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Rama09 [41]
3 years ago
7

A bowler who always left the same three pins standing could be considered a(n) ____ bowler.

Physics
1 answer:
RideAnS [48]3 years ago
4 0
A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.
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How much force, in g cm/s2
scoray [572]
In this question force is measured in  g cm/s2 so we know that to get the answer we times g by cm/s2 
50 × 20 = 1000
3 0
3 years ago
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In your own words, explain what conservation of energy means. Also, give an example of the conservation of energy using somethin
Genrish500 [490]
Energy can not be created or destroyed but can change from one form to another.

example: as a roller coaster cart loses height the more speed it gains, the potential energy is transferred into kenetic energy
8 0
2 years ago
A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se
Rudik [331]

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

x = 32t - 38\frac{t^3}{3} \\\\

The velocity of the particle is calculated as the change in the position of the  particle with time;

v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s

Acceleration is the change in velocity with time;

a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2

4 0
2 years ago
A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 29.8 kg an
jonny [76]

Answer:

28.81 m

Explanation:

Ff = -123

m * a  = -123

(29.8+10.3) * a = -123

a = -123/40.1 = -3.07

We know,

v^2 = u^2 + 2as

0^2 = 13.3^2 + 2*(-3.07)*s

s = 176.89/6.14 = 28.81

[ If there's a problem with the solution, pleaase let me know ]

6 0
3 years ago
A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands
pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}

v=\infty

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}

v=34

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0
3 years ago
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