What happens when a body of cold air approaches a body of warm air
1 answer:
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Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ + t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis
Answer:
50s .
Explanation:
Here the power of the microwave is 0.2kW . And as we know that ,
- Rate of doing work is called power .
So from the definition , we have ;
- Here the work done is equal to the energy consumed by the microwave i.e. 10000 J .So we can write it as ,
Cross multiply ,
Simplify ,
Answer:
Acceleration=
Speed=0.67 m/s
Explanation:
We are given that
Horizontal force=F=20 N
Mass of box=m=40 kg
We know that
Acceleration=
Using the formula
Acceleration of box=
The acceleration of the box=
Initial velocity=u=0
Force=F=30 N
Distance=s=0.3 m
Substitute the values
Hence, the speed of the box after it has been pulled a distance of 0.3 m=0.67 m/s