Answer:
a = 5.05 x 10¹⁴ m/s²
Explanation:
Consider the motion along the horizontal direction
= velocity along the horizontal direction = 3.0 x 10⁶ m/s
t = time of travel
X = horizontal distance traveled = 11 cm = 0.11 m
Time of travel can be given as
inserting the values
t = 0.11/(3.0 x 10⁶)
t = 3.67 x 10⁻⁸ sec
Consider the motion along the vertical direction
Y = vertical distance traveled = 34 cm = 0.34 m
a = acceleration = ?
t = time of travel = 3.67 x 10⁻⁸ sec
= initial velocity along the vertical direction = 0 m/s
Using the kinematics equation
Y = t + (0.5) a t²
0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²
a = 5.05 x 10¹⁴ m/s²
.Answer:
The value of the work done is 
.
Explanation:
When a charged particle having charge 
is moving through an electric field 
, the net force (
) on the charge is
and the work done (
) by the particle is
Given, 
.
Substitute the value of electric field in equation (1) and then substitute the result in equation (2).
![W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}](https://tex.z-dn.net/?f=W%20%26%3D%26%20%5Cint%5Climits%5E7_0%20%7Bq%5Cdfrac%7BA_%7B0%7D%7D%7Bx%5E%7B1%2F2%7D%7D%7D%20%5C%2C%20dx%20%5C%5C%26%3D%26%20qA_%7B0%7D%20%5Cint%5Climits%5E7_0%20%7Bx%5E%7B-1%2F2%7D%7D%20%5C%2C%20dx%20%5C%5C%26%3D%26%202qA_%7B0%7D%5Bx%5E%7B1%2F2%7D%5D_%7B0%7D%5E%7B7%7D%5C%5C%26%3D%26%205.29%20qA_%7B0%7D)
Answer:
Firstly they are, by design, easy to use in most scientific and engineering calculations; you only ever have to consider multiples of 10. If I’m given a measurement of 3.4 kilometres, I can instantly see that it’s 3′400 metres, or 0.0034 Megametres, or 3′400′000 millimetres. It’s not even necessary to use arithmetic, I just have to remember the definitions of the prefixes (“kilo” is a thousand, “megametre” is a million, “milli” is a thousandth) and shift the decimal point across to the left or the right. This is especially useful when we’re considering areas, speeds, energies, or other things that have multiple units; for instance,
1 metre^2 = (1000millimetre)^2 = 1000000 mm^2.
If we were to do an equivalent conversion in Imperial, we would have
1 mile^2 = (1760 yards)^2
and we immediately have to figure out what the square of 1760 is! However, the fact that SI is based on multiples of 10 has the downside that we can’t consider division by 3, 4, 8, or 12 very easily.
Secondly they are (mostly) defined in terms of things that are (or, that we believe to be) fundamental constants. The second is defined by a certain kind of radiation that comes from a caesium atom. The metre is defined in terms of the second and the speed of light. The kelvin is defined in terms of the triple point of water. The mole is the number of atoms in 12 grams of carbon-12. The candela is defined in terms of the light intensity you get from a very specific light source. The ampere is defined using the Lorentz force between two wires. The only exception is the kilogram, which is still defined by the mass of a very specific lump of metal in a vault in France (we’re still working on a good definition for that one).
Thirdly, most of the Imperial and US customary units are defined in terms of SI. Even if you’re not personally using SI, you are probably using equipment that was designed using SI.
Answer:
1.8 m/s²
36 N
34.8 N
Explanation:
For the monkey :
m₁ = mass of monkey = 4.50 kg
T₁ = Tension force in the rope on monkey's side
a = acceleration
From the force diagram, force equation for the motion of monkey is given as
m₁ g - T₁ = m₁ a
(4.50 x 9.8) - T₁ = 4.5 a
T₁ = 44.1 - 4.5 a eq-1
For the bunch of bananas :
m₂ = mass of bunch of bananas = 3 kg
T₂ = tension force in the rope on the side of banana
From the force diagram, force equation for the motion of bananas is given as
T₂ - m₂ g = m₂ a
T₂ - (3 x 9.8) = 3 a
T₂ = 29.4 + 3 a eq-2
m = mass of the pulley = 1.50 kg
r = radius of the pulley = 0.090 m
α = angular acceleration of pulley = a/r
Torque equation for the pulley is given as
(T₁ - T₂ )r = I α
(T₁ - T₂ )r = I (a/r)
T₁ - T₂ = (0.5 m r²) (a/r²)
T₁ - T₂ = (0.5) ma
using eq-1 and eq-2
44.1 - 4.5 a - (29.4 + 3 a) = (0.5) ma
44.1 - 4.5 a - (29.4 + 3 a) = (0.5) (1.50) a
a = 1.8 m/s²
Using eq-1
T₁ = 44.1 - 4.5 a
T₁ = 44.1 - 4.5 (1.8)
T₁ = 36 N
using eq-2
T₂ = 29.4 + 3 a
T₂ = 29.4 + 3 (1.8)
T₂ = 34.8 N
20 joule is your answer
Answer:
here
mass m =100kg
distance d=50m
acceleration due to gravity a =10m/s²
work =force×displacement
= ma/d=100×10/50=20joule
