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3 years ago

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The angular momentum of a flywheel having a rotational inertia of 0.200 kg·m2 about its central axis decreases from 3.80 to 0.40

0 kg·m2/s in 1.50 s. (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? N·m (b) Assuming a uniform angular acceleration, through what angle does the flywheel turn? rad (c) How much work is done on the wheel? J (d) What is the average power of the flywheel? W

1 answer:

babymother [125]3 years ago

3 0

Explanation:

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Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to

marta [7]

Answer:

$\frac{ 112.5}{15+m_{A}}=v_{f}$

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed $v_{iA}$ of 0 m/s and a mass $m_{A}$ and the IMAX camera with an initial speed $v_{ic}$ of 7.5 m/s and a mass $m_{c}$ of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

$P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}$

By the law of conservation we know that $P_{i} =P_{f}$

For $P_{f}$ (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

$P_{i} =P_{f}=112.5\\$

$112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}$

3 0

3 years ago

A car slows down at -5.00 m/s^2 until it comes to a stop after travelling 15.0 m. How much time did it take to stop?

Xelga [282]

In physics, there are already derived equation that are based on Newton's Law of Motions. The rectilinear motions at constant acceleration have the following equations:

x = v₁t + 1/2 at²
a = (v₂-v₁)/t

where
x is the distance travelled
v₁ is the initial velocity
v₂ is the final velocity
a is the acceleration
t is the time

Now, we solve first the second equation. Since it mentions that the car comes eventually to a stop, v₂ = 0. Then,

-5 = (0-v₁)/t
-5t = -v₁
v₁ = 5t

We use this new equation to substitute to the first one:
x = v₁t + 1/2 at²
15 = 5t(t) + 1/2(-5)t²
15 = 5t² - 5/2 t²
15 = 5/2 t²
5t² = 30
t² = 30/5 = 6
t = √6 = 2.45

Therefore, the time it took to travel 15 m at a deceleration of -5 m/s² is 2.45 seconds.

5 0

3 years ago

The microwave in your kitchen operates at 2,358,000 Hz. If the speed of light is 300,000,000 m/s, what is the wave’s wavelength?

Novosadov [1.4K]

Answer:

wavelength = speeed / frequency

= 300,000,000 / 2358000

= 300 / 2.358

= 127.2 m

5 0

4 years ago

Will give 45 points to who ever solves this

Lubov Fominskaja [6]

Answer:

1. 230 kg...

W = m * g

W= 230 kg * 9.81 m/s^2

<u>W= 2256,3 N</u>

m= 230kg , W = 2256,3 N , g= 9.81 m/s^2

2. 887 N

W= m * g

887 N = m * 9.81 m/s^2

<u>m= 90,42 kg</u>

m= 90,42 kg, W = 887 N , g= 9.81 m/s^2

3. 420 kg

W= m * a

w= 420 kg * 9.81 m/s^2

<u>w=</u><u> </u><u>4120,2 N</u>

m= 420 kg , W = 4120,2 N , g= 9.81 m/s^2

4. Determine the gravity on Pluto where a 15 kg object weighs 55.5N.

w = m * g

55.5 N = 15 kg * g

<u>a= 3,7 m/s^2</u>

m = 15 kg , W= 55.5 N , g= 3,7 m/s^2

7 0

2 years ago

Read 2 more answers

A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, c

Aleks04 [339]

Answer:

The velocities of the skaters are $v_{1} = 3.280\,\frac{m}{s}$ and $v_{2} = 0.024\,\frac{m}{s}$ , respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$ (1)

Second skater

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$ (2)

Where:

$m_{1}$ - Mass of the first skater, in kilograms.

$m_{2}$ - Mass of the second skater, in kilograms.

$v_{1,o}$ - Initial velocity of the first skater, in meters per second.

$v_{1}$ - Final velocity of the first skater, in meters per second.

$v_{b}$ - Launch velocity of the meter, in meters per second.

$v_{2}$ - Final velocity of the second skater, in meters per second.

If we know that $m_{1} = 70\,kg$ , $m_{b} = 0.043\,kg$ , $v_{b} = 32\,\frac{m}{s}$ , $m_{2} = 58.5\,kg$ and $v_{1,o} = 3.30\,\frac{m}{s}$ , then the velocities of the two people after the snowball is exchanged is:

By (1):

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$

$m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}$

$v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}$

$v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)$

$v_{1} = 3.280\,\frac{m}{s}$

By (2):

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$

$v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}$

$v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}$

$v_{2} = 0.024\,\frac{m}{s}$

5 0

3 years ago