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tatyana61 [14]
3 years ago
14

A 7.2 kg bowling ball is accelerated with a force of 112,5 N.

Physics
1 answer:
Dominik [7]3 years ago
6 0

Answer:

15.625 m/s

Explanation:

112.5/7.2 = 15.625 m/s

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An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5.0 ms. What is the magnitude of this
IrinaK [193]
 <span>s=2.7 centimeters = 0.027 meters 
t=3.9 milliseconds = 0.0039 seconds 

s=(1/2)a*t^2 
so 
a=(2.7*2)/(0.0039)^2 
= 355,029.58 m/s^2 

a=355,029.58 m/s^2 = 355.02958 km / s^2</span>
7 0
3 years ago
A car goes East from 2 m/s to 16m/s in 3.5s. What is the Car's acceleration?
BaLLatris [955]
A=v1+v2 / t
2+16/3.5=?
8 0
3 years ago
Velocity vector and acceleration vector in a uniform circular motion are related as.
mr_godi [17]

They are related as \bold{\underline {v}\,.\,\underline a }= \bold{0}

  • In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
  • This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
  • For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
  • Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
  • In mathematics, 2 vectors (\underline p , \underline q) that are perpendicular to each other have a quality that their dot product (\underline p\,.\, \underline q) equal to zero vector (\bold 0) which is written as \undeline p\,.\, \underline q = \bold 0.
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#SPJ4

8 0
11 months ago
In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text
Aleks [24]

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

 gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second ,  change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

=  4.859 x 10⁻² oscillations .

5 0
3 years ago
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
3 years ago
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