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kati45 [8]
3 years ago
13

mini stereo speaker cables are made of copper wire surrounded by robert electricity travel down the copper portion of the cable

while the rubber protects you from being zapped by the electricity which physical property explains the difference in ability of copper and rubber to transmit electricity
Physics
1 answer:
Tom [10]3 years ago
7 0

Here current is flowing through the copper wire so this shows that copper is good conductor of electricity.

It is having less resistance as it conducts the current easily

Now a rubber coating on it will protect us from electric shock

So this property shows that rubber is a bad conductor of electricity

It is having large electrical resistance due to which it will conduct no current

Rubber : - No transmittance of electricity

copper :- good transmittance of electricity

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You are in a car and go around a corner very fast. What happens to you?
Fiesta28 [93]

Answer:

You may tip the car over or crash.

Explanation:

Going really fast at high speeds and turn a corner might make the car crash into an object or fall over.

3 0
3 years ago
A ball bearing of radius of 1.5 mm made of iron of density
Serjik [45]

Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times 64.7196

\sf \implies \eta =  2 \times 7.191

\sf \implies \eta =  14.382 \: poise

6 0
3 years ago
Define the difference between the rigid body problem and a single particle problem.
agasfer [191]
. In single particle problem whole mass is concentrated at a single point so it has a single displacement, single velocity and single acceleration. while, in rigid body mass is distributed
3 0
4 years ago
a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her
Rama09 [41]
Acceleration = velocity/time
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5 0
4 years ago
Read 2 more answers
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
3 years ago
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