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lawyer [7]
3 years ago
9

Should a flat grassland or a hillside have deeper soil? Explain your answer.

Physics
1 answer:
Fudgin [204]3 years ago
7 0
A hillside of course my friend
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A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles?
jenyasd209 [6]
40 000 kg.m/s...P=MV then 20 000=M2V therefore Pnew=2MV i.e Pnew= 2Pold ...Pnew=2Pold...Pnew=2(20 000)...Pnew=40 000
8 0
3 years ago
Read 2 more answers
A mysterious rocket-propelled object of mass 47.5kg is initially at rest in the middle of the horizontal, frictionless surface o
NARA [144]

Answer:

Distance d=8.63m

Explanation:

mass of object m=47.5 kg

Initial velocity vi=0 m/s

Magnitude of force on object F(t)=(17.0 N/s)t

To find

Distance

Solution

The rocket accelerates due to variable force so we apply Newtons second law but the acceleration will not be constant because the force is not constant

We use  ax=Fx/m to find acceleration but it must be integrate to find velocity and then the distance that rocket travels

So

ax=Fx/m\\gives\\ax(t)=(17.0N/s)t/(47.5kg)\\ax(t)=(0.35789m/s^{3})t\\ As\\Vx(t)=V_{o}+\int\limits^{}_{} { \lim_{0 \to \ t} ax(t) } \, dt=(0.178945m/s^{3} )t^{2}\\   X-X_{o}=\int\limits^{}_{} { \lim_{0 \to \ t} v(t) } \, dt=(0.05964m/s^{3})t^{3}\\   as\\t=5.25s\\So\\X-X_{o}=(0.05964m/s^{3})(5.25s)^{3}\\X-X_{o}=8.63m

Distance d=8.63m    

8 0
3 years ago
Please explain how I would do this!!!
nevsk [136]

Answer:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities.

Explanation:

NMNMBNMNMN

6 0
3 years ago
After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 36.0 m horizontally from the end o
Cloud [144]

Answer

v = 18.99 m/s

\theta = 33.20^0

Explanation:

given,

displacement = 36 m

velocity = 19 m/s

angle =33.2 °

a) v_x = v cos (\theta)

   v_x = 19 cos (33.2^0)

   v_x = 15.89\ m/s

   v_y = v sin (\theta)

   v_y = 19 sin (33.2^0)

   v_y = 10.40\ m/s

magnitude

   v = \sqrt{15.89^2+10.40^2}

         v = 18.99 m/s

b) direction

         tan \theta = \dfrac{10.40}{15.89}

         \theta = tan^{-1}(0.654)

         \theta = 33.20^0

3 0
3 years ago
At full power, how long would it take for the car to accelerate from 0 to 58.0 mph ? neglect friction and air resistance.
Advocard [28]

For the given problem, we calculate the required time by using the formula P = W/t. 

P = [(1/2)mv_f^2 – (1/2)mv_i^2]/ t 

The car accelerates from 0 -58 mph, so the power engine will be 

P = (1/2) x m x 58^2 / t = 1682 m / t 

According to the problem, the engine produces full power so the time required can be calculated as  

420.5 m / 1.40 = 1682 m / t 

t = 5.6 seconds

3 0
4 years ago
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