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Ivan
2 years ago
8

A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What

is the acceleration of the car during this 4.0-second interval?
Physics
1 answer:
pychu [463]2 years ago
5 0
(6-16)/4.0=-2.5 m/s²
Acceleration of the car is -2.5 m/s²
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The actions of an employee are not attributable to the employer if the employer has not directly or indirectly encouraged the em
zepelin [54]

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4 0
2 years ago
A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency
allochka39001 [22]

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

\nu=\dfrac{v}{\lambda}

\nu=\dfrac{251\ m/s}{0.051\ m}

\nu=4921.56\ Hz

(2) Angular frequency of the wave is given by :

\omega=2\pi\nu

\omega=2\pi\times 4921.56\ Hz

\omega=30923.07\ rad/s

(3) The period of oscillation is given by T as :

T=\dfrac{1}{\nu}

T=\dfrac{1}{4921.56}

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

5 0
2 years ago
A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri
vodka [1.7K]
We know that tangential acceleration is related with radius and angular acceleration according the following equation:  
at = r * aa  
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)  
So the radius is r = d/2 = 1.2/2 = 0.6 m  
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5 0
1 year ago
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through
Svet_ta [14]

Answer:

The speed is \sqrt{2}v_{0}.

(a) is correct option.

Explanation:

Given that,

Potential difference V= V_{0}

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If it were accelerated instead

Potential difference V'=2V_{0}

We need to calculate the speed

Using formula of initial work done on proton

W = q V

We know that,

\Delta W=\Delta K.E

q V=\dfrac{1}{2}mv^2

Put the value into the formula

q V_{0}=\dfrac{1}{2}mv_{0}^2

v_{0}^2=\dfrac{2qV_{0}}{m}....(I)

If it were accelerated instead through a potential difference of 2 V_{0}, then it would gain a speed will be given as :

Using an above formula,

v_{0}'^2=\dfrac{2qV_{0}}{m}

Put the value of V_{0}

v_{0}'^2=\dfrac{2q\times2V_{0}}{m}

v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}

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Hence, The speed is \sqrt{2}v_{0}.

6 0
1 year ago
A 4.24 kg marble slab has a volume of 1564 cm what is it’s density in g/cm?
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2.71 g/cm3

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i got the answer on ck-12 wrong until it showed me the answer.

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