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2 years ago

I need help with the blanks

Physics

1 answer:

Alexeev081 [22]2 years ago

6 0

The top
2- motion,acceleration, force
4- pulls,Earth
5- forces
6- objects
7- matter
8- stops, potential

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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

san4es73 [151]

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.

3 0

3 years ago

Help me. It's question 2 I need the answer for

Alex73 [517]

Ok, let me see if I can help

Sound is caused by vibrations. These can pass through a solid, liquid, and gas. But not through vacuum because there are no particles

6 0

2 years ago

Given uranium 235 go through an alpha, beta, beta, alpha, gamma, neutron and alpha. What do you have now?

Licemer1 [7]

Answer:

b. 88, 222

Explanation:

235U₉₂ ----→ Alpha --------→ 231P₉₀ ----→- beta -----→ 231Q₉₁ ------→-beta -------→231R₉₂--------→-alpha ------→-227S₉₀ ------→ gamma -----→-227S₉₀ ----------→ neutron ------→-226T₉₀-----------→ alpha --------→222 X ₈₈

Atomic No is 88 , atomic weight = 222 .

3 0

2 years ago

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h

vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

$R_A=2R_B$

Also Potential of both sphere is same

$V_A=V_B$

$V=\frac{kQ}{R}$

thus

$k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}$

$\frac{Q_A}{Q_B}=\frac{R_A}{R_B}$

$\frac{Q_A}{Q_B}=\frac{2}{1}=2$

$\frac{Q_B}{Q_A}=\frac{1}{2}$

(b)Ratio of $\frac{E_B}{E_A}$

Electric Field is given by $E=\frac{kQ}{R^2}$

thus $E_A=\frac{kQ_A}{R_A^2}$ ----1

$E_B=\frac{kQ_B}{R_B^2}$ ----2

Divide 2 by 1

$\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}$

$\frac{E_B}{E_A}=\frac{1}{2}\times 4=2$

8 0

3 years ago

A barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a va

Alisiya [41]

Answer:

(a). 14.4 lbf/in^2.

(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

Explanation:

So, from the question above we are given the following parameters which are going to help us in solving this particular Question;

=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"

=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."

With these knowledge, let us delve right into the solution;

(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].

The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.

(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.

Length of mercury = 27.8 in.

AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

7 0

2 years ago