Explanation: acorrding to my calculation of my 8th grade education with the proxsimity of the y=mx+b you will get -397758.4837y483 now you multiply that times pi which is 3.145926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 and you get 42066669 divide that by 3 and you get D

5 0

3 years ago

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A fireman clings to a vertical ladder and directs the nozzle of a hose horizontally toward a burning building. The rate of water

Aleks [24]

Answer:

A torque of 102.5375 Nm must be exerted by the fireman

Explanation:

Given:

The rate of water flow = 6.31 kg/s

The speed of nozzle = 12.5 m/s

Now, from the Newton's second law we have

The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction

thus we have,

F = 6.31 kg/s x 12.5m/s

F = 78.875 N

Now,

The torque (T) exerted by water force about the fireman's will be

T = (F x d)

T = 78.875 N x 1.30 m

T = 102.5375 Nm

hence,

<u>A torque of 102.5375 Nm must be exerted by the fireman</u>

4 0

3 years ago

Fig. 28-9 shows the cross-section of a hollow cylinder of inner radius a = 5 cm and outer radius b = 7 cm. A uniform current den

Dmitrij [34]

Answer:

1.507×10⁻⁴ T

Explanation:

a = 5 cm = 0.05 m

b = 7 cm = 0.07 m

j = 1 A/cm²

Distance from magnetic field = d = 10 cm = 0.1 m

μ₀ = Vacuum permeability = 4π×10⁻⁷ H/m

Magnetic of hollow cylinder

$\oint B.ds=\mu_0 I\\\Rightarrow B2\pi d=\mu_0 I\\\Rightarrow B2\pi d=\mu_0J\pi(b^2-a^2)\\\Rightarrow B=\frac{\mu_0J\pi(b^2-a^2)}{2\pi d}\\\Rightarrow B=\frac{4\pi\times 10^{-7}\times 1\times 10^{-4}\pi(0.07^2-0.05^2)}{2\pi 0.1}\\\Rightarrow B=1.507\times 10^{-4}\ T$

∴ Magnitude of the magnetic field at a distance d = 10 cm from the axis of the cylinder is 1.507×10⁻⁴ T

7 0

4 years ago

What is true of a mechanical wave? @3.2.P.B5 @1

statuscvo [17]

I think it is B. Mechanical waves carry energy through a medium.

7 0

3 years ago

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