What type of telescope is shown in Figure 24-2

Find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the

Natali [406]

Answer:

The potential energy of the hiker is $1.13\times 10^6\ J$ .

Explanation:

Given that,

Mass of the hiker, m = 61 kg

Height above sea level, h = 1900 m

We need to find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington. The potential energy is given by :

$E=mgh$

g is the acceleration due to gravity

$E=61\ kg\times 9.8\ m/s^2\times 1900\ m\\\\E=1.13\times 10^6\ J$

So, the potential energy of the hiker is $1.13\times 10^6\ J$ . Hence, this is the required solution.

8 0

3 years ago

What word does this define..<br><br>The distance traveled per unit of time

nexus9112 [7]

Answer:..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Volosity???????

Explanation: THINK THATS IT PROBABLY MISPELLED IT SORRY IN ADVANCE IF WRONG

8 0

3 years ago

Read 2 more answers

Under which circumstances would protesters’ right to assembly most likely be protected?

ICE Princess25 [194]

Answer:

Explanation:

Mob violence is never protected. Organizers are always careful about not letting things get out of hand. Not D

That remark sort of applies to C as well. Property destruction is never protected.

B is something assemblies are allowed to do. So B.

A falls into the same category as C and D. You may not like the fact that he/she is doing something illegal, but you can't take away their privacy.

7 0

2 years ago

two punds of water vapor at 30 psia fill the 4ft3 left chmaber of a partitioned system. The right chmaber has twice the volume o

tamaranim1 [39]

Answer:

pressure of water will be 49.7 atm

Explanation:

given data

pressure = 30 psi = 2.04 atm

water = 2 pound = 907.18

mole of water vapor = 907.19 /2 = 50.4 mole

volume = 4 ft³ = 113.2 L

temperature = 40 F = 277.59 K

to find out

pressure of water

solution

we will apply here ideal gas condition

that is

PV = nRT .......................1

put here all value and here R = 0.0821 , T temperature and V volume and P pressure and n is no of mole

and we get here temperature

PV = nRT

2.04 × 113.2 = 50.4×0.0821×T

solve it and we get

T = 55.8 K

so we have given right chamber has twice the volume of the left chamber i.e

volume = twice of volume + volume

volume = 2(113.2) + 113.2

volume = 339.6 L

so from equation 1 pressure will be

PV = nRT

P(339.6) = 50.4 × ( 0.0821) × (277.59)

P = 3.3822 atm = 49.7 atm

so pressure of water will be 49.7 atm

7 0

3 years ago

Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a st

zubka84 [21]

Answer:

R=3818Km

Explanation:

Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.

Since you travel 2π radians in 24H, the angle can be calculated as:

$\alpha =\frac{2*\pi *t}{24H}$ t being expressed in hours.

$\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad$

From the triangle formed by A,B and the center of the planet, we know that:

$cos(\alpha )=\frac{r}{r+H}$ Solving for r, we get:

$r=\frac{H*cos(\alpha) }{1-cos(\alpha) } =3818Km$

6 0

3 years ago