If magnesium has 13 neutrons, what is its mass for the isotope

To sterilize a 48.0-g glass baby bottle, we must raise its temperature from 16.0°C to 94.0°C. How much heat transfer is required

FinnZ [79.3K]

Answer:

3144.96 J

Explanation:

Heat transfer is,

$q=mc\Delta t$

As we know that the specific heat of the glass is,

$c=0.840J/g^{\circ}C$

Given that the mass of bottle is 48 g.

And the initial temperature is 16°C and the final temperature is 94°C.

Therefore,

$q=48\times 0.840(94^{\circ}-16^{\circ})\\q=48\times 0.840\times 78^{\circ}\\q=3144.96 J$

7 0

4 years ago

When temperature decreases but actual water vapor content remains the same, what happens to relative humidity?

ruslelena [56]

Answer:

d. Relative humidity increases.

Explanation:

The expression of relative humidity in terms of absolute humidity, absolute pressure and saturation pression at measured temperature is:

$\phi = \frac{\omega \cdot P}{(0.622+\omega)\cdot P_{sat}}$

When temperature decreases, the saturation pressure decreases also and, consequently, relative humidity increases. Therefore, the right answer is option D.

8 0

3 years ago

Read 2 more answers

All of this sun and heat is making you thirsty You decide to walk down the street to the food stand to buy some lemonade As you

raketka [301]

Answer:

I would say erosion

5 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$