Answer:
the distance traveled by the sports car is 90 m
Explanation:
Given;
mass of the sports car, m = 500 kg
initial velocity of the sports car, u = 0
final velocity of the sport car, v = 30 m/s
time of motion of the car, t = 6 s
The distance traveled by the sports car is calculated as;
Therefore, the distance traveled by the sports car is 90 m
Sometimes they are the same in physics. it depends how the question is worded but speed comes before velocity. If you take the derivative of speed you will end up with its velocity. The same goes when you take the integral of velocity.
There's no reason why the center of gravity must be in a place where
there is a any mass.
The center of gravity is simply a LOCATION ... the place where the
amount of mass in any direction from it is the same amount.
For that matter, whenever you know the location of the center of gravity
for ANY object, you can always go in there and scoop our a tiny spherical
hole at that place. Then the center of gravity won't move, but it will be in an
empty space, 'outside the body' of the object.
Here are a few more points to ponder:
-- The center of gravity of a basketball, beach ball, tennis ball, or any other
inflated ball is the center of the ball, where there is no part of the skin.
-- The center of gravity of a party balloon is somewhere inside the balloon,
where there is no rubber. If the balloon is spherical, then its center of gravity
is the center of the sphere.
-- The center of gravity of a square is the center of the square, not on any
of its sides.
-- The center of gravity of a triangle is the centroid of the triangle, not on any
of its sides.
Explanation:
a) Given in the y direction (taking down to be positive):
Δy = 50 m
v₀ = 0 m/s
a = 10 m/s²
Find: t
Δy = v₀ t + ½ at²
50 m = (0 m/s) t + ½ (10 m/s²) t²
t = 3.2 s
b) Given in the x direction:
v₀ = 12 m/s
a = 0 m/s²
t = 3.2 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²
Δx = 38 m
Answer:
The width of the slit is 0.167 mm
Explanation:
Wavelength of light, 
Distance from screen to slit, D = 88.5 cm = 0.885 m
The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m
We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :
where
a = width of the slit
a = 0.000167 m
a = 0.167 mm
So, the width of the slit is 0.167 mm. Hence, this is the required solution.
