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Andrews [41]
3 years ago
15

During takeoff, an airplane climbs with a speed of 130 m/s at an angle of 47 degrees above the horizontal. The speed and directi

on of the airplane constitute a vector quantity known as the velocity. The sun is shining directly overhead. How fast is the shadow of the plane moving along the ground? (That is, what is the magnitude of the horizontal component of the plane's velocity?)
Physics
1 answer:
Reptile [31]3 years ago
5 0
130cos(47deg) = 88.6598 m/s
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A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga
pashok25 [27]
A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
p_i V_i = nRT_i
where
p_i=1.0 atm=1.01 \cdot 10^5 Pa is the initial pressure of the gas
V_i is the initial volume of the gas
n=2.3 mol is the number of moles
R=8.31 J/K mol is the gas constant
T_i=240^{\circ}C=513 K is the initial temperature of the gas

By re-arranging this equation, we can find V_i:
V_i =  \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3

2) Now the gas cools down to a temperature of
T_f = 14^{\circ}C=287 K
while the pressure is kept constant: p_f = p_i = 1.01 \cdot 10^5 Pa, so we can use again the ideal gas law to find the new volume of the gas
V_f =  \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
W=p \Delta V= p(V_f -V_i)
by using the data we found at point 1) and 2), we find
W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J
where the negative sign means the work is done by the surrounding on the gas.
5 0
3 years ago
Can anyone help me please​
ArbitrLikvidat [17]

Answer:

Gravity.

Rocket ships.

Ball.

Basketball.

Explanation:

Gravity has to do a lot with air. It puts the planets in there area.

Rocket Ship has to do a lot with air. If i'm right, they calculate the area, weather, about the air.

A ball gets throwed in the air, which gravity comes into place.

Basketball is also a similar example to a ball.

7 0
3 years ago
Volcanic eruptions are natural phenomena in the . They sometimes erupt, spilling out lava and dust. The lava solidifies, forming
solniwko [45]

Answer:

B. Geosphere

A. Biosphere

A. Atmosphere

Explanation:

Volcanic eruptions occurs within the Geosphere. The geosphere is the rock solid earth make up of rocks that extends into the deep interior.

Magma formed deep within the crust rises to elevated parts and finally erupts as lava on the surface. When they cool, they solidify to form volcanic rocks.

The volcanic eruptions affects the biosphere significantly. The biosphere is the portion of the earth where all life forms exists.

Gases and ash spewed during an eruption into the atmosphere causes severe changes to weather and leads to pollution. The atmosphere is the gaseous envelope round the earth.

3 0
3 years ago
Formular for coefficient of friction<br>​
Bumek [7]

Answer:

hope you like it mark as brainliest

Explanation:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact.

The coefficient of friction (fr) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together. It is represented by the equation: fr = Fr/N.

7 0
3 years ago
In a charging process, 4 × 1013 electrons are removed from one small metal sphere and placed on a second identical sphere. Initi
Alina [70]

Answer:

The distance between the two spheres is 914.41 X 10³ m

Explanation:

Given;

4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;

1 e = 1.602 X 10⁻¹⁹ C

4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C

V = Ed

where;

V is the electrical potential energy between two spheres, J

E is the electric field potential between the two spheres N/C

d is the distance between two charged bodies, m

V = \frac{K*q}{d^2}*d = \frac{K*q}{d}

d = \frac{K*q}{V}

where;

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063

d = 914.41 X 10³ m

Therefore, the distance between the two spheres is 914.41 X 10³ m

3 0
4 years ago
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