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Arte-miy333 [17]
3 years ago
14

A 1400 kg car drives up a 16 m high hill. The car’s speeds at the bottom and top are 21 m/s and 13 m/s, respectively. If the car

's engine does 80 kJ of work then what is the work done by friction?
Physics
1 answer:
Roman55 [17]3 years ago
4 0

Answer:

Explanation:

mass, m = 1400 kg

height, h = 16 m

initial velocity, u = 21 m/s

final velocity, v = 13 m/s

Work done by engine, We = 80 kJ

Let the work done by the friction force is Wf.

Use the work energy theorem

net work done = change in kinetic energy

work done by engine + work done by friction force + work done by the gravitational force = Change in kinetic energy

80000 + Wf - m x g x h = 0.5 m ( v² - u²)

80000 + Wf - 1400 x 9.8 x 16 = 0.5 x 1400 x ( 169 - 441 )

- 139520 + Wf = - 190400

Wf = 50880 J

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A on edge 2020

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A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
3 years ago
When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?
Oduvanchick [21]
First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
a = r \times ( \frac{1}{n1}  -  \frac{1}{n2} )
We insert the values

a = 2.18 \times  {10}^{ - 18}  \times ( \frac{1}{ {1}^{2} } -  \frac{1}{ {6}^{2} }  )
= 2.12 \times {10}^{ - 18}
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
wavelength =  \frac{h \times c}{a}
Finally we insert the values
\frac{6.626 \times  {10}^{ - 34}  \times 3 \times  {10}^{8} }{2.12 \times  {10}^{ - 18} }  = 9.376 \times  {10}^{ - 8}
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3 years ago
Two electric charges A and B were placed facing each other at a distance of separation "r". The common electrostatic force betwe
lutik1710 [3]

Answer:

From the formula of force:

F =  \frac{kAB}{ {r}^{2} }  \\

since AB and k are constants:

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x is a constant of proportionality

• when force is 4N, separation distance is 1

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therefore, equation becomes

F =  \frac{4}{ {r}^{2} }  \\

when r is doubled, r becomes 2. find F:

F =  \frac{4}{ {2}^{2} }  \\  \\ F =  \frac{4}{4}  \\  \\ { \underline{force \: is \: 1N}}

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