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azamat
3 years ago
14

An 8-passenger Learjet has a force of gravity of 6.6x10^4 [down] acting on it as it travels at a constant velocity of 6.4x10^2 k

m/h [W]. If the forward thrust provided by the engines is 1.3x10^4 N [W], determine: a.) the upward lift on the plane b.) the force due to air resistance on the plane
Physics
1 answer:
IgorC [24]3 years ago
7 0
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tatuchka [14]
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on june 21 some earth locations have 24 hours of daylight these locations are all between the latitudes of
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66° N and 90° N

the area of the artic circle in the northern hemisphere

4 0
3 years ago
An atomic physicist determines that an atom is composed of 8 positively charged particles and has a mass of 15 amu. which is the
stellarik [79]
<span> 1.the atom has 7 neutrons and 8 protons....

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5 0
3 years ago
Read 2 more answers
A bowling ball of mass m = 1.7 kg drops from a height h = 14.2 m. A semi-circular tube of radius r = 6.2 m rest centered on a sc
marshall27 [118]

Answer:

W_net = mg + 2mgh/r

Explanation:

The forces applied in this motion of the bowling ball are both gravitational and centripetal forces.

Now, gravitational force is; F_g = mg

While centripetal force is; F_c = mv²/r

Since we want to express the net force in terms of the variables in the statement and we are not given "v", let's find an expression of v with the variables given.

Now, from Newton's equation of motion, at initial velocity of 0, v² = 2gh.

Thus;

F_c = 2mgh/r

Where;

m is ball mass

r is tube radius

h is fall height

Thus, the net force will be;

F_net = F_g + F_c

Now, Net force would be equal to the net weight that will be read on the scale.

Thus;

W_net = F_net = F_g + F_c

W_net = mg + 2mgh/r

8 0
3 years ago
The voltage across a 5-uF capacitor is: v (t )equals 10 cos open parentheses 6000 t close parentheses space straight V. What is
mamaluj [8]

Answer:

- 0.3sin6000t A

Explanation:

Voltage, v = 10 cos 6000t V

Capacitance = 5-uF

Current flowing through, i(t)

i(t) = c * d/dt (V)

c = 5-uF = 5 * 10^-6 F

i(t) = (5 * 10^-6) * d/dt(10 cos 6000t)

d/dt(10 cos 6000t) = (10 * 6000) * (-sin 6000t)

Hence,

i(t) = (5*10^-6) * (10*6000) * (-sin 6000t)

i(t) = 5*10^-6 * 6*10^4 * - sin6000t

i(t) = 30 * 10^-2 * - sin6000t

i(t) = 0.3*-sin6000t

i(t) = - 0.3sin6000t Ampere

4 0
2 years ago
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