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3 years ago

What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol

Physics

1 answer:

san4es73 [151]3 years ago

3 0

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

$Q=\frac{\Delta H}{n}$

where,

Q = energy absorbed = 27.2 kJ

$\Delta H$ = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

Now put all the given values in the above expression, we get:

$27.2kJ=\frac{6.01kJ/mol}{n}$

$n=0.221mol$

Now we have to calculate the mass of ice.

$\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}$

Molar mass of ice = 18.02 g/mol

$\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g$

Thus, the mass of ice melted can be, 3.98 grams.

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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

2 years ago

The _____ deflects these winds to the right in the northern hemisphere and to the Southern Hemisphere

almond37 [142]

Answer:

Coriolis Effect

Explanation:

The Coriolis effect is responsible for the deflection of winds to the right in the Northern hemisphere and to the right in the Southern hemisphere. It is an effect that occurs because of the rotation of the earth around its axis.

The implication of this is that in areas of low pressure in the Northern hemisphere, winds tend to blow in anticlockwise direction, and in areas of high pressure, it blows in a clockwise direction. The opposite of this happens in the Southern hemisphere.

6 0

3 years ago

Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the

Svet_ta [14]

Answer:

b) $x_{cm}$ = 4.88 cm , c) $x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃) and d)

$x_{cm}$ ’= 1.88 cm

Explanation:

The definition of mass center is

$x_{cm}$ = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

D₂ = d₁ + d₂

D₂ = 1.1 + 1.9

D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

D₃ = D₂ + d₂

D₃ = 3.0 + 3.9

D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

M = m₁ + m₂ + m₃

M = 24 + 12 + 56

M = 92 g

$x_{cm}$ = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

$x_{cm}$ = 1/92 (448.8)

$x_{cm}$ = 4,878 cm

$x_{cm}$ = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

$x_{cm}$ ’= 1/M (m₁ d₁ + m₂ d₂ + m₃ d₃)

$x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃)