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4 years ago

Light waves are not mechanical waves. The Sun transmits light waves to Earth through __ _. *

Physics

2 answers:

Natasha2012 [34]4 years ago

3 0

Electromagnetic waves?

Gemiola [76]4 years ago

3 0

Light waves are not mechanical waves. The Sun transmits light waves to Earth through __________ _________. *
*Wave Length*

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An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop the loop maneuver. The acceleration

Natali5045456 [20]

The radius of the loop is 18.9 km

Explanation:

When the airplane is at the top of the loop, the pilot experiences two forces:

The force of gravity, acting downward, of magnitude $mg$
The normal reaction exerted by the seat on the pilot, also acting downward, N

Since the plane is moving in a circular motion, the net force on the pilot must be equal to the centripetal force, therefore we can write:

$mg+N = m\frac{v^2}{r}$

where

m is the mass of the pilot

$g=9.8 m/s^2$ is the acceleration of gravity

N is the normal reaction

v = 430 m/s is the speed of the plane

r is the radius of the loop

Here we are told that the pilot feel weightless at the top of the loop: this means that the normal reaction is zero,

N = 0

Therefore the equation becomes

$mg=\frac{mv^2}{r}$

And so we can find the radius of the loop:

$r=\frac{v^2}{g}=\frac{430^2}{9.8}=18.9 \cdot 10^3 m = 18.9 km$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

The average human has a density of 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling. (a) Without making any swimming movem

Gala2k [10]

Answer:

20% of the human body float above the surface in the dead sea.

Explanation:

Given data,

The human density after inhaling, d₁ = 945 kg/m³

The human density after exhaling, d₂ = 1020 kg/m³

Therefore the average human density, d = 982.5 kg/m³

The density of the dead sea, D = 1230 kg/m³

The percentage of density of human body to the dead sea is,

P % = (982.5 / 1230) x 100 %

= 80 %

Therefore, the human body has 80% of dead sea density.

Hence, 20% of the human body float above the surface in the dead sea.

8 0

4 years ago

Tarzan is running with a horizontal velocity, along level ground. While running, he encounters a 2.21 m vine of negligible mass,

Lena [83]

Answer:

a) $a_c = 1.09m/s^2$

b) T = 720.85N

Explanation:

With a balance of energy from the lowest point to its maximum height:

$m*g*L(1-cos\theta)-1/2*m*V_o^2=0$

Solving for $V_o^2$ :

$V_o^2=2*g*L*(1-cos\theta)$

$V_o^2=2.408$

Centripetal acceleration is:

$a_c = V_o^2/L$

$a_c = 2.408/2.21$

$a_c = 1.09m/s^2$

To calculate the tension of the rope, we make a sum of forces:

$T - m*g = m*a_c$

Solving for T:

$T =m*(g+a_c)$

T = 720.85N

3 0

4 years ago

Someone please help me will give BRAILIEST!!!!!

Readme [11.4K]

P= F/A so F= p*A. You're welcome

5 0

3 years ago

A four-wheel-drive vehicle is transporting an injured hiker to the hospital from a point that is 30 km from the nearest point on

zepelin [54]

Answer:

$D=200\ km$

Explanation:

distance on terrain, $d_t=30\ km$

distance on the road, $d_r=70\ km$

speed on terrain, $v_t=30\ km.hr^{-1}$

speed on road, $v_r=130\ km.hr^{-1}$

time taken on the terrain,

$t_t=\frac{d_t}{v_t}$

$t_t=\frac{30}{30}$

$t_t=1\ hr$

time taken to cover the distance on the road:

$t_r=\frac{d_r}{v_r}$

$t_r=\frac{70}{130}$

$t_r=\frac{7}{13}\ hr$

Now the distance covered on terrain in the total time:

$D= v_r\times (t_r+t_t)$

$D= 130\times (\frac{7}{13}+1)$

$D=130\times \frac{20}{13}\$

$D=200\ km$

is the distance the vehicle must target on the road to minimize the time taken in going off the road.

3 0

3 years ago