Acceleration=(change in speed)/(time for the change). 43/0.28 = 153.6 m/s^2.
Answer:
The answer is
D. 1230j
Explanation:
When a bullet is shot out of a gun the person firing experiences a backward impact, which is the recoil force, while the force propelling the bullet out of the gun is the propulsive force
given data
Mass of gun M=10kg
Mass of bullet m=20g----kg=20/1000 =0.02kg
Propulsive speed of bullet = 350m/s
Hence the moment of the bullet will be equal and opposite to that of the gun
mv=MV
where V is the recoil velocity which we are solving for
V=mv/M
V=0.02*350/10
V=7/10
V=0.7m/s
The energy contained in the bullet can be gotten using
KE=1/2m(v-V)²
KE=1/2*0.02(350-0.7)²
KE=1/2*0.02(349.3)²
KE=1/2*0.02*122010.49
KE=1/2*2440.20
KE=1220.1J
roughly the energy is 1230J
Answer:
Explanation:
We shall apply Pascal's Law in fluid mechanics
According to it , pressure is transmitted in liquid from one point to another without any change .
25 cm diameter = 12.5 x 10⁻² m radius
Area = 3.14 x (12.5 x 10⁻²)²
= 490.625 x 10⁻⁴ m²
Pressure by vehicle
Force / area
13000 / 490.625 x 10⁻⁴
= 26.497 x 10⁴ Pa
5 cm diameter = 2.5 x 10⁻² radius
area = 3.14 x (2.5 x 10⁻²)²
= 19.625 x 10⁻⁴ m²
If we assume required force F on this area
Pressure = F / 19.625 x 10⁻⁴ Pa
According to Pascal Law
F / 19.625 x 10⁻⁴ = 26.497 x 10⁴
F = 19.625 x 26.497
= 520 N
Explanation:
a stanza can contain 4-6 lines or more than that, that's all upon the writer.
hope this helps you
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