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shutvik [7]
1 year ago
11

A small block of mass 200 g starts at rest at A, slides to B where its speed is vB=8.0m/s,vB=8.0m/s, then slides along the horiz

ontal surface a distance 10 m before coming to rest at C. (See below.) (a) What is the work of friction along the curved surface? (b) What is the coefficient of kinetic friction along the horizontal surface?
Physics
1 answer:
Nana76 [90]1 year ago
7 0

Answer

given,

mass of the block = 200 g = 0.2 Kg

Velocity at A = 0 m/s

Velocity at B = 8 m/s

slide to the horizontal distance = 10 m

height of the block be = 4 m

potential energy of the block

    P = m g h

    P = 0.2 x 9.8 x 4

    P =7.84 J

kinetic energy

    KE = \dfrac{1}{2}mv^2

    KE = \dfrac{1}{2}\times 0.2 \times 7.84^2

    KE =6.14 J

Work = P - KE

work = 7.84 - 6.14

work = 1.7 J

b) v² = u² + 2 a s

   0 = 8² - 2 x a x 10

   a = 3.2 m/s²

ma - μ mg = 0

 \mu = \dfrac{a}{g}

 \mu = \dfrac{3.2}{9.8}

 \mu = 0.327

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shusha [124]

Answer: at when distance r = infinity.

Explanation: The formulae for the electric potential of an electric charge to an arbitrary point is given by the formulae below

V = q/4πεr

V = electric potential (volts)

q = magnitude of electric charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)

Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).

Same goes for electric potential energy between two charges, the formulae is given below as

W = q1 *q2/4πεr

W= electric potential energy

q1 = magnitude of first charge.

q2 = magnitude of second charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.

Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)

6 0
1 year ago
Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. Wha
devlian [24]

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

K_{max}=\frac{hc}{\lambda}-W

Here h is the Planck's constant, c is the speed of light, \lambda is the wavelength of the light and W the work function of the element:

W=\frac{hc}{\lambda}-K_{max}\\W=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{495*10^{-9}m}-0.5eV\\W=2.01eV

Now, we calculate the wavelength for the new maximum kinetic energy:

W+K_{max}=\frac{hc}{\lambda}\\\lambda=\frac{hc}{W+K_{max}}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{2.01eV+1.5eV}\\\lambda=3.54*10^{-7}m=354*10^{-9}m=354nm

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

8 0
2 years ago
An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,
Ne4ueva [31]
A)  f = 1.8 rev/s = 2 Hz 
<span>T = 1 / f = 0.55s

B)  not really sure..srry

C)  </span><span>T = 2 pi √ ( L / g ) </span>
<span>0.57 = 2 x 3.14 x √ ( 0.2 / g )
</span><span>
g = 25.5 m/s²
</span>
Hope this helps a little at least.. :)

5 0
1 year ago
What is the absolute value of the horizontal force that each athlete exerts against the ground?
alexandr402 [8]
Refer to the diagram shown below.

When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.

At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.

Answer:  
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.

6 0
1 year ago
Which of these is true of a chemical reaction? (a) energy is always transferred from reactants to products. (b) Energy is transf
lorasvet [3.4K]
The answer is B: energy is transferred, but it can go to the products or the reactants.
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