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Mariana [72]
3 years ago
8

To get an idea of how much thermal energy is contained in the world's oceans, estimate the heat liberated when a cube of ocean w

ater, 3 km on each side, is cooled by 4 K. (Approximate the ocean water as pure water for this estimate.)
Physics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

Q = 4.52 10¹⁷ J

Explanation:

Thermal energy can be calculated with  

      Q = m c_{e} ΔT

in this case it indicates that we approximate seawater to pure water with  

    c_{e} = 4186 J/ kg K  

with the density

    ρ = m / V  

    m = ρ V  

    V = L³  

we substitute  

   m = ρ L³  

   Q = ρ L3 c_{e} ΔT

calculate  

   Q = 1000 (3 103) 3 4186 4  

   Q = 4.52 10¹⁷ J

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3 years ago
A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?
katrin [286]

Answer:

<em>k = 25.18 N/m</em>

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}

If the period is known, we can find the value of the constant by solving for k:

\displaystyle k=m\left(\frac{2\pi}{T}\right)^2

Substituting the given values m=5 Kg and T=2.8 seconds:

\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2

k = 25.18 N/m

5 0
3 years ago
Calculate the pressure exerted on the heel of a boy’s foot if the boy weighs 80 N and he lands on one heel,which has an area of
cricket20 [7]

Pressure at a given surface is given as ratio of normal force and area

so here force due to heel of the shoes is given as 80 N

and the area of the heel is given as 16 cm^2

so we can say

P = \frac{F}{A}

here we have

F = 80 N

A = 16 cm^2 = 16 * 10^{-4} m^2

P = \frac{80}{16 * 10^{-4}}

P = 5 * 10^4 N/m^2

so pressure at the surface due to its heel will be 5 * 10^4 N/m^2

3 0
3 years ago
List the jovian planets in order of increasing distance from the sun?
trapecia [35]
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Hope this helps.
5 0
3 years ago
The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the ce
NISA [10]

Answer:

The potential at the center of the sphere is -924 V

Explanation:

Given;

radius of the sphere, R = 0.22 m

electric field at the surface of the sphere, E = 4200 N/C

Since the electric field is directed towards the center of the sphere, the charge is negative.

The Potential is the same at every point in the sphere, and it is given as;

V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R} -------equation (1)

The electric field on the sphere is also given as;

E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}

|q |=  4 \pi \epsilon _o} R^2E

Substitute in the value of q in equation (1)

V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed  \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V

Therefore, the potential at the center of the sphere is -924 V

8 0
3 years ago
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