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Aneli [31]
3 years ago
15

As altitude increases in the troposphere and stratosphere, the air temperature does what?

Physics
1 answer:
natima [27]3 years ago
3 0

Answer:

1

Explanation:

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2) A ray of light in air is approaching the boundary with water at an
KiRa [710]

Answer:

Explanation:

ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface

          n₁sinθ₁ = n₂sinθ₂

          1 sin52 = 1.33sinθ₂

                  θ₂ = arcsin(sin52 / 1.33)

                  θ₂ = 36°

as measured from the perpendicular to the surface

5 0
2 years ago
A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.
Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same  provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\

assume the bullet goes to right side and the gravitational acceleration =10 ms^{-2}

so now the weight of the rifle=\frac{25}{10}  

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}

this is a negative velocity to the right side. that means the rifle recoils to the left side

3 0
2 years ago
A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw
navik [9.2K]

Answer:  75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7  x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 =  200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

7 0
2 years ago
Read 2 more answers
A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acce
erastova [34]

force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N

8 0
2 years ago
Read 2 more answers
A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when
Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

v^2-u^2=2ad

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

a=-8.0 m/s^2 is the acceleration

Solving for u, we find the initial velocity:

u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s

6 0
3 years ago
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