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2 years ago

Which of the following is a source of thermal energy that is absorbed by Earth's atmosphere?

Physics

1 answer:

Mazyrski [523]2 years ago

8 0

Answer:

100% of the energy entering earth's atmosphere comes from the sun. ~50% of the incoming energy is absorbed by the earth's surface i.e. the land and oceans.

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A projectile is launched with an initial speed of 40 mys from the foor of a tunnel whose height is 30 m. What angle of elevation

KiRa [710]

Answer:

We are given the trajectory of a projectile:

y=H+xtan(θ)−g2u2x2(1+tan2(θ)),

where H is the initial height, g is the (positive) gravitational constant and u is the initial speed. Since we are looking for the maximum range we set y=0 (i.e. the projectile is on the ground). If we let L=u2/g, then

H+xtan(θ)−12Lx2(1+tan2(θ))=0

Differentiate both sides with respect to θ.

dxdθtan(θ)+xsec2(θ)−[1Lxdxdθ(1+tan2(θ))+12Lx2(2tan(θ)sec2(θ))]=0

Solving for dxdθ yields

dxdθ=xsec2(θ)[xLtan(θ)−1]tan(θ)−xL(1+tan2(θ))

This derivative is 0 when tan(θ)=Lx and hence this corresponds to a critical number θ for the range of the projectile. We should now show that the x value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace tan(θ) with Lx in the second equation from the top and solve for x.

H+L−12Lx2−L2=0.

This leads immediately to x=L2+2LH−−−−−−−−√. The angle θ can now be found easily.

7 0

2 years ago

Difference between basic units and derived unit

Molodets [167]

Answer:

Basic units=those units of measurement which do not depend upon any other units are called basic units. Example length, mass, time, current etc. 

Derived units=those units which are expressed in terms of fundamental units are called derived units. Example area, volume, density etc. 

4 0

3 years ago

A hula hoop is rolling along the ground with a translational speed of 26 ft/s. It rolls up a hill that is 16 ft high. Determine

nikklg [1K]

Answer:12.8 ft/s

Explanation:

Given

Speed of hoop $v=26\ ft/s$

height of top $h=16\ ft$

Initial energy at bottom is

$E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2$

Where m=mass of hoop

I=moment of inertia of hoop

$\omega$ =angular velocity

for pure rolling $v=\omega R$

$I=mR^2$

$E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2$

$E_b=mv^2=m(26)^2=676m$

Energy required to reach at top

$E_T=mgh=m\times 32.2\times 16$

$E_T=512.2m$

Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

$\Delta E=676m-512.2m=163.8m$

Therefore

$163.8 m=mv^2$

$v=\sqrt{163.8}$

$v=12.798\approx 12.8\ ft/s$

7 0

2 years ago

Name:

Brums [2.3K]

Answer:

1.a) 1 kJ

1.b) 4 kJ

ratio 1:4

1.c) 4 times as before

2.a) 3.33 m/s2

Explanation:

1.a) bicycle's velocity =Displacement/time

=100/20 m/s

=5 m/s

bicycler's KE =1/2 *mass*(velocity)^2

=1/2*80*5^2

=1000 J = 1 kJ

1.b) bicycle's new velocity =200/20 m/s

=10 m/s

bicycler's new KE =1/2*80*10^2

=4000 J = 4 kJ

Ratio= KE 1 :KE new

= 1 :4

1.c) when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it

ex : speed is triple the prior value , then the KE is as 3^2 times as before. that is 9 times

2.a) car acceleration = (20-0)/6 m/s2

= 3.33 m/s2

4 0

2 years ago

What is the height of a building that an object is dropped from if it has a mass of 3 kg and hits the ground with a velocity of

Marizza181 [45]

Answer:

125 m

Explanation:

m = 3kg

v = 50m/s

u = 0m/s

a = +g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

50² = 0² + 2(10)(H)

2500 = 20H

H = 2500/20

H = 125m

8 0

2 years ago