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Elodia [21]
3 years ago
6

What’s the difference between distance and displacment

Physics
1 answer:
Brut [27]3 years ago
3 0
Distance is how far you are and displacment is separate like for someone oh From somewere
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Before the invention of a wheel turning on an axle, ancient people moved heavy loads by placing rollers under them. (Modern peop
matrenka [14]

Answer:K.E.=36.277+1.0429=37.319 J

Explanation:

Velocity of Top most point of wheel is twice the Velocity of centre of mass of wheel

i.e V_{wheel}=2V_{cm}=0.319 m/s

Thus angular velocity is given by

\omega =\frac{V_{cm}}{R}

\omega =0.465 rad/s

Kinetic\ Energy=Rotational + transtational

K.E.=\frac{1}{2}m_{stone}V^2+2\left [ \frac{1}{2}m_{cyl}\left [ \frac{V}{2}\right ]^2\right ]+2\left [ \frac{1}{2}I\omega ^2\right ]

K.E.=\frac{V^2}{2}\left [ m_{stone}+\frac{m_{roller}}{2}\right ]+I\omega ^2

K.E.=\frac{0.319^2}{2}\left [ 672+\frac{82}{2}\right ]+\frac{82\times 0.343^2\times 0.465^2}{2}

K.E.=36.277+1.0429=37.319 J

3 0
4 years ago
An object is placed on the axis of a converging mirror of focal length 200mm. The image produced is inverterd and has a magnific
sp2606 [1]

Answer:1.7 is the answer

Explanation:im smart

6 0
3 years ago
The rate of change of an object's velocity is called _______.
tamaranim1 [39]

Answer:

The rate of change of an object's velocity is called acceleration.

4 0
3 years ago
Read 2 more answers
Which of the following has greater kinetic energy?
vladimir1956 [14]

Answer:

A

Explanation:

6 0
3 years ago
A 4.00-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 27.0 N is require
NikAS [45]

Answer:

a)

135Nm⁻¹

b)

0.925 Hz

c)

1.2ms⁻¹

d)

0 m

e)

6.7ms⁻²

f)

\pm 0.2 m

Explanation:

a)

F = force required to hold the object at rest connected with stretched spring = 27 N

x = stretch in the spring from equilibrium position = 0.2 m

k = force constant of the spring

force required to hold the object at rest is same as the spring force , hence

F = k x

k = \frac{F}{x}

inserting the values

k = \frac{27}{0.2} = 135 Nm⁻¹

b)

frequency of the oscillations is given as

f =\frac{1}{2\pi }\sqrt{\frac{k}{m}}

inserting the values

f =\frac{1}{2(3.14) }\sqrt{\frac{135}{4}}\\f = 0.925 Hz

c)

A = Amplitude of oscillations = 0.2 m

w = angular frequency

Angular frequency is given as

w = 2\pi f = 2 (3.14) (0.925) = 5.8 rads⁻¹

Maximum speed of oscillation is given as

v_{max} = Aw

v_{max} = (0.2)(5.8)\\v_{max} = 1.2 ms⁻¹

d)

maximum speed of the object occurs at the equilibrium position, hence

x = 0 m

e)

Maximum acceleration of oscillation is given as

a_{max} = Aw^{2}

a_{max} = (0.2)(5.8)^{2}\\a_{max} = 6.7ms⁻²

f)

maximum acceleration occurs when the object is at extreme positions, hence

x = \pm 0.2 m

4 0
3 years ago
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