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Kipish [7]
3 years ago
13

A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μkmgcosθ=mgs

inθ−ma, where g=9.80meter/second2, a=3.60meter/second2, θ=27.0∘, and m is not given. Which of the following represents a simplified expression for μk?
tanθ− ag
To avoid making mistakes, the expression should not be simplified until the numerical values are substituted.
gsinθ−agcosθ
The single equation has two unknowns and cannot be solved with the information given.
Physics
1 answer:
BaLLatris [955]3 years ago
4 0

Answer:

\mu_k=\dfrac{gsin\theta -a}{gcos\theta}

Explanation:

g = Acceleration due to gravity = 9.80 m/s²

a = Acceleration= 3.6 m/s²

\theta = Angle = 27°

The equation is

\mu_kmgcos\theta=mgsin\theta -ma

Mass gets cancelled

\\\Rightarrow \mu_kgcos\theta=gsin\theta -a

Rearranging for \mu_k

\\\Rightarrow \mu_k=\dfrac{gsin\theta -a}{gcos\theta}

The simplified expression is

\mathbf{\mu_k=\dfrac{gsin\theta -a}{gcos\theta}}

*the options are incomplete. The above answer is the required solution

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If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w
marishachu [46]

Answer:

2.13 x 10^-19 J or 0.53 eV

Explanation:

cut off wavelength, λo = 700 nm = 700 x 10^-9 m

λ = 400 nm = 400 x 10^-9 m

Use the energy equation

E = \frac{h c}{\lambda _{o}}+K

Where, K be the work function

\frac{h c}{\lambda} = \frac{h c}{\lambda _{o}}+K

K =hc\left ( \frac{1}{\lambda } -\frac{1}{\lambda _{0}}\right )

K =6.63\times 10^{-34}\times 3\times 10^{8}\left ( \frac{1}{4\times 10^{-7} } -\frac{1}{7\times 10^{-7}}\right )

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3 years ago
A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of
Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.  

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

K_1 = K_2 + P + W

then, the work done by the engine is:

W = K_1 - K_2 - P

The formulas for the Kinetic and potential energy are:  

K=\frac{1}{2}mV^2\\P=mgh

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh

Replacing the values:

W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

t=\frac{3200m}{20m/s}=160s

the power delivered by the car's engine was:

power=\frac{work}{time}=\frac{1690400J}{160s}=10565W

8 0
4 years ago
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