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Eduardwww
2 months ago
15

the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce

ntimeter, how far does the minute hand travel during a 42 minute period
Physics
1 answer:
Ostrovityanka [3.2K]2 months ago
4 0

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

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A uniform magnetic field makes an angle of 30o with the z axis. If the magnetic flux through a 1.0 m2 portion of the xy plane is
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Response:

(b) 10 Wb

Clarification:

Given;

angle of the magnetic field, θ = 30°

initial area of the plane, A₁ = 1 m²

initial magnetic flux through the plane, Φ₁ = 5.0 Wb

The equation for magnetic flux is;

Φ = BACosθ

where;

B denotes the magnetic field strength

A represents the area of the plane

θ is the inclination angle

Φ₁ = BA₁Cosθ

5 = B(1 x cos30)

B = 5/(cos30)

B = 5.7735 T

Next, calculate the magnetic flux through a 2.0 m² section of the same plane:

Φ₂ = BA₂Cosθ

Φ₂ = 5.7735 x 2 x cos30

Φ₂ = 10 Wb

<pHence, the magnetic flux through a 2.0 m² area of the same plane is 10 Wb.

Option "b"

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1. Susie wondered if the height of a hole punched in the side of a quart-size milk carton would affect how far from the containe
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Constant: All other factors aside from the independent variable, such as the liquid volume.
Control: None that I recognize.
Number of groups: 4
Trials per group: 4
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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
inna [3103]

Answer:

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu ≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + z^{2} k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + v^{2}k

The normal vector N is computed as

N = S_{u}XS_{v}

Where:

S_{u} = =

S_{v} = =

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, v^{2}>. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + uv^{2}

Thus

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu ≈ 3077.34

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The appropriate choice is C.

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