Upthrust is the upward force exerted by fluids on the surface of an object immersed in fluids. Thrust:- It is the force acting perpendicular to the surface. Upthrust:- It is the upward force exerted by the fluid on the surface of an object immersed in liquid.
I hope it's help you
The question is incomplete! The complete question along with answer and explanation is provided below.
Question:
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
Given Information:
Mass = m = 0.5 kg
Horizontal distance = d = 40 cm = 0.4 m
Vertical distance = h = 7 cm = 0.07 m
Normal force = Fn = 1 N
Required Information:
Potential energy = PE = ?
Work done = W = ?
Answer:
Potential energy = 0.343 Joules
Work done = 0.39 N.m
Explanation:
The potential energy is given by
PE = mgh
where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.
PE = 0.5*9.8*0.07
PE = 0.343 Joules
As you can see in the attached image
sinθ = opposite/hypotenuse
sinθ = 0.07/0.4
θ = sin⁻¹(0.07/0.4)
θ = 10.078°
The horizontal component of the normal force is given by
Fx = Fncos(θ)
Fx = 1*cos(10.078)
Fx = 0.984 N
Work done is given by
W = Fxd
where d is the horizontal distance
W = 0.984*0.4
W = 0.39 N.m
Answer:
Explanation:
Given
Cannon is fired with a velocity of
Using Equation of motion
where
after time
So after 3.3 s cannon ball is at a height of 185.89 m
Answer:
2.6×10⁻³ N
Explanation:
From coulomb's law,
F = kq'q/r²................ Equation 1
Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.
From the question,
since q' = q
Then,
F = kq²/r²..................... Equation 2
Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,
Constant: k = 9×10⁹ Nm²/kg².
Substitute into equation 2
F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²
F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)
F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)
F = 2.6×10⁻³ N.
Answer:
When you blow into a tuba the air vibrates very slowly.
Explanation:
Tuba is a buzz instrument ie sound is produced in it with the help of lip vibration . It is the lowest pitched musical instrument in the brass family .
Due to absence of resonance in it , it produces music of lowest pitch , So when one blows into it the air column of the instrument vibrates very slowly producing low pitched sound.