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BartSMP [9]
3 years ago
13

Suppose that during a storm the force of the wind blowing against a skyscraper can be expressed by the vector , where each measu

re in the ordered triple represents the force in newtons. what is the magnitude of this force?
Physics
2 answers:
just olya [345]3 years ago
8 0

Answer: A. 3457 N

Explanation:

Andrei [34K]3 years ago
3 0

Answer:

Here we do not have the vector, but I will try to give a kinda general solution to this type of problem.

If the vector is written as (a, b, c) we have that the force in the x-axis is of a Newtons, in the y-axis is of b Newtons, and in the z-axis is of c Newtons.

Then, we can calculate the total magnitude of this force as:

F = √( a^2 + b^2 + c^2)

wich gives us the total magnitude of the force, but not a direction or anything like that, this is just a scalar.

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What characterizes static stretching? A. having a partner hold limbs in a stretch position B. assuming and holding a stretch pos
balandron [24]

Answer:

It is B

Trust me it is B

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3 years ago
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BRAINLYIST ASAPPP write and claim evidence and reasoning statement that explains why mercury and Venus do not have any moons
umka2103 [35]

answer:They are too close to the sun!

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adoni [48]
The air would contract therefore the answer is the second choice.
5 0
4 years ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
marishachu [46]

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0
3 years ago
A truck accelerating at 0.0083 meters/secondÆ covers a distance of 5.8 × 10Ê meters. If the truck's mass is 7,000 kilograms, wha
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work done = force * distance moved (in direction of the force)

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force=58.1N

58.1*(5.8*10^4)
=3,369,800 J
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3 years ago
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