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Alja [10]
3 years ago
5

I need help!!!! Please and thank u!

Physics
1 answer:
alexandr1967 [171]3 years ago
5 0

Answer:

False

True

False

True

False

True

Explanation:

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A bowling ball rolls 33\,\text m33m33, start text, m, end text with an average speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m ​ 2
asambeis [7]

The ball rolled for 13.2 s

<h3>Further explanation</h3>

Speed is scalar and no direction

\tt avg~speed=\dfrac{total~distance}{elapsed~time}=\dfrac{\Delta x}{\Delta t}

A bowling ball rolls 33 m, with average speed = 2.5 m/s

So elapsed time :

\tt t=\dfrac{33~m}{2.5`m/s}=13.2~s

4 0
3 years ago
To measure the strength of an earthquake, you can use either a _____ scale or _____ scale.
allsm [11]
To measure the strength of an earthquake, you can use either a Richter scale or Mercalli scale. Richter scale uses the amplitued of the wave and the distance from the source. Mercalli scale uses observations of people and is not considered to be scientific as Richter scale.
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3 years ago
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Two friends are carrying a crate of mass 200 kg up a flight of stairs. The crate has length 1.25 m and height 0.500 m, and its c
Marysya12 [62]

Ans; see attached file for calculation and answer

Explanation:

4 0
3 years ago
Suppose a plot of inverse wavelength vs frequency has slope equal to 0.119, what is the speed of sound traveling in the tube to
strojnjashka [21]

Answer:

8.40 m/s

Explanation:

Slope of the plot is 0.119

Slope of a plot is given by the change in y direction divided by the change in x direction

Here, the y axis represents inverse wavelength and the x axis represents frequency.

f = Frequency (Hz, assumed)

v = Phase velocity (m/s, assumed)

λ = Wavelength (m, assumed)

So, slope

m=\frac{\frac{1}{\lambda}}{f}

Now,

\lambda=\frac{v}{f}\\\Rightarrow \lambda^{-1}=\frac{f}{v}

\\\Rightarrow m=\frac{\frac{f}{v}}{f}\\\Rightarrow 0.119=\frac{1}{v}\\\Rightarrow v=\frac{1}{0.119}\\\Rightarrow v=8.40\ m/s

The speed of sound travelling in the tube is 8.40 m/s

5 0
3 years ago
In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the
Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0
3 years ago
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