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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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Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?
Schach [20]
Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.
6 0
2 years ago
When a 12 newton horizontal force is applied to a box on a horizontal
sdas [7]

Answer:

ur mum

Explanation:

6 0
3 years ago
If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?
mamaluj [8]

Answer:

Explanation:

KE=\frac{1}{2}mv^2

784=\frac{1}{2}(40)v^2

784=20v^2

39.2=v^2

v=6.26m/s

4 0
3 years ago
A submarine travels 25 km/h north for 3.2 hours. What is its displacement?
Vika [28.1K]

displ = velocity x time

25 x 3.2 = 75+5 km north.

7 0
2 years ago
Read 2 more answers
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

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

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
2 years ago
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