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Semenov [28]
3 years ago
12

An object is traveling on a circle with a radius of 6 feet. If in 80 seconds a central angle of 9/4 radians is swept out, then f

ind both the angular speed and the linear speed of the object. Give exact answers. DO NOT provide a decimal approximation.
Physics
1 answer:
trapecia [35]3 years ago
8 0

Answer:

The angular speed of the object is 0.0281 rad/s

The linear speed of the object is 0.169 ft/s

Explanation:

Given;

radius of the circle, r = 6 ft

time of motion of the object around the circle, t = 80 s

central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad

The angular speed of the object is calculated as;

\omega = \frac{\theta }{t} = \frac{2.25 \ rad}{80 \ s} = 0.0281 \ rad/s

The linear speed of the object is calculated as;

v = ωr

v = 0.0281 rad/s   x    6ft

v = 0.169 ft/s

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The amount of space an object takes up is its
Likurg_2 [28]

Volume.

Hope this helps.

r3t40

4 0
3 years ago
Assume that the Moon is in circular orbit around the Earth. The distance between the Moon and the Earth is approximately 3.56 ×
stira [4]

Answer:

Acceleration = 311.2 Km/hr²

Explanation:

Given: Radius of the Orbit  r= 3.56 × 10⁶ km

Period of the orbit = 28 days = 672 hrs

Sol: We have Fc = MV²/r

⇒M ac = MV²/r

⇒ac = V²/r

First we have to Speed V so for this we have to find the circumference ( distance covered by the moon in one orbit)

⇒ Circumference= 2 π r

= 2 × 3.13149 × 3.56 × 10⁶ km

= 22,368,139.69 Km

Now Speed = Distance /time

Speed = 22,368,139.69 Km / 672 hrs

Speed V = 33,285.92 Km/Hr

Now

ac = V²/r = (33,285.92 Km/hrs)² / 3.56 × 10⁶ km

ac = 311.2 Km/hr²

6 0
3 years ago
An electron is released from rest at the negative plate of a parallel-plate capacitor. If the distance across the plate is 2.0 m
ddd [48]

Answer:

Velocity of electron will be 1.325\times 10^6m/sec

Explanation:

We have given distance across the plate d = 2 mm =2\times 10^{-3}m

Potential difference V = 6 volt

We know that potential difference at any distance is given by

V = Ed , here V is potential difference, E is electric field and d  is distance

So E=\frac{V}{d}=\frac{6}{2\times 10^{-3}}=3000N/C

Charge on electron e=1.6\times 10^{-19}C

We know that expression of velocity is given by v=\sqrt{\frac{2qEd}{m_e}}, here q is charge on electron, E is electric field and d is distance

So v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 3000\times 3000}{9.11\times 10^{-31}}}=1.325\times 10^6m/sec

8 0
4 years ago
The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases
Lady bird [3.3K]

Answer: 1.79\ rad/s^2

Explanation:

Given

Initial angular speed is \omega_1=26.2\ rad/s

Final angular speed is \omega_2=36.5\ rad/s

Time period t=5.75\ s

Magnitude of the fan's acceleration is given by

\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}

Insert the values

\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2

Thus, fan angular acceleration is 1.79\ rad/s^2

8 0
3 years ago
Read 2 more answers
What is the charge on an object that has 1.09x10^13 excess electrons?
Nadya [2.5K]

Answer:

-1.74\cdot 10^{-6} C

Explanation:

The electron is the particle that rotates around the nucleus of the atom; it has a negative electric charge equal to :

e=-1.6\cdot 10^{-19}C

which is known as fundamental charge.

For an object containing N excess electrons, the total charge of the object is

Q=Ne

In this problem, the number of excess electrons in the object is:

N=1.09\cdot 10^{13}

Therefore, by plugging it the numbers, we can find the value of Q, the total charge of the object:

Q=(1.09\cdot 10^{13})(-1.6\cdot 10^{-19})=-1.74\cdot 10^{-6} C

5 0
4 years ago
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