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3 years ago

1. The process of producing energy by utilizing heat trapped inside the earth's

Physics

2 answers:

babymother [125]3 years ago

6 0

Answer:

Explanation:

The process is called Geo-Thermal energy because it's an inexhaustible source of energy

earnstyle [38]3 years ago

3 0

Explanation: The process of producing energy by utilizing heat trapped inside the earth surface is called Geo thermal energy. ... Explanation: The heat is apparent from the increase in temperature of the earth with increasing depth below the surface.

Answer ) Geo thermal energy.

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Morse code and string phone cup

polet [3.4K]

Answer:

Following are the responses to the given question:

Explanation:

Following are the difference in the speed and accuracy:

The biggest difference was its tone between both the cup method as well as the Morse code. One should talk using cup system and sequence system and let its covering be understood in Morse code, just a series of beeps weren’t tangible... Secondly, a method of cup involves a range string. Rukun Negara is often distributed by cable, but this is a greater range. It reaches concurrently in those other areas.

In the case of using the string method for the block, its distance and stiffness of the string are limited and the barriers are eliminated. It will certainly be convenient and clearer if one utilizes Sign language.

If two experts are acting like such a machine on both sides, their information between both the 2 persons is smoother and quicker as they will not have to wait for data from a third person which may create another pause at the time.

Morse code is faster, eventually. For some very small periods when sensitive information is not transmitted, strings or cup methods must be reserved.

8 0

3 years ago

What is the current in a copper wire if 650C of charge passes through it in 6 minutes

expeople1 [14]

Answer:

3. 1.8A

Explanation:

Given the following data;

Quantity of charge, Q = 650C

Time = 6 minutes to seconds = 6 * 60 = 360 seconds.

To find the current l;

Quantity of charge = current * time

Substituting into the equation, we have;

650 = current * 360

Current = 650/360

Current = 1.8 Amperes

6 0

3 years ago

You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then lo

IgorC [24]

ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

STEP-BY-STEP EXPLANATION:

F(h) is Horizontal Force = 200 N

V is Speed = 2.4 m/s

The total weight increase by 42%

coefficient of rolling friction decrease by 19%

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F(h) = F(f)

F(h) = mg* u

m is mass

g is gravitational acceleration = 9.8 m/s^2

200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u

F(h) = (0.81μ) (1.42 m g)

= (0.81) (1.42) (μ m g)

= (0.81) (1.42) (200)

= 230 N

4 0

3 years ago

An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and i

ASHA 777 [7]

Answer:

$\delta = 0.385\,m$ (Compression)

Explanation:

The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:

$\delta = \frac{P\cdot L}{A \cdot E}$

Where:

$P$ - Load experimented by the bar, measured in newtons.

$L$ - Length of the bar, measured in meters.

$A$ - Cross section area of the bar, measured in square meters.

$E$ - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.

The cross section area of the bar is now computed: ( $D_{o} = 0.04\,m$ , $D_{i} = 0.03\,m$ )

$A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})$

Where:

$D_{o}$ - Outer diameter, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}]$

$A = 5.498 \times 10^{-4}\,m^{2}$

The total contraction of the bar due to compresive load is: ( $P = -180\times 10^{3}\,N$ , $L = 0.1\,m$ , $E = 85\times 10^{9}\,Pa$ , $A = 5.498 \times 10^{-4}\,m^{2}$ ) (Note: The negative sign in the load input means the existence of compressive load)