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Kay [80]
3 years ago
14

Ewxrdctfvgybhjnkml,;kjffff

Physics
1 answer:
grin007 [14]3 years ago
6 0

Answer:

You have to Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100.

Explanation: And here's the answer in Binary for some fun: 01000110 01101001 01101110 01100100 00100000 01111000 00101100 00100000 01111001 00101100 00100000 01100001 01101110 01100100 00100000 01111010 00100000 01110011 01110101 01100011 01101000 00100000 01110100 01101000 01100001 01110100 00100000 01111000 11000010 10110011 00101011 01111001 11000010 10110011 00101011 01111010 11000010 10110011 00111101 01101011 00101100 00100000 01100110 01101111 01110010 00100000 01100101 01100001 01100011 01101000 00100000 01101011 00100000 01100110 01110010 01101111 01101101 00100000 00110001 00100000 01110100 01101111 00100000 00110001 00110000 00110000 00101110

Separate Answer: 01001000 01100101 01101100 01101100 01101111 00100000 01110100 01101000 01100101 01110010 01100101

You might be interested in
The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a
emmasim [6.3K]

The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\  I_2= 0.125 \ foot-candles

Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles

To learn more about the distance refer to the link;

brainly.com/question/26711747

#SPJ1

6 0
2 years ago
Can yall Please fill out them blanks i need help it's due Today.
hjlf

Answer:

tech ed class

Explanation:

4 0
3 years ago
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p
Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0
2 years ago
PLEASE ANYONE SOLVE THIS NOW FAST PLEASE IM IN A HURRY
m_a_m_a [10]

Hello,

<u>Solution for A:</u>

Force = 3.00N

Mass = 0.50 Kgs

Time = 1.50 Seconds

According to newton's second law of motion;

Force = Mass times Acceleration(a)

3.00 = 0.50 * a

a = 3.00/0.50 = 6.00 m/s^2

We know that acceleration = Velocity / time

So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2

<u>Solution for B:</u>

The net force = 4.00N - 3.00N = 1.00N to the left

Force = 1.00N

Mass = 0.50Kg

Time = 3.00 Seconds

Again; F = MA (Where F is force, M is mass and A is acceleration)

1.00N = 0.5 * A

A = 1/0.5 = 2 m/s^2

Velocity = Acceleration * Time = 2 * 3 = 6 m/s

3 0
3 years ago
A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca
UkoKoshka [18]

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial  energy to gravitational potencial energy, so:

E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J

then we have only gravitational potencial energy when the ball is at its maximun height.

E_g=m*g*h

because all the energy was transformed Eg=Ee

h=\frac{0.081J}{9.8m/s^2*m}

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m

6 0
3 years ago
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