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dalvyx [7]
3 years ago
9

A cement truck of mass 14,000 kg moving 10m/s slams into a wall and comes to a halt in .2s. What is the force of impact on the t

ruck?
Physics
1 answer:
skelet666 [1.2K]3 years ago
6 0

And it’s x10 is 100000

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Look at the diagrams. Each model the arrangement of particles in a substance.
Kisachek [45]

Answer:

a

Explanation:

because it has  compact  molecules  

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2 years ago
Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its su
skad [1K]

Answer:

g_n=\dfrac{2}{9}g

Explanation:

M = Mass of Earth

G = Gravitational constant

R = Radius of Earth

The acceleration due to gravity on Earth is

g=\dfrac{GM}{R^2}

On new planet

g_n=\dfrac{G2M}{(3R)^2}\\\Rightarrow g_n=\dfrac{2GM}{9R^2}

Dividing the two equations we get

\dfrac{g_n}{g}=\dfrac{\dfrac{2GM}{9R^2}}{\dfrac{GM}{R^2}}\\\Rightarrow \dfrac{g_n}{g}=\dfrac{2}{9}\\\Rightarrow g_n=\dfrac{2}{9}g

The acceleration due to gravity on the other planet is g_n=\dfrac{2}{9}g

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3 years ago
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Which process best describes part of a scientific investigation?
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3 years ago
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A string, stretched between two fixed posts, forms standing-wave resonances at 325 Hz and 390 Hz. What is the largest possible v
Pavel [41]

Answer:

65

Explanation:

The resonant frequencies for a fixed string is given by the formula  nv/(2L).  

Where n is the multiple .

v is speed in m/s .

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e  fn+1 – fn = 390 -325

                      =  65

3 0
3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
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