Magnitude of the force of tension: 139 N
Explanation:
The surface of the ramp here is assumed to be the positive x-direction.
To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.
There are three forces acting along the x-direction:
- The force of tension,
, acting up along the plane - The force of friction,
, acting down along the plane - The component of the weight in the x-direction,
, acting down along the plane
We know that the magnitude of the weight is

So its x-component is

The net force along the x-direction can be written as

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

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Answer:
9 times
Explanation:
Kinetic energy is:
KE = ½ mv²
When we triple the velocity, the kinetic energy increases by a factor of 9.
9KE = ½ m(3v)²
Answer:
The answer to the question is
The roller coaster will reach point B with a speed of 14.72 m/s
Explanation:
Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h
Where m = mass
g = acceleration due to gravity = 9.81 m/s²
h = starting height of the roller coaster
we have the given variables
h₁ = 36 m,
h₂ = 13 m,
h₃ = 30 m
v₁ = 1.00 m/s
Total energy at point 1 = 0.5·m·v₁² + m·g·h₁
= 0.5 m×1² + m×9.81×36
=353.66·m
Total energy at point 2 = 0.5·m·v₂² + m·g·h₂
= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m
The total energy at 1 and 2 are not equal due to the frictional force which must be considered
Total energy at point 2 = Total energy at point 1 + work done against friction
Friction work = F×d×cosθ = (
× mg)×60×cos 180 = -117.72m
0.5·m·v₂² + 127.53·m = 353.66·m -117.72m
0.5·m·v₂² = 108.41×m
v₂² = 216.82
v₂ = 14.72 m/s
The roller coaster will reach point B with a speed of 14.72 m/s
given that snow is projected at an angle of 40 degree
It range is given as a = 19 ft

now we can use the formula of horizontal range





<u>so its initial speed must be 7.6 m/s</u>
Answer:
2.295 eV
Explanation:
maximum wavelength, λ = 542 nm = 542 x 10^-9 m
The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

where, h is the Plank's constant and c be the speed of light
h = 6.634 x 10^-34 Js
c = 3 x 10^8 m/s


Wo = 2.295 eV
Thus, the work function of this metal is 2.295 eV.