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BARSIC [14]
4 years ago
14

Jamal plugged his radio into the wall. The radio's plug had copper wires surrounded by rubber. The rubber protects Jamal from___

___?
Physics
2 answers:
Mariana [72]4 years ago
3 0
The rubber protects him from being electrocuted by the flow of current going through the plug.
Hope this helped!!
aliya0001 [1]4 years ago
3 0
The rubber surrounding the wires protects Jamal from electrocution. 
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A school track team member ran for a total of 15.85 miles in practice over 5 days. How many miles did he average per day?
liraira [26]

{\underline{\pink{\textsf{\textbf{ Answer : }}}}}

➡ 15.85/5

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6 0
3 years ago
You are standing on a sidewalk. There is a car in the distance. The horn on the car sounds. You hear it at a pitch that correspo
AnnyKZ [126]

Answer: The answer: The car is moving away from you.

Both A and C are true as Car can be moving in line away from you or has component of velocity in opposite direction.

Explanation:The decrease in the frequency of the sound is the result of Doppler's effect. A/c to Doppler's effect the frequency of received sound of source is changed if it is moving relative to the receiver, i.e. the distance between them is changing due to motion.

The general formula of Doppler's Effect is attached as the picture.

In this formula v_D is the velocity of Detector i.e the receiver relative to wind. While v_s is the velocity of source relative to wind and v is the velocity of sound.

The Doppler's effect is not effected by the velocity of wind as the wind itself could not change the distance between the two objects i.e. you and the car. Wind velocity can change the speed of sound and its wavelength but the change does not effect the frequency.

Hence if we assume the car to be moving with velocity v_c and you are stationary

f'=f_s*\frac{v}{v-v_c}

hence the frequency is reduced.

4 0
4 years ago
Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De
My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0
3 years ago
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