Answer:
<em><u>Given: </u></em>
m1 = 7 kg
V1 = 12 m/s
m2 = 25 kg
V2 = 6 m/s
<em><u>To find:</u></em>
Combined speed of two balls stick together after collision V = ?
<em><u>Solution:</u></em>
<em>According to law of conservation of momentum,</em>
m1V1 + m2V2 = (m1+m2)V
7×12 + 25×6 = (7+25)V
84 + 150 = 32V
V = 234/32
V = 7.31 m/s
Combined speed of two ball is 7.31 m/s
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Answer:
we know that that frequency= n/t
= 70/20
= 3.5 Hertz
Explanation:
i think that the answer
Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa
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