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Elanso [62]
3 years ago
14

Calculate the speed of a bowlin ball that moves 8 meters in 4 seconds

Physics
2 answers:
Mariana [72]3 years ago
8 0

Answer:

2m/s

Explanation:

The formula for speed is [ s = d/t ].

s = 8/4

s = 2m/s

Best of Luck!

disa [49]3 years ago
8 0

Answer:

\boxed {\boxed {\sf 2 \ m/s}}

Explanation:

Speed is the distance traveled per unit of time. It is calculated using the following formula:

s= \frac {d}{t}

The bowling ball travels a distance of 8 meters in 4 seconds.

  • d= 8 m
  • t= 4 s

Substitute these values into the formula.

s= \frac{8 \ m}{ 4 \ s}

Divide.

s= 2 \ m/s

The speed of the bowling ball is <u>2 meters per second.</u>

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The difference between the frequency fff and the frequency ωωomega is that fff is measured in cycles per second or hertz (abbrev
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Answer:

Radians

Explanation:

The angular speed is a measure of the rotation speed of a body. It is defined as the angle rotated by a unit of time. Thus, It refers to the angular displacement per unit time and is designated by the Greek letter \omega. Its unit in the International System is radian per second (rad / s).

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3 years ago
The rotting of wood is an example of which of the following reactions?
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Decomposition is your answer.
8 0
3 years ago
A spacecraft is in orbit around Mars. Suppose that it was previously observed to have speed 5 km/sec while at a distance r = 500
Lina20 [59]

Answer:

V' = 4.56Km/s

Explanation:

We can use the formula of Orbital Speed, who say,

V = \sqrt{\frac{GM}{r}}

The relative velocity is given by,

V' = \sqrt{\frac{GM}{r'}}

We need to find the relation between the two speeds,

\frac{V'}{V} = \sqrt{\frac{GM}{r}*\frac{r'}{GM}}

\frac{V'}{V} = \sqrt{\frac{r}{r'}}

V' = V \sqrt{\frac{r}{r'}}

Substituting,

V' = (5)*\sqrt{\frac{5000}{6000}}

V' = 4.56Km/s

6 0
3 years ago
An astronaut notices that a pendulum which took 2.45 s for a complete cycle of swing when the rocket was waiting on the launch p
klio [65]

Answer:

2.84 g's with the remaining 1 g coming from gravity (3.84 g's)

Explanation:

period of oscillation while waiting (T1) = 2.45 s

period of oscillation at liftoff (T2) = 1.25 s

period of a pendulum (T) =2π. \sqrt{\frac{L}{a} }

where

  • L = length
  • a = acceleration

therefore the ration of the periods while on ground and at take off will be

\frac{T1}{T2} =(2π \sqrt{\frac{L}{a1} } ) /  (2π\sqrt{\frac{L}{a2} })

where

  • a1 = acceleration on ground while waiting
  • a2 = acceleration during liftoff

\frac{T1}{T2} = \frac{\sqrt{\frac{L}{a1} }}{\sqrt{\frac{L}{a2} }}

squaring both sides we have

(\frac{T1}{T2})^{2} = \frac{\frac{L}{a1} }{\frac{L}{a2} }

(\frac{T1}{T2})^{2} = \frac{a2}{a1}

assuming that the acceleration on ground a1 = 9.8 m/s^{2}

(\frac{T1}{T2})^{2} = \frac{a2}{9.8}

a2 = 9.8 x (\frac{T1}{T2})^{2}

substituting the values of T1 and T2 into the above we have

a2 = 9.8 x (\frac{2.45}{1.25})^{2}

a2 = 9.8 x 3.84

take note that 1 g = 9.8 m/s^{2} therefore the above becomes

a2 = 3.84 g's

Hence assuming the rock is still close to the ground during lift off, the acceleration of the rocket would be 2.84 g's with the remaining 1 g coming from gravity.

5 0
3 years ago
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