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2 years ago

(Serious Please) patulong

Physics

1 answer:

Mnenie [13.5K]2 years ago

7 0

Answer:

a. Potential energy is highest at Part A

The kinetic energy is highest at Part C and Part D

b. The potential energy is lowest at Part C and Part D

c. The roller coater has equal amount of potential and kinetic energy at Part B, Part D and part F

2) Yes, the mechanical energy is the same from point A to F according to the first law of thermodynamics

Explanation:

The total mechanical energy is constant where the roller coaster moves by only the initial velocity, and the the force of gravity

Total mechanical energy, M.E. = Kinetic energy, K.E. + Potential energy, P.E.

M.E. = K.E. + P.E. = Constant

Therefore, we have;

a. Potential energy is the energy stored in a body, due to its position or elevation, state or arrangement

The higher the elevation, the higher the potential energy, therefore, the highest amount of potential energy is gained when the roller coaster is at the highest point in the motion = Part A

From M.E. = K.E. + P.E. = Constant, the highest kinetic energy is given at the point the roller coaster has the lowest potential energy, which corresponds with the lowest points = Part C and Part D

b. Potential energy, which is the energy of body due to its position or state is lowest at the lowest points = Part C and Part D

c. The value of potential energy, P.E. due to elevation, can be found as follows

P.E. = Mas, m × Gravity, g × Height, h

Therefore, the potential energy will be half the maximum value where the height, h = (Maximum height)/2 and given that M.E. = K.E. + P.E., the kinetic energy, will increase by the same amount, and we have;

K.E. = P.E. at the half the maximum height of the track = Part B, Part D and part F

2) The mechanical energy is the input energy, which according to the first law of thermodynamics cannot be created and destroyed an it is therefore, constant and it is the same from point A to F

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Answer:

kinematic viscosity in SUS is = 671.64 SUS

Explanation:

given data

kinetic viscosity = 145 mm^2/s

we know

1 mm = 0.1 cm

so kinetic viscosity in cm is $\nu =145 (0.1)^{2} =1.45 cm^{2}/s$

other unit of kinetic viscosity is centistokes

$1 cm^{2}/s = 100 cst$

so 1.45 cm^2/s will be 145 cst

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3 years ago

An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground

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Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m

AC = distance of person on the ladder from the bottom end = x

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$F_{g}$ = weight of the person = 710 N

F = force by the wall on the ladder

N = normal force by ground on the ladder = ?

Using equilibrium of force along the vertical direction

N = $F_{g}$ + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f =static frictional force on the ladder

Static frictional force is given as

f = μ N

f = (0.55) (950)

f = 522.5 N

Force equation along the horizontal direction is given as

F = f

F = 522.5 N

using equilibrium of torque about point A

F Sin50 (AD) = W Cos50 (AB) + ( $F_{g}$ Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

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Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

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Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

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$v^{2}=8.86\times10^{11}$

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(Serious Please) patulong​

(Serious Please) patulong