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4 years ago

How to reduce friction of moving a heavy load .

1 answer:

3 0

● Use lubricants such as oil.
● Place the load on a piece of cloth as it can ease the friction
● Place the load on a cart with wheels so it can be converted to rolling friction and can be pushed easily.
● Place the load on an inclined plane

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(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of en

inysia [295]

Answer:

4.42 x 10⁷ W/m²

Explanation:

A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

E = A/η

E = 500/0.90

E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

P = E /t

P = 555.55/4.00

P = 138.88 Watt

d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m

Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Intensity of the beam is given as

I = P /A

I = 138.88 / (3.14 x 10⁻⁶)

I = 4.42 x 10⁷ W/m²

6 0

3 years ago

Read 2 more answers

A conducting bar moves along a circuit with a constant velocity. A constant magnetic field is perpendicular to the bar and circu

lana [24]

Answer:

0.500 T

Explanation:

Since the change in time and the number of coils are both 1, I set the problem up to be 1.3=(1.5(x)-13(x)). I then plugged in numbers for x until I got the answer to be 1.3 V.

6 0

3 years ago

If you're going 80 mph how long does it take to go 80 miles

Elza [17]

1 mile. Is this a joke lol

6 0

4 years ago

Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How

zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

$angle = \frac{1 Parsec}{distance}$

so we have

$distance = \frac{1 Parsec}{angle}$

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

$d_a = \frac{1 Parsec}{1} = 1 Parsec$

distance of star B is given as

$d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec$

So star A is closer than star B

7 0

3 years ago

At what wavelength would a star radiate the greatest amount of energy if the star has a surface temperature of 60,000 K?

kompoz [17]

Answer:

$\lambda=4.81\times 10^{-8}\ m$

Explanation:

We have,

The surface temperature of the star is 60,000 K

It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

$\lambda T=2.89\times 10^{-3}$

Here,

$\lambda$ = wavelength

$\lambda=\dfrac{2.89\times 10^{-3}}{60000}\\\\\lambda=4.81\times 10^{-8}\ m$

So, the wavelength of the star is $4.81\times 10^{-8}\ m$ .

7 0

3 years ago