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2 years ago

How does subjustion change the ocean florr this is acctualy in science

1 answer:

7 0

Subductions causes the ocean floor to sink into deep-ocean trenches and the earths plates move apart from molten rock or magma ana the earths crusts rises a departs.

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All the planets orbit the Sun

aivan3 [116]

Answer:

Yes, they do

Explanation:

7 0

3 years ago

Read 2 more answers

Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv

deff fn [24]

Answer:

electric field E = - k Q (1 /r(r-a)), force F = - k Q qo / r (r-a) and force for r>>a F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

λ = q / x

As it is constant, we can write it based on differentials

λ = dq / dx

dq = λ dx

We already have all the terms, let's integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

E = k ∫ λ dx / x²

E = k la (- 1 / x)

Let's get the negative sign from the parentheses

E = - k λ (1 / x)

E = - k λ (1 /(r-a) -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

E = - k Q/a [a / r (r-a)]

E = - k Q (1 /r(r-a))

b) We calculate the force.

F = E qo

F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

F = - k Qqo / r² (1 - a/r)

F ≈ - k Q qo / r²

6 0

3 years ago

A 5.0-ohm resistor, a 10.0-ohm resistor, and a 15.0-ohm resistor are connected in parallel with a battery. The current through t

Kaylis [27]

Answer:

E = 3456 J

Explanation:

The electrical energy expended in a resistor can be easily calculated by using the following formula:

$E = Pt$

where,

E = Energy Expended = ?

I = current through 5 ohm resistor = 2.4 A

R = Resistance = 5 ohms

P = Electrical Power = VI

Since,

V = IR (Ohm's Law)

Therefore,

P = (IR)(I) = I²R = (2.4 A)²(5 ohms) = 28.8 Watt

t = time taken = (2 min)(60 s/1 min) = 120 s

Therefore,

E = (28.8 Watt)(120 s)

E = 3456 J

8 0

3 years ago

A brick is lying on a table in a state of static equilibrium. If the mass of the brick is 7.52 kilograms, what is the normal for

frez [133]

In mechanics, the normal force is the component, perpendicular to the surface (surface being a plane) of contact, of the contact force exerted on an object . We calculate as follows:

∑F along x = 0 = F - Fn

Fn = F = mg = 7.52(9.81) = 73.77 N <------OPTION B

5 0

2 years ago

A 2.4-m high 200-m2 house is maintained at 22Â°C by an air-conditioning system whose COP, is 3.2. It is estimated that the kitch

kotegsom [21]

Answer:

The amount that is "vented" out by "the fans" is $0.50 for 10 hours.

Option: a

Explanation:

"Energy discharged by air in every hour" can be determined by,

$\mathrm{Q}=\mathrm{m}_{\mathrm{air}} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}$

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat (units J/kg∙K)

$\mathrm{m}_{\mathrm{air}}=\rho \mathrm{v}$

$\text { Density of air } \rho=1.20 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Density of air } \rho=1.20 \times 200 \times 2.4$

$\text { Density of air } \rho=576 \mathrm{kg}$

∆T = 10 hours

$\text { Specific Heat Capacities of Air. The nominal values used for air at } 300 \mathrm{K} \text { are } \mathrm{C_P}=1.00 \mathrm{kJ} / \mathrm{kg} . \mathrm{K}$