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Sauron [17]
2 years ago
14

A conveyor belt at a recycling plant launches bottles and bottle caps into the air, so that an automatic image recognition devic

e can count them. They land same height at which they are launched. Glass bottles are launched at an angle of 35.0 degrees above horizontal and land after 0.500 seconds.
a) Solve for the initial speed at which the bottles are launched.
b) Solve for the horizontal displacement at which the bottle land
Physics
1 answer:
dem82 [27]2 years ago
6 0

(a) The initial speed at which the bottles are launched is 4.27 m/s.

(b) The horizontal displacement at which the bottle land is 1.75 m.

<h3>Initial speed of the bottle</h3>

The initial speed of the bottle is calculated as follows;

T = \frac{2usin\theta}{g}

where;

  • T is time of flight
  • u is the initial speed

2usinθ = Tg

u = Tg/(2sinθ)

u = (0.5 x 9.8)/(2 x sin35)

u = 4.27 m/s

<h3>Horizontal displacement of the bottle</h3>

X = u²sin(2θ)/g

X = (4.27² x sin(70))/(9.8)

X = 1.75 m

Learn more about projectile here: brainly.com/question/12870645

#SPJ1

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A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that incre
Evgen [1.6K]

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20  \ \Omega

Answer:

The current is   I =  0.0007 41 \ A

Explanation:

From the question we are told that

   The  area is  A = 8.00 \ cm^2  = 8.0 *10^{-4} \  m^2

   The initial magnetic field at t_o = 0 \ seconds  is B_i = 0.500 \ T

   The magnetic field at t_1 = 0.99 \ seconds is  B_f  = 1.60 \ T

     The resistance is  R = 1.20  \ \Omega

Generally the induced emf is mathematically represented as

      \epsilon  =  A * \frac{B_f - B_i }{ t_f - t_o }

=>   \epsilon  =  8.0 *10^{-4} * \frac{1.60  - 0.500 }{ 0.99- 0  }

=>   \epsilon  = 0.000889 \ V

Generally the current induced is mathematically represented as

     I = \frac{\epsilon}{R }

=>  I = \frac{0.000889}{ 1.20 }  

=>  I =  0.0007 41 \ A  

6 0
3 years ago
Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of be
ANTONII [103]

Answer:

W=315 x 10⁵ J

Explanation:

Given that

F= 2.5 x 10⁵ N

d= 90 m

K.E.=5.4 x 10⁷ J

We know that work done by all force is equal to the change in kinetic energy

Lets take work done by catapult is W

W + F.d= K.E.

W= 5.4 x 10⁷ -  2.5 x 10⁵  x 90 J

W= (540 - 25 x 9) 10⁵ J

W=315 x 10⁵ J

5 0
3 years ago
A discharge lamp rated at 25 W (1 W = 1 J/s) emits yellow light of wavelength 580 nm. How many photons of yellow light does the
Scrat [10]

Answer:

the number of photons of yellow light does the lamp generate in 1.0 s is 7 x 10^{19}

Explanation:

given information:

power, P = 25 W

wavelength. λ - 580 nm = 5.80 x 10^{-7} m

time, t = 1 s

to calculate the number of photon(N), we use the following equation

N = λPt/hc

where

λ = wavelength (m)

P = power (W)

t = time interval (s)

h = Planck's constant (6.23 x 10^{-34} Js)

c = light's velocity (3 x 10^{8} m/s^{2})

So,

N = λPt/hc

   = (5.80 x 10^{-7})(25)(1)/(6.23 x 10^{-34})(3 x 10^{8} m/s^{2})

    = 7 x 10^{19}

3 0
3 years ago
How would improvement in use of renewable energy sources impact climate change sea-level rise?
bonufazy [111]

Answer:

Almost immeasurably small.

Explanation:

The STORY is that humans are BAD for the environment and have caused a HUGE change in the amount of CO₂ in the atmosphere.

Let's look at the reports and draw our own conclusions.

Current CO₂ levels are 409.8 parts per million (PPM)

at the beginning of the Industrial revolution in the 1700's, the presumed beginning of the huge increase in CO₂ the level was about 280 PPM

For perspective lets assume we capture the whole atmosphere and squish it down to 2400 one liter bottles of air

That's 100 cases of 24 bottles per case.

We now separate all the air components into their own bottles

Nitrogen is 78% of our air, so we subtract 78 cases from our 100 leaving 22

Subtracting Oxygen at 21% of air leaves 1 case of liter bottles left

Of those 24 bottles, Argon makes up 0.93% of air so we subtract 22 bottles

The remaining two bottles contain all of the other gasses in our air, One of those bottles contains CO₂.

If we take the CO₂ levels from the 1700's at about 280 PPM as a baseline and assume ALL of the increase is human caused, that is (410 -280) / 280 = 46 % of the total.

The human caused addition of CO₂ would be 460 mililiters out of 2400 liters over the course of 250 years 

The claim is, that less than half of a liter of CO₂ out of 2400 liters of air is responsible for heating not only the gas in all the other bottles but also the surface of the earth itself.

Personally, it boggles my mind.

And it says NOTHING of a far more powerful greenhouse gas that is far more prevalent in the atmosphere...water vapor.

Water vapor is about 1% of air at sea level and about 0.4% overall. It was not considered in the above analysis because water vapor can condense out and is not a constant in the air.

Notice that there is about 100 times the amount of water vapor in the air as compared to CO₂. Water vapor also has between 4 and 8 times the greenhouse effect that CO₂ does.

Makes one wonder why we choose to pick on CO₂.

7 0
3 years ago
In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t
Vitek1552 [10]

Answer:

So the ratio will be \frac{T_L}{T_H}=-0.171

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So Q=450j

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency \eta =\frac{W}{Q}=\frac{290}{450}=0.6444

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine =\frac{0.6444}{0.55}=1.171

Efficiency of Carnot engine is \eta =1-\frac{T_L}{T_H}

1.171 =1-\frac{T_L}{T_H}

\frac{T_L}{T_H}=-0.171

3 0
3 years ago
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