What do solar cells convert into electricity? wind SERE water sunlight plants

It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used

stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

$a_{c}$ = rω²

ω² = $a_{c}$ / r

ω = √( $a_{c}$ / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a $a_{c}$ = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²

ω = 0.0500647 ≈ 0.05 rad/s

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM

1 rpm = 60 rph

so

ω = 0.478082 × 60

ω = 28.6849 revolutions per hour

Therefore, the required revolution per hour is 28.6849

7 0

2 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

2 years ago

22) A driver traveling at 80 km/h brakes her 2000 kg truck to stop for a red light. How much internal energy is produced

blsea [12.9K]

Answer:

$E=4.9\times 10^5\ J$

Explanation:

Given that,

The speed of a driver, v = 80 km/h = 22.22 m/s

The mass of the truck, m = 2000 kg

We need to find how much internal energy is produced. The internal energy produced by a truck is its kinetic energy and it is given by :

$E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\cdot22.22^{2}\cdot2000\\\\E=493728.4\ J$

or

$E=4.9\times 10^5\ J$

So, $4.9\times 10^5\ J$ of internal energy is produced.

3 0

2 years ago

A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a

Anton [14]

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

5 0

2 years ago

Two technicians are discussing electronic leveling systems. Technician A says that a weight should be placed in the vehicle as p

Ronch [10]

Answer:

C. Both A and B

Explanation:

Electronic leveling systems: These are automatic lifting jacks use for raising wheels and vehicle when parked or when work is to be done on the vehicle engine underneath or on wheels. unlike the manual jacks it is very quick and efficient.

Both technician right, for Technician A, a weight may be use during the procedure for balancing and leveling when raising, and for Technician B, the sensor in the electronic leveling systems is adjustable to adjust the height so as to maintain balancing and levelling.

8 0

2 years ago