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2 years ago

7

Use the chemical reaction to answer the question.

2 answers:

balu736 [363]2 years ago

7 0

Answer:

I believe this answer is -They maintain their properties-

Explanation:

Let's see what others say. :)

daser333 [38]2 years ago

6 0

The answer should be the first one, maintaining their properties

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3 years ago

. For transverse waves, compare the direction that the particles move to

Elza [17]

Answer:

The particles in transverse waves move perpendicular to the direction of how the wave moves.

Explanation:

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3 years ago

The expression known as Froude's number is given as Upper F equals StartFraction v squared Over gl EndFraction . This number wa

bearhunter [10]

Answer: $v=5.529\ m/s$

Explanation:

Given

number is given as $=\frac{v^2}{gl}$

and $number=2.6$

$l=1.2\ m$

g=acceleration due to gravity $(9.8\ m/s^2)$

calculating for $v$

$v^2=gl(2.6)$

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8 0

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

the temperature of a 2.0-kg increases by 5*c when 2,000 J of thermal energy are added to the block. What is the specific heat of

nata0808 [166]

To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
c = 2000/ 10
c = 200

Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1

3 0

3 years ago