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ValentinkaMS [17]
3 years ago
12

Two identical objects in outer space have a head-on collision and stick together. If, before the collision, one had been moving

at 2 m/s and the other at 1 m/s, their combined speed after the collision would be?
Physics
1 answer:
julia-pushkina [17]3 years ago
5 0

Answer:

1.5 m/s

Explanation:

Momentum is conserved and conservation of momentum is

p₁ + p₂ = p'₁ + p'₂

or

m₁v₁ + m₂v₂ = m₁v'₁  + m₂v'₂

In our problem, after collision v'₁ will be equal to v'₂.

Since objects are identical m₁ = m₂

m(v₁+ v₂) = 2m x v'₁

(2m/s + 1m/s) = 2v'₁

v'₁ = v'₂ = 1.5 m/s

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kvasek [131]

Answer:

8.37×10⁻⁴ N/C

Explanation:

Electric Field: This is the ratio of electrostatic force to electric charge. The S.I unit of electric field is N/C.

From the question, the expression for electric field is given as,

E = F/Q.......................... Equation 1

Where E = Electric Field, F = force experienced by the charged balloon, Q = Charge on the balloon.

Given: F = 8.2×10⁻² Newton, Q = 9.8×10 Coulombs = 98 Coulombs

Substitute these values into equation 1

E = 8.2×10⁻² /98

E = 8.37×10⁻⁴ N/C

Hence the Electric Field of the charged balloon =  8.37×10⁻⁴ N/C

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