Answer:
Supercells
Explanation:
supercells are rotating thunderstorms that has a well-defined radar circulation called a mesocyclone. They can sometimes produce destructive hail, severe winds, frequent lightning, and flash floods.
Answer:
4th answer
Explanation:
The gradient of a distance-time graph gives the speed.
gradient = distance / time = speed
Here, the gradient is a constant till 30s. So it has travelled at a constant speed. It means it had not accelarated till 30s. and has stopped moving at 30s.
<h2>
Answer:</h2>
D. (1m, 0.5m)
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Explanation:</h2>
The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;
x = (m₁x₁ + m₂x₂ + m₃x₃) / M ----------------(i)
y = (m₁y₁ + m₂y₂ + m₃y₃) / M ----------------(ii)
Where;
M = sum of the masses
m₁ and x₁ = mass and position of first mass in the x direction.
m₂ and x₂ = mass and position of second mass in the x direction.
m₃ and x₃ = mass and position of third mass in the x direction.
y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.
From the question;
m₁ = 6kg
m₂ = 4kg
m₃ = 2kg
x₁ = 0m
x₂ = 3m
x₃ = 0m
y₁ = 0m
y₂ = 0m
y₃ = 3m
M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg
Substitute these values into equations (i) and (ii) as follows;
x = ((6x0) + (4x3) + (2x0)) / 12
x = 12 / 12
x = 1 m
y = (6x0) + (4x0) + (2x3)) / 12
y = 6 / 12
y = 0.5m
Therefore, the center of mass of the system is at (1m, 0.5m)
Answer:
t = 2.2 s
Explanation:
Given that,
A person observes a firework display for A safe distance of 0.750 km.
d = 750 m
The speed of sound in air, v = 340 m/s
We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.
So, the required time is 2.2 seconds.
Answer:
215955.06 m/s^2
Explanation:
length of barrel, s = 0.89 m
initial velocity of the bullet, u = 0 m/s
Final velocity of the bullet, v = 620 m/s
Let a be the acceleration of the bullet in the barrel
Use third equation of motion, we get
a = 215955.06 m/s^2
Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.
