An airplane whose air speed is 527 mi/hr covers a distance of 907 miles in 2.25 hrs. How strong was the head wind against it?

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}$

Since

$L_{su}= m_{su}*v_{su}*r_1$

then

$v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}$

$v_{su} = 19.44 m/s$

7 0

4 years ago

Which of the following best explains why snow predictions by meteorologists are sometimes incorrect?

den301095 [7]

Answer:

C. Interference from the sun causes data to be collected inaccurately.

Explanation:

Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.

6 0

3 years ago

uppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 0.8 ou

Brrunno [24]

Answer:

x= 9.53 ounces

Explanation:

Given that

Mean ,μ= 9 ounces

Standard deviation ,σ=0.8 ounces

He wants to sell only those potatoes that are among the heaviest 25%.

P=25% = 0.25

When P= 0.25 then Z=0.674

Lest take x is the the minimum weight required to be brought to the farmer's market.

We know that

x = Z . σ + μ

x= 0.674 ₓ 0.8 + 9 ounces

x= 9.53 ounces

3 0

3 years ago

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

4 0

3 years ago

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp

IRINA_888 [86]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?

Answer:

601000 N

Explanation:

Given that :

Acceleration due to gravity at lunar outpost = 1.6m/s²

Supported Weight of supplies = 1 * 10^5 N

Acceleration due to gravity on the earth surface = 9.8m/s²

Maximum weight of supplies as measured on EARTH :

Ratio of earth gravity to lunar post gravity:

(Earth gravity / Lunar post gravity) ;

(9.8 / 1.63) = 6.01

Hence, maximum weight of supplies as measured on EARTH should be :

6.01 * (1 × 10^5)

6.01 × 10^5

= 601000 N

3 0

3 years ago