Answer:
The correct choice is 8o.
Explanation:
The given weight is hanging at the mid point of the rope and the buildings are at the same level, Obtain the given equation by equating the vertical components of force,
2Tsin(φ)=W
where φ is the angle made by rope with the horizontal.
Given T=1800 N and W=500 N, the value of φ is calculated as follows,
2*1800*sin(φ)=500
sin(φ)=0.1388
φ=7.90
φ≈8o
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3200 N
1600X2= 3200
It doubles so you multiply 3200 by 2
Answer:
Mumbai
Explanation:
Because Mumbai is a coastal city
due to high humidity,vigorous rusting damages all the iron structures
∴ rusting is a major problem in Mumbai
Answer:
hence the y - component of the velocity is 50m/s
This question is incomplete, the complete question is;
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass
;
a) F_throw = 8.083 N
b) F_throw = 9.181 N
c) F_throw = 2.284 N
d) F_throw = 16.014 N
e) None of these is correct
Answer:
the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
Explanation:
Given that;
m = 0.408 kg
d = 1.909 m
u = 0 { from rest}
t = 0.439 s
Now using Kinetic equation
d = ut + 1/2 at²
we substitute
1.909 = (0 × 0.439) + 1/2 a(0.439)²
1.909 = 0 + 0.09636a
1.909 = 0.09636a
a = 1.909 / 0.09636
a = 19.8111 m/s²
Now force applied will be;
F = ma
we substitute
F = 0.408 × 19.8111
F = 8.0828 ≈ 8.083 N
Therefore the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer