillith (from brainly)

7 0

3 years ago

How much force must be applied to move a 3.0 kg toy train that would accelerate to 7.0 m/sec2?

stiv31 [10]

Answer: 21 N

Explanation:

Force=mass x Accceleration

So 21=3x7

5 0

3 years ago

Read 2 more answers

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

rjkz [21]

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:

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

3 0

3 years ago

At a 1500 m race, Ken ran at an average speed of 200 m/min. How long did it take for Ken to finish the race? (meter = m, minute

Anettt [7]

$speed = \frac{distance}{time} \\ = > \frac{200m}{min} = \frac{1500m}{t} \\ = > \frac{200m}{60s} = \frac{1500m}{t} \\ = > \frac{10m}{3s} = \frac{1500m}{t} \\ = > t = 1500m \times \frac{3s}{10m} \\ = > t = 150 \times 3s \\ = > t = 450s \:$

This is the answer.

Hope it helps!!

3 0

3 years ago