Answer:
HELLO! I WILL EXPLAIN WHY GOLF BALLS HAVE DIMPLES.
Explanation:
DIMPLES ON A GOLF BALL CREATE A THIN BOUNDARY LAYER OF AIR THAT CLINGS TO THE BALL'S SURFACE. THIS ALLOWS THE AIR TO FOLLOW THE BALL'S SURFACE A LITTLE FARTHER AROUND THE BACKSIDE OF THE BALL, THEREBY DECREASING THE SIZE OF THE WAKE.
I AM SO SORRY ABOUT THE CAP LOCKS I THINK MY CAP LOCKS BUTTON IS PUSHED, AND IT IS NOT POPPING UP FOR SOME REASON.
ANYWAYS, HAVE A GREAT DAY!!!!!!
Answer:
1977.696 J
Explanation:
Given;
Weight of the box = 28.0 kg
Force applied by the boy = 230 N
angle between the horizontal and the force = 35°
Therefore,
the horizontal component of the force = 230 × cosθ
= 230 × cos 35°
= 188.405 N
Coefficient of kinetic friction, μ = 0.24
Force by friction, f = μN
here,
N = Normal force = Mass × acceleration due to gravity
or
N = 28 × 9.81 = 274.68 N
therefore,
f = 0.24 × 274.68
or
f = 65.9232 N
Now,
work done by the boy, W₁ = 188.405 N × Displacement
= 188.405 N × 30
= 5652.15 J
and,
the
work done by the friction, W₂ = - 65.9232 N × Displacement
= - 65.9232 N × 30 m
= - 1977.696 J
[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]
Answer:0.253Joules
Explanation:
First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.
F = ke where;
F is the force
k is spring constant = 34N/m
e is the extension = 0.12m
F = 34× 0.12 = 4.08N
To get work done,
Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.
Work done = Force × Distance
Since F = 4.08m, distance = 0.062m
Work done = 4.08 × 0.062
Work done = 0.253Joules
Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules
Let's calculate the average acceleration. It is the rate of changing speeds. Hence, we need to calculate the difference of speeds. 10-6=4 m/s. The rate is now
.
In general, the formula for the mean acceleration between two times 1 and 2 is given by:
where v1 and v2 are the speeds at the respective points and T is the time interval between them.
Answer:
Correct answer: C. 50 cm
Explanation:
Given data:
The distance of the object from the top of the concave mirror o = 50.0 cm
The magnitude of the concave mirror focal length 25.0 cm.
Required : Image distance d = ?
If we know the focal length we can calculate the center of the curve of the mirror
r = 2 · f = 2 · 25 = 50 cm
If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.
We conclude that the image distance is 50 cm.
We will now prove this using the formula:
1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50
1/d = 1/50 => d = 50 cm
God is with you!!!
