A 0.62kg basketball is dropped from a height of 1.7 meters. How much work is done ?

An airplane starts from rest and accelerates at 10.8 m/s2 . what is its speed at the end of a 400 m long runway?

Cloud [144]

The final speed of an airplane is v = 92.95 m/s

The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.

Solution-

Here given,

Acceleration a= 10.8 m/s2 .

Displacement (s)= 400m

Then to find final speed of airplane v=?

Therefore from equation of motion can be written as,

v²=u²+ 2as

where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.

v²= 2×10.8 m/s²×400m

v²=8640m/s

v=92.95m/s

hence the final speed of airplane v =92.95m/s

To know more about speed

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5 0

1 year ago

Which is a consequence of the third law of thermodynamics Energy is not always conserved. Engines cannot discharge waste heat. H

miss Akunina [59]

Heat engines are less than 100% efficient because absolute zero cannot be reached

8 0

3 years ago

Read 2 more answers

A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0

3 years ago

Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This

vova2212 [387]

Answer:

$1.554\times 10^{32}\ \text{kg}$

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = $6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2$

Radius of orbit is given by

$R=\dfrac{vT}{2\pi}$

We have the relation

$\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}$

The mass of each star is $1.554\times 10^{32}\ \text{kg}$

6 0

3 years ago

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

suter [353]

Answer:

actual efficiency is 47.78 %

Carnot efficiency is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy = 7.0 × 10^9 cal (4.184 J/1 cal) = 29.29 × 10^9 J

actual efficiency = 1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency = 0.4778

actual efficiency is 47.78 %

and

Carnot efficiency is = 1 - ( 703 / 2173 )

so Carnot efficiency is = 0.67648

Carnot efficiency is 67.65 %

and

power output = work / time

power output = 1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0

3 years ago