Answer: v = 4.4 m/s
Explanation:
In the absence of friction, the total mechanical energy will be constant
KE₀ + PE₀ = KE₁ + PE₁
0 + mg(6) = ½mv₁² + mg(5)
½mv₁² = mg(6 - 5)
v = √(2g(1)) = 4.4 m/s
The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
Explanation:
Consider the vertical motion of ball,
We have equation of motion v = u + at
Initial velocity, u = u sin θ
Final velocity, v = 0 m/s
Acceleration = -g
Substituting
v = u + at
0 = u sin θ - g t

This is the time of flight.
Consider the horizontal motion of ball,
Initial velocity, u = u cos θ
Acceleration, a =0 m/s²
Time,
Substituting
s = ut + 0.5 at²

This is the range.
In this problem
u = 30 m/s
g = 9.81 m/s²
θ = 45° - For maximum range
Substituting

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.
So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
When you ask for "joules per second", you're asking for "watts".
The rate of energy "transfer" is 'power'. In this case, the light bulb
transfers energy out of the electrical circuit and into the space
around it, in the form of light and heat radiation.
Electrical power = (voltage) x (current) =
(6 volts) x (0.5 ampere) =
3 watts = 3 joules per second.
Answer:
Pascal's law (also Pascal's principle[1][2][3] or the principle of transmission of fluid-pressure) is a principle in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.[4] The law was established by French mathematician Blaise Pascal in 1653 and published in 1663.[5][6]
Answer:
The correct answer is B)
Explanation:
When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.
The formula for calculating the velocity of a point on the edge of the wheel is given as
= 2π r / T
Where
π is Pi which mathematically is approximately 3.14159
T is period of time
Vr is Velocity of the point on the edge of the wheel
The answer is left in Meters/Seconds so we will work with our information as is given in the question.
Vr = (2 x 3.14159 x 1.94m)/2.26
Vr = 12.1893692/2.26
Vr = 5.39352619469
Which is approximately 5.39
Cheers!