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3 years ago

The grounded conductor and the grounding conductor are one and the same conductor and perform the same function.

Physics

1 answer:

STALIN [3.7K]3 years ago

5 0

Answer:

Grounded used to carry the current in the normal conditions. while on the other side grounding is refer to that safety wire that is connected to the earth and currently does not transmit through it.

As grounding wire is referred to as safety wire hence it carries current only in an emergency like a short circuit.

Explanation:

A grounded conductor is referred to as one of the wire that needed in an electric circuit. it is basically a neutral conductor. It used to carry the current in the normal conditions. while on the other side grounding is refer to that safety wire that is connected to the earth and currently does not transmit through it.

As grounding wire is referred to as safety wire hence it carries current only in an emergency like a short circuit.

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You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 _. If each light ha

UNO [17]

V = IR

By completing the equation, i found that the total power equation is : 4.8,
Which means that it's not exceed the power rating.

So i believe the answer would be : The string will remain lit

hope this helps

6 0

3 years ago

Calculate the pressure the fluid exerted on your diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is

erica [24]

Complete question:

A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.

Answer:

Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

Explanation:

Given;

density of water, ρ = 1000 kg/m³

diver's position below the surface of the water, h = 10 m

acceleration due to gravity, g = 9.8 m/s²

Let the atmospheric pressure, P₀ = 101325 Pa

The pressure 10 m below the surface of the water is calculated as;

P = P₀ + ρgh

P = 101325 Pa + (1000 x 9.8 x 10)Pa

P = 199325 Pa

P = 1.99 x 10⁵ Pa.

Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

5 0

2 years ago

Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre

castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:

7.5

×m

...........................(

)

in Above equation in x and t. Separating the variables and integrating,

∫

/7.5

×=

∫

−

4.7619

Here C =constant of integration.

0 at t

, we get: C

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619 =

−

4.7619

(

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= −

4.7619

−

4.7619

= −

3.7619

or x

7.15

(b) To find the jogger's acceleration in m

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

7.5×

Now differentiating above equation w.r.t time we get:

−

0.675

the joggers acceleration is :

−

0.675

−

4.34

6 in equation (2). We get:

−

4.7619

(

−

(

0.04

6 )

10=

−

4.7619

0.832

6 0

3 years ago

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

Strike441 [17]

Answer:

θ = 13.7º

Explanation:

According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

$\Delta K + \Delta U = W_{nc} (1)$

ΔK, is the change in kinetic energy, as follows:

$\Delta K = \frac{1}{2}* m* (v_{f} ^{2} - v_{0} ^{2}) (2)$

ΔU, is the change in the gravitational potential energy.
If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º

8 0

3 years ago

Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by

bixtya [17]

Answer:

v = 0.99 c = 2.99 x 10⁸ m/s

Explanation:

From the special theory of relativity:

$t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2} } }\\$

where,

v = speed of travel = ?

c = speed of light = 3 x 10⁸ m/s

t = time measured on earth = 90 years

t₀ = time measured in moving frame = 6 months = 0.5 year

Therefore,

$90\ yr = \frac{0.5\ yr}{\sqrt{1-\frac{v^2}{c^2} } } \\\\\\\sqrt{1-\frac{v^2}{c^2}} = \frac{0.5\ yr}{90\ yr}\\ 1-\frac{v^2}{c^2} = 0.00003086\\\frac{v^2}{c^2} = 1-0.00003086\\$

8 0

3 years ago