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Licemer1 [7]
3 years ago
5

Newtons law of gravitation is called a universal law why​

Physics
1 answer:
mihalych1998 [28]3 years ago
7 0

Answer:

it is applicable on all bodies having mass and the bodies Will be governed by the same law

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Cosmic background radiation is electromagnetic radiation from the sky with no discernible source. The origin of this radiation depends on the region of the spectrum that is observed. 
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Mickey had three substances. He hit each of them with a hammer. A red
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A & D

Explanation:

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a soccer player kicks a ball. According to Newton's third law of motion, the player's foot exerts a force on the ball. The ball
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The forces are exerted on different objects so they are not balanced forces.

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3 years ago
two barges full of salted toad guts have a collision. the red barge has a mass of 150000kg and is traveling northwest at 0.25m/s
solniwko [45]

The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

Final velocity(v3)  of the red barge is calculated by following formula

m1×v1+ m2×v2= (m1+m2)v3

Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500= 1150000×v3

v3= 0.311 m/s

<h3>What is elastic collision velocity? </h3>
  • The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

#SPJ9

6 0
1 year ago
A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the
ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

a_{B} =a_{D}  +(a_{B/D} )_{t}

2.5j=1.5j  +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

2\alpha =1\\\alpha =0.5rad/s^{2}CW

b) The acceleration of point A is:

a_{A} =a_{D}  +(a_{A/D} )_{t}

(aA/D)t = ADαj

a_{A} =a_{D}  +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}

The acceleration of point E is:

(aE/D)t = -EDαj

a_{E} =a_{D}  -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}

7 0
4 years ago
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