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VARVARA [1.3K]
2 years ago
12

A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on

e-half his normal weight. How fast is he going
Physics
1 answer:
n200080 [17]2 years ago
6 0

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

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Oxygen (O2) is best described as which of the following: An element A compound A mixture
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Answer:

in it's natural state it's an element

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What form of energy are microwaves?
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A car traveling at 23 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration?
kykrilka [37]
We could determine the acceleration using this formula
\boxed{a= \dfrac{v_{1}-v_{0}}{t} }

Given from the question v₀ = 23 m/s, v₁ = 0 (the car stops), t = 5 s
plug in the numbers
a= \dfrac{v_{1}-v_{0}}{t}
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8 0
3 years ago
Problem #3: 3.6. A diode has wdo = 0.4 µm and φj = 0.85 V. (a) What reverse bias is required to triple the depletion-layer width
Alchen [17]

Answer:

a) 6.8 Volt

b) 1.21Цm

Explanation:

We are given from the question that

   The zero -bias depletion layer width(W_{do}) is 0.4Цm

  The built in voltage φj  is 0.85V

Now to calculate the reverse voltage( V_{R}) that would be required  to triple the depletion - layer width.

  The depletion - layer width (W_{d}) of the diode has the formula

                          W_{d} = W_{do} \sqrt{1 + \frac{V_{R} }{Qj} }

    For three times of   W_{d} we have

        3W_{d} = W_{do} \sqrt{1 +\frac{V_{R} }{Qj} }

         =>      \frac{V_{R} }{Qj} = 3^{2} -1

        => V_{R} = 8Qj

        Substituting value of φj

       We have

                         V_{R} = 8(0.85V)

                               =   6.8 V

The required bias voltage  V_{R} is  6.8 V

   The solution for the b part of the question is uploaded on first image

7 0
3 years ago
57. Estimate Potential Energy A boulder with a
Oksanka [162]

Answer: 4.9 x 10^6 joules

Explanation:

Given that:

mass of boulder (m) = 2,500 kg

Height of ledge above canyon floor (h) = 200 m

Gravita-tional potential energy of the boulder (GPE) = ?

Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).

GPE = mgh

GPE = 2500kg x 9.8m/s2 x 200m

GPE = 4900000J

Place result in standard form

GPE = 4.9 x 10^6J

Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules

3 0
3 years ago
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