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FrozenT [24]
4 years ago
7

What is the force of gravity between a 500kg bal and a 100,000kg basket 5m apart?

Physics
2 answers:
klemol [59]4 years ago
8 0

Answer:

10 Mil. J

Explanation:

multiplyed balls and divided by 5

zalisa [80]4 years ago
5 0

Answer:

19,600,000 N

Explanation:

G times m1 times m2 over distance squared

9.8*500*100,000 = 490,000,000

490,000,000 divided by (5*5) = 19,600,000

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Where are you likely to find the oldest oceanic crust?
Stells [14]

Answer:

it would be letter E. near oceanic ridges

Explanation:

new ocean crust is formed at the mid ocean ridges

4 0
3 years ago
Pls answerrr it's urgent!!!!!
lilavasa [31]

Explanation:

the average velocity of the car is 15 m/s example I have this on a test

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2 years ago
A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure
Alex787 [66]
V = t^2 - 9t + 18

position, s
s = t^3 /3 - 4.5t^2 +18t + C

       t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1

Average velocity: distance / time

   distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
   Average velocity = 27.67 / 8 = 3.46 m/s

t = 5 s

     v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
     speed = |-2| m/s = 2 m/s
 
Moving right
     V > 0 => t^2 - 9t + 18 > 0
     (t - 6)(t - 3) > 0

     => t > 6 and t > 3 => t > 6 s => Interval (6,8)

    => t < 6 and t <3 => t <3 s => interval (0,3)

    

Going faster and slowing dowm

acceleration, a = v' = 2t - 9
     a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
     Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
     


4 0
3 years ago
Water falls without splashing at a rate of 0.370 l/s from a height of 2.90 m into a 0.690-kg bucket on a scale. if the bucket is
raketka [301]
<span>R = rate of flow = 0.370 L/s H = height = 2.9 m T= time = 3.9 s V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2 G value = 9.8 m/s2 Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N Wa = weight of accumulated water after 3.9 s Fi = force of impact of water on the bucket S = reading on the scale = Wa + Wb + Fi mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg Therefore, Wa = 1.443 x 9.8 = 14.1414 N Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt) = 0.37 x 7.539 = 2.78943 N S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
8 0
3 years ago
If C is 1kg and D is 100kg, and the initial velocities of both balls are 5m/s, how would the magnitude of the forces exerted by
kupik [55]

Answer:

Explanation:

The forces exerted by each mass is best understood in terms of their momentum.

Momentum is a sort of compelling force or impulse. It is given as:

                  Momentum  =  mass x velocity

Let us consider the momentum of the balls;

Substance C;

    Mass  = 1kg

     Velocity  = 5m/s

    Momentum of C = 1 x 5  = 5kgm/s

Substance D:

      Mass  = 100kg

      Velocity  = 5m/s

     Momentum of D  = 100kg x 5m/s  = 500kgm/s

Body D has a higher momentum compared to Body C. This suggests that body D will exert a higher force than C when they collide.

The higher the momentum, the more the force of impact it has.

3 0
4 years ago
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