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2 years ago

Share your thoughts about this statement by John Wesley "Electricty is the soul of universe"

Physics

2 answers:

yawa3891 [41]2 years ago

5 0

Answer:

ooo

Explanation:

ikadub [295]2 years ago

4 0

Answer:

<em>I think this statement is true at least in modern day times. The world runs on nothing but technology. People use more technology instead of old school textbooks and papers. Imagine living in a world without technology… that means no cars, no trains, no devices, no machines like your stove or printer, no lights and lots more. John Wesley is 100 percent correct with this statement. Electricity is indeed the powerhouse of the universe. </em>

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True or False:

bulgar [2K]

Answer:

False

True

False

8 0

3 years ago

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

$d = v*t$

Where:

v: is the tangential speed of the disk

t: is the time = 30 s

The tangential speed can be found as follows:

$v = \omega*r$

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

$v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s$

Now, the distance traveled by the disk is:

$d = v*t = 5.24 m/s*30 s = 157.2 m$

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

3 0

2 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

3 years ago

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

worty [1.4K]

Answer:

$v_{min} \approx 17.153\,\frac{m}{s}$

Explanation:

The roller coaster begins with maximum kinetic energy and no gravitational potential energy. The gravitational potential energy reaches its maximum when roller coaster is upside down at the top of the circle. The physical model for the roller coaster is constructed by means of the Principle of Energy Conservation:

$\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h$

The minimum velocity is:

$v_{min} = \sqrt{2\cdot g \cdot h}$

Let assume that radio of curvature is measured in meters. Hence:

$v_{min} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(15\,m)}$

$v_{min} \approx 17.153\,\frac{m}{s}$

8 0

3 years ago

you toss a bowling ball straight up into the air with a speed of 2.1. how long does it take the bowling ball to reach its highes

omeli [17]

I am going to assume 2.1 metres per second and that we're rounding acceleration due to gravity to -10 metres per second squared. At the highest point, velocity is going to be 0. v= intial velocity + acceleration*time, sub in 0 for velocity, 2.1 for initial velocity and -10 for acceleration to get 0= 2.1-10t. Now solve for t. t=0.21 seconds.

3 0

3 years ago

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