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2 years ago

10

The effects of agriculture and industry can be more harmful with the further loss of wetlands.

1 answer:

Anna71 [15]2 years ago

7 0

Yes, it is true that the effects of agriculture and industry can be more harmful with the further loss of wetlands since industry usually requires large amounts of land.

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Which diagram best represents the electric field around a negatively charged conducting sphere? (See pic)

dalvyx [7]

The answer is D !!!!!!!

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3 years ago

A 50-kg copper block initially at 140Â°c is dropped into an insulated tank that contains 90 l of water at 10Â°c. Determine the f

xxMikexx [17]

Answer:

$T_f=24.71$

Explanation:

From the question we are told that:

Mass of block $m=50$

Temperature of block $T_b =140 \textdegree C$

Volume of water $V= 90L$

Temperature of water $T_w=10 \textdegree C$

Density of water $\rho=1000kg/m^3$

Specific heat of water $C_w=4.18KJ/kg-k$

Specific heat of copper $C_p=0.96KJ/kg-k$

Generally the equation for equilibrium stage is mathematically given by

$mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)$

$50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)$

$48(140-T_f)=376.2(T_f-10)$

$140-T_f=7.8375(T_f-10)$

$140-T_f=7.8375T_f-78.375$

$-8.8375T_f=-218.375$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=24.71$

6 0

3 years ago

In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or

Margaret [11]

Answer:

123 J transfer into the gas

Explanation:

Here we know that 123 J work is done by the gas on its surrounding

So here gas is doing work against external forces

Now for cyclic process we know that

$\Delta U = 0$

so from 1st law of thermodynamics we have

$dQ = W + \Delta U$

$dQ = W$

so work done is same as the heat supplied to the system

So correct answer is

123 J transfer into the gas

8 0

3 years ago

A box is moving along the x-axis and its position varies in time according to the expression:

Colt1911 [192]

Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. $x = 6 * 3.2^2 = 61.44 m$

b) at t = 3.2 + Δt. $x = 6*(3.2 + \Delta t)^2$

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

$v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}$

$v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}$

$v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}$

$v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}$

$v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}$

$v = 38.4 + \Delta t$

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

3 0

3 years ago

Which property is closely related to its temperature

Lilit [14]

Answer:

Mass – The single most important property that determines other properties of the star. Luminosity – The total amount of energy (light) that a star emits into space. Temperature – surface temperature, closely related to the luminosity and color of the star

Explanation:

8 0

3 years ago

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