Answer:
740, mm of Hg
Explanation:
The pressure of argon , in mm of Hg = difference in the level of mercury on either side of the manometer.
mercury column in the open end of the manometer is 22 mm below that in the side connected to the argon and 762 mmHg, end is open to atmospheric pressure.
therefore, The pressure of argon , in mm of Hg =762 -22 = 740, mm of Hg
B force i belive becuase when is done it force
The stretching force acting on the second wire, given the data is 588 N
<h3>Data obtained from the question</h3>
- Radius of fist wire (r₁) = 3.9×10⁻³ m
- Force of first wire (F₁) = 450 N
- Radius of second wire (r₂) = 5.1×10⁻³ m
- Force of second wire (F₂) =?
<h3>How to determine the force of the second wire</h3>
F₁ / r₁ = F₂ / r₂
450 / 3.9×10⁻³ = F₂ / 5.1×10⁻³
cross multiply
3.9×10⁻³ × F₂ = 450 × 5.1×10⁻³
Divide both side by 3.9×10⁻³
F₂ = (450 × 5.1×10⁻³) / 3.9×10⁻³
F₂ = 588 N
Learn more about spring constant:
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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
Sometimes in the same direction but most of them go in the opposite direction