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Lorico [155]
3 years ago
10

Please please please please please please please please please help me for these 2 questions​

Physics
2 answers:
Genrish500 [490]3 years ago
5 0
The answer to your question is 75j
weqwewe [10]3 years ago
4 0

Answer:

75j

Explanation:

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Adrian 80.0kg jumps out of an airplane at an altitude of 1.00×10^3m. After his parechute deploys, the adrian lands with a veloci
Vera_Pavlovna [14]

-- Before Adrian left the airplane, his gravitational potential energy was

(mass) x (gravity) x (height) = (80kg) x (9.81m/s²) x (1,000m) = 784,800 joules


-- When he reached the ground, his kinetic energy was

(1/2) x (mass) x (speed)² = (40kg) x (5m/s)² = 1,000 joules


-- Between the airplane and the ground, the Adrian lost

(784,800 joules) - (1,000 joules) = 783,800 joules 

Where did all that energy go ?
Energy never just disappears.  If it's missing, it had to go somewhere.

The Adrian used  783,800 joules of energy to push air our of his way
so that he could continue his parachute jump, and reach the ground
in time to be home for dinner. 
4 0
3 years ago
How much negative charge has been removed from a positively charged electroscope if it has a charge of 7.5×10^-11
vova2212 [387]

I think 4.69 x 108 charges have been removed

5 0
3 years ago
Hi,please i need help with this one in physics.hurry and correct i need it.<br>Topic:Equilibrium.​
Rudik [331]

The half-meter rule (easy math) is 0.5 meters or 50 centimeters since a meter is 1 meters long, which is equivalent to 100 centimeters. Therefore, we shall apply the 50 cm rule.

A 50 cm rule's center of mass is now 25 cm away.

Additionally, according to the data, the object is pivoted at 15 cm, while the 40 g object is hung at 2 cm from the rule's beginning. Using a straightforward formula, we can compare the two situations: the distance from the pivot to the center of the mass times the mass of the 40 g object divided by 2 cm must equal the distance from the pivot to the center of the mass times mass of the 10 x g object

The result of the straightforward computation must be 52g.

Most simplified version:

the center of mass of the rule is at the 25 cm mark

⇒ 40 g * (15 cm - 2 cm)

⇒ = M * (25 cm - 15 cm)

#SPJ2

5 0
2 years ago
Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat
QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

                               j_{H_2O}=\frac{D}{l} \bigtriangleup c

where

  • j_{H_2O} is the mass flux of the water which is to be calculated.
  • D is diffusion coefficient which is given as 0.25 cm^2/s
  • l is the thickness of the film which is 0.15 cm thick.
  • \bigtriangleup c is given as

                                \bigtriangleup c= \frac{P_{sat}-P_a}{RT}

In this

  • P_{sat} is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
  • P_a is the air pressure which is given as 0.5 times of P_{sat}
  • R is the universal gas constant as 8.314 kJ/kmol-K
  • T is the temperature in Kelvin scale which is 20+273= 293K

By substituting values in the equation

                                    \bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3

Converting \bigtriangleup c into cm^3/cm^3

As 1 mole of water 18 cm^3 so

                               \bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6}  cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6}  cm^3/cm^3

Putting this in the equation of mass flux equation gives

                            j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6}  cm/s

For calculation of water level drop in a day, converting mass flux as

                     j_{H_2O}=14.4 \times 10^{-6}  \times 24 \times 3600  cm/day\\ j_{H_2O}=1.24  cm/day

So the water level will drop by about 1.24 cm in 1 day.

7 0
4 years ago
What is the processes called when matter, that is a liquid, transforms into a solid?
AURORKA [14]
Oh yeah I just got cheated lol lol lol lol
4 0
3 years ago
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