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OverLord2011 [107]

3 years ago

8

Which element is this?

2 answers:

hammer [34]3 years ago

7 0

Answer:Silicon

Explanation:

You add up the protons and neutrons and then look at the element paper

Alexxx [7]3 years ago

6 0

I’m pretty sure it’s silicone

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A water heater warms 35 l of water from a temperature of 22.7 c to a temperature of 83.7

fgiga [73]

The amount of energy needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
$C_s$ is its specific heat capacity
$\Delta T$ is the increase in temperature

The water volume is $V=35 L= 35 dm^3 = 0.035 m^3$ , since its density is $d=1000 kg/m^3$ , the mass of this sample of water is
$m=dV=(1000 kg/m^3)(0.035 m^3)=35 kg$

The water specific heat capacity is $C_s = 4.18 kJ/kg ^{\circ}C$

and the increase in temperature is $\Delta T=83.7 ^{\circ}C-22.7 ^{\circ}C=61^{\circ}C$

Therefore, the amount of energy needed is
$Q=mC_s \Delta T=(35 kg)(4.18 kJ/kg ^{\circ}C)(61^{\circ}C)=8924 kJ = 8.92 \cdot 10^6 J$

8 0

4 years ago

Which of the following are homogeneous solutions?

Sindrei [870]

Salt water
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Vinegar

3 0

4 years ago

A prediction or educated guess that can be tested is ?

Ksivusya [100]

An educated guess is a “hypothesis”

7 0

3 years ago

Which statements best describe the process of radioactive decay? Check all that apply.

Ksenya-84 [330]

An unstable isotope changes until it reaches a different element that is stable.

An unstable isotope changes until it reaches a different isotope of the same element that is stable.

i just did this on ed the other answer was wrong. these ones are right tho

6 0

3 years ago

Read 2 more answers

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

mina [271]

Answer:

(a) v = 463.97 m/s, $a_{c} = 0.0337\ m/s^{2}$

(b) v' = 196.02 m/s, $a'_{c} = 0.0143\ m/s^{2}$

Solution:

As per the question:

Time period of the rotation of earth, T = 24 h = $24\times 3600 = 86400\ s$

Radius of the earth, R = $6.38\times 10^{6}\ m$

Angle, $\theta = 65.0^{\circ}$

Now,

Angular velocity is given by:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{86400} = 7.27\times 10^{- 5} \rad$

(a) To calculate the speed and the centripetal acceleration at the equator:

Linear velocity or the speed, v = $R\omega$

$v = 6.38\times 10^{6}\times 7.27\times 10^{- 5} = 463.967\ m/s$

Centripetal Acceleration, $a_{c} = \frac{v^{2}}{R}$

$a_{c} = \frac{463.967^{2}}{6.38\times 10^{6}} = 0.0337\ m/s^{2}$

(b) To calculate the speed and acceleration at an altitude of $65.0^{\circ}$ N:

The horizontal component of the radius, R' = $Rcos\theta$

$R' = 6.38\times 10^{6}cos65.0^{\circ} = 2.69\times 10^{6}\ m$

Now,

For the speed, v' = $R'\omega = 2.69\times 10^{6}\times 7.27\times 10^{- 5} = 196.02\ m/s$

For the centripetal acceleration,

$a'_{c} = \frac{v'^{2}}{R'}$

$a'_{c} = \frac{196.02^{2}}{2.69\times 10^{6}} = 0.01428\ m/s^{2}$

8 0

4 years ago