Answer:
v1 = 15.90 m/s
v2 = 8.46 m/s
mechanical energy before collision = 32.4 J
mechanical energy after collision = 32.433 J
Explanation:
given data
mass m = 0.2 kg
speed = 18 m/s
angle = 28°
to find out
final velocity and mechanical energy both before and after the collision
solution
we know that conservation of momentum remain same so in x direction
mv = mv1 cosθ + mv2cosθ
put here value
0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)
3.6 = 0.1765 V1 + 0.09389 v2 ................1
and
in y axis
mv = mv1 sinθ - mv2sinθ
0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)
0 = 0.09389 v1 - 0.1768 v2 .......................2
from equation 1 and 2
v1 = 15.90 m/s
v2 = 8.46 m/s
so
mechanical energy before collision = 1/2 mv1² + 1/2 mv2²
mechanical energy before collision = 1/2 (0.2)(18)² + 0
mechanical energy before collision = 32.4 J
and
mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²
mechanical energy after collision = 32.433 J
Answer:
total surface area is 432
Explanation:
Given data
base = 6
diagonals = 8
altitude = 12
to find out
total surface area
solution
we know total surface area of prism is
total surface area = lateral surface area + 2base area ..............1
so
first we calculate base perimeter i.e = 2 length + 2 width
so perimeter = 2(8) + 2(6) = 25
and area = length * width = 8*6 = 48
so lateral surface area is perimeter * height i.e
lateral surface area = 28* 12
lateral surface area = 336
put this value in equation 1 we get
total surface area = lateral surface area + 2base area
total surface area = 336 + 2(48)
total surface area is 432
The gravitational field is the Force divided by the mass
Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.
g = F / m
g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg
Note that the unit N/kg is equivalent to m/s^2
Explanation:
PRIMERO ENCUENTRAS EL PESO DEL PARACAIDISTA
= 65 kg(
) = 637 N
CON LA FÓRMULA DE LA ENERGÍA POTENCIAL
U = 637 N(6000 m - 3500 m) = 1592500 J
Answer:
0.911 atm
Explanation:
In this problem, there is no change in volume of the gas, since the container is sealed.
Therefore, we can apply Gay-Lussac's law, which states that:
"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"
Mathematically:
where
p is the gas pressure
T is the absolute temperature
For a gas undergoing a transformation, the law can be rewritten as:
where in this problem:
is the initial pressure of the gas
is the initial absolute temperature of the gas
is the final temperature of the gas
Solving for p2, we find the final pressure of the gas:
