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3 years ago

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A student measured the mass of ice in a glass container with a tight lid. He allowed the ice to melt, and then found the mass of

the container, its lid, and its contents again. What conclusion could he draw based on the masses he measured?
No matter how tight the lid is, some mass is lost when a solid melts.

The liquid formed from a melted solid has less mass than the solid has.

The liquid formed from a melted solid has the same mass as the solid has.

The liquid formed from a melted solid has a greater mass than the solid has.

1 answer:

vfiekz [6]3 years ago

7 0

Answer:

The liquid formed from a melted solid has the same mass as the solid has.

Explanation:

As long as no water can escape, the mass of the ice before melting must equal the mass of the liquid water after.

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An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical

Ostrovityanka [42]

Answer:

The change in gravitational potential energy of the climber-Earth system is $\Delta PE = 396900 \ J$

Explanation:

From the question we are told that

The mass of the hiker is $m = 75 \ kg$

The time taken is $T = 2 \ hr = 2 * 3600 = 7200 \ s$

The vertical elevation after time T is $H = 540 \ m$

The change in gravitational potential is mathematically represented as

$\Delta PE = mgH$

here g is the acceleration due to gravity with value $g = 9.8 \ m/s^2$

substituting values

$\Delta PE = 75 * 9.8 * 540$

$\Delta PE = 396900 \ J$

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3 years ago

An object located near the surface of Earth has a weight of a 245 N

gayaneshka [121]

Answer:

ثر أنواع التربة خصوبة التربحمراء .

ج- السوداء

Explanation:

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7 0

3 years ago

List the location in order from lear to greatest pressure

kramer

Lear vv
F
B
C
D
A
E
G
greatest pressure ^^

I’m really sorry if I’m wrong

7 0

3 years ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

3 years ago

A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the

harina [27]

Answer:

$R=2.78\ \Omega$

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

$P=I^2R$

Put all the values to find R.

$R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega$

So, the resistance is equal to $2.78\ \Omega$ .

7 0

3 years ago