Answer:
T₂ = 111.57 °C
Explanation:
Given that
Initial pressure P₁ = 9.8 atm
T₁ = 32°C = 273 + 32 =305 K
The final pressure P₂ = 11.2 atm
Lets take the final temperature = T₂
We know that ,the ideal gas equation
If the volume of the gas is constant ,then we can say that


Now by putting the values in the above equation ,we get


T₂ = 384.57 - 273 °C
T₂ = 111.57 °C
Answer:
a) Velocity = 4.2m/s
b) Acceleration = 2.94m/s^2
c) Force exerted on the floor= 1401.4×10^3N
Explanation:
a) Velocity,V=sqrt(2×9.8×0.900)
V= 4.2m/s
b) Vf2= V^2+2ay2
a= 4.2^2 - 0/2×3
a= 17.64/6= 2.94m/s^2
c) Newton's 2nd law indicates:
Fnet= F - mg=ma
F= m(g+a)
F=110(9.8+2.94)
F=110×12.94
F= 1401.4N
Answer:
d = 2021.6 km
Explanation:
We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them
Airplane 1
Height y₁ = 800m
Angle θ = 25°
cos 25 = x / r
sin 25 = z / r
x₁ = r cos 20
z₁ = r sin 25
x₁ = 18 103 cos 25 = 16,314 103 m
= 16314 m
z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m
2 plane
Height y₂ = 1100 m
Angle θ = 20°
x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m
z₂ = 20 103 without 25 = 8.452 103 m = 8452 m
The distance between the planes using the Pythagorean Theorem is
d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2
Let's calculate
d² = (18126-16314)² + (1100-800)² + (8452-7607)²
d² = 3,283 106 +9 104 + 7,140 105
d² = (328.3 + 9 + 71.40) 10⁴
d = √(408.7 10⁴)
d = 20,216 10² m
d = 2021.6 km
Answer:
66.98 db
Explanation:
We know that

L_T= Total signal level in db
n= number of sources
L_S= signal level from signal source.

= 66.98 db
Answer:
0.3677181864 m
Explanation:
u = Velocity = 1.5 m/s
= Angle = 20°
y = -20 cm
Velocity components


Acceleration components




Time taken is 0.26088 seconds

The distance the beetle travels on the ground is 0.3677181864 m