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Marianna [84]
3 years ago
14

1.concave lens

Physics
1 answer:
topjm [15]3 years ago
3 0

What to do here tell please.......

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A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00
svetoff [14.1K]

Answer:

a) 200 m

b) 100 m/s

c) 709 m

d) -118.2 m/s

e) 26.24 s

Explanation:

The rocket flies upward with constant acceleretion.

The equation for position under constant acceleration is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

V0 = 0

Y(t) = 1/2 * 25 * t^2

Y(t) = 12.5 * t^2

And speed under constant acceleration:

Vy(t) = Vy0 + a * t

Vy(t) = 25 * t

It burns for 4 s and runs out of fuel

Y(4) = 12.5 * 4^2 = 200 m

V(4) = 25 * 4 = 100 m/s

Form t = 4 the rocket will coast, it will be in free fall, affected only by gravity

It will be under constant acceleration. These new equations will have different starting constants.

Y(t) = Y4 + Vy4 * (t - 4) + 1/2 * g * (t - 4)^2

Vy(t) = Vy4 + g * (t - 4)

When it reaches its highest point it will have a speed of zero.

0 = Vy4 + g * (t - 4)

0 = 100 - 9.81 * (t - 4)

100 = 9.81 * (t - 4)

t - 4 = 100 / 9.81

t = 10.2 + 4 = 14.2 s

At that moment it will have a height of:

Y(14.2) = 200 + 100 * (14.2 - 4) - 1/2 * 9.81 * (14.2 - 4)^2 = 709 m

The rocket will fall and hit the ground:

Y(t) = 0 = 200 + 100 * (t - 4) - 1/2 * 9.81 * (t - 4)^2

0 = 200 + 100 * t - 400 - 4.9 * (t^2 - 8 * t +16)

0 = -4.9 * t^2 + 139.2 * t -278.4

Solving this equation electronically:

t = 26.24 s

At that time the speed will be:

Vy(t) = 100 - 9.81 * (26.24 - 4) = -118.2 m/s

6 0
3 years ago
Assuming the bar has no weight where does the fulcrum (the top point of the tringle) need to be positioned for the two sides to
Inessa05 [86]

Fulcrum need to be positioned balanced with weight on both the sides following law of lever.

What is the physical law of the lever?

  • It is the foundation for issues with weight and balance. According to this rule, a lever is balanced when the weight multiplied by the arm on one side of the fulcrum, which serves as the pivot point for the device, equals the weight multiplied by the arm on the opposing side.
  • The lever is balanced, in other words, when the sum of the moments about the fulcrum is zero.
  • The situation in which the positive moments (those attempting to turn the lever clockwise) equal the negative moments is known as this (those that try to rotate it counterclockwise).
  • Moving the weights closer to or away from the fulcrum, as well as raising or lowering the weights, can alter the balance point, or CG, of the lever.

Learn more about the Fulcrum with the help of the given link:

brainly.com/question/16422662

#SPJ4

3 0
1 year ago
A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their g
REY [17]

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t     (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.  

3 0
3 years ago
Find the cost of excavating a space 84 ft long, 42 ft wide, and 9 ft deep at a cost of $39/yd3. (simplify your answer completely
m_a_m_a [10]

The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864

Information about the problem:

  • Space long= 84 ft
  • Space wide= 42 ft
  • Space deep= 9 ft
  • Cost by yard3 = $39/yd3
  • Total cost= ?

To solve this problem, we have to state the equation using the information of the problem:

Calculating the volume of the total space:

space volume = space long * space wide * space deep

space volume = 84 ft * 42 ft * 9 ft

space volume = 31752 ft3

By converting the volume from ft3 to yd3, we have:

31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3

Calculating the cost of excavating the volume space:

Total cost = space volume * cost by yard3

Total cost = 1176 yd3 * $39/yd3

Total cost = $45864

<h3>What is volume?</h3>

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

Learn more about volume at: brainly.com/question/12628341

#SPJ4

4 0
1 year ago
We divide the electromagnetic spectrum into six major categories of light, listed below. Rank these forms of light from left to
Zigmanuir [339]

Answer:

gamma rays , X rays,  ultraviolet , visible light , infrared,  radio waves

Explanation:

The electromagnetic spectrum is the set of electromagnetic radiations distributed in their different frequencies or wavelengths, which in turn are related to their energy.   If we go from the smallest wavelengths known up to now (because according to physics the electromagnetic spectrum is infinite and continuous) to the longest, the electromagnetic spectrum covers the following radiations:  

Gamma rays, X-rays, ultraviolet, visible light (all the colors we are able to see), infrared, radio waves and microwaves.  

Let's make a brief of them:

-Gamma rays: With a wavelength in the order of 10^{-12}m, is a type of ionizing radiation capable of penetrating matter quite deeply and is able to cause serious damage to the nucleus of the cells. Inaddiito, these rays are used to sterilize medical equipment and food.

-X rays: With a wavelength between 1m and 10km. It is invisible to the human eye, capable of crossing opaque bodies and of being an ionizing radiation.

-Ultraviolet light: Whose wavelength is approximately between 100 nm and 380 nm; is a type of electromagnetic radiation that is not visible to the human eye.

-Visible light: This part of the spectrum is located between ultraviolet light and infrared light (380 nm - 780 nm).  It should be noted, the fact the only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

-Infrared: This type of radiation is not visible to the human eye, since its wavelengths are outside the visible spectrum (between 700 nm and 1 mm).  

These waves can be divided into:  

<u>- Near infrared</u> or long wave infrared: it is the least sensitive to color and is easily absorbed by water.  

<u>- Medium or medium wave infrared:</u> it is also insensitive to color and easily absorbed by water and many types of plastics and paints.  

-<u> Far infrared or short wave infrared: </u>it is more penetrating than the long wave and is good for heating metals, these waves also can pass through clear materials.  

This light has many uses, including heating lamps in physiotherapy and medical treatments, heat sensing devices, among others.

-Radio waves: These are a type of electromagnetic radiation with wavelengths between 10 m to 10,000 m. This type of electromagnetic waves is very well reflected in the ionosphere, the layer of the atmosphere through which they travel directly or using repeaters.  In addition, they are very useful to transport information, being important in telecommunications. They are used not only for conventional radio transmissions but also in mobile telephony and TV.  

5 0
3 years ago
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