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sergey [27]
3 years ago
15

How does the force affect the rocks during collisions? *

Physics
1 answer:
Jet001 [13]3 years ago
3 0

In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.
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8. If the mass of a balloon is 300 kg and the lift force provided by the atmosphere is 3300 N, what
zmey [24]

Answer:

Toward the north, 300 N

Explanation:

Consider the upward direction as positive and downward direction as negative.

Given:

Mass of the balloon is, m=300\ kg

Lift force on the balloon in the upward direction is, F_L=3300\ N

Take acceleration due to gravity as 10 N/kg.

Now, weight of the balloon acting downward is given as the product of its mass and acceleration due to gravity. Therefore,

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3 0
3 years ago
Interactive LearningWare 4.1 reviews the approach taken in problems such as this one. A 1800-kg car is traveling with a speed of
Lubov Fominskaja [6]

Answer:

F= 4788 N

Explanation:

Because the car moves with uniformly accelerated movement we apply the following formula:

vf²=v₀²+2*a*d Formula (1)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s    

vf: final speed in m/s  

a: acceleration in m/s²

Data

d=36.9 m

v₀=14.0 m/s m/s    

vf= 0  

Calculating of the acceleration of the car

We replace dta in the formula (1)

vf²=v₀²+2*a*d

(0)²=(14)²+2*a*(36.9)

-(14)²= (73.8) *a

a= - (196) /  (73.8)

a= - 2.66 m/s²

Newton's second law of the car in direction  horizontal (x):

∑Fx = m*ax Formula (2)

∑F : algebraic sum of the forces in direction x-axis (N)

m : mass (kg)

a : acceleration  (m/s²)

Data

m=1800 Fkg

a= - 2.66 m/s²

Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :

We replace data in the formula (2)

-F= (1800 kg) * ( -2.66 m/s² )

F= 4788 N

6 0
3 years ago
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