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3 years ago

Which of the following equations describe the 2nd law of motion?

Physics

2 answers:

Paladinen [302]3 years ago

7 0

Answer:

b. force= mass *acceleration

Explanation:

Sever21 [200]3 years ago

4 0

Answer:

b. F = m × a is the correct answer.

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What mass of silver (in grams) is solidified when 749 joules of heat are released by a sample of molten silver at its freezing p

Anastasy [175]

heat released Q = 749 joules

heat of fusion of silver L = 109 J/g

Here phase of silver is changing from liquid to solid

so temperature will remain same

all heat will be released due to its phase change

and in this case we use Q=mL

where m is the mass of silver in gram

Q= mL

749 = m * 109

m = 749/109

m = 6.87 gram

4 0

3 years ago

Read 2 more answers

A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 35 2. How

allsm [11]

Answer:

Explanation:

6/35=0.17

8 0

3 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution decrease and the vapor pressure of the solution decrease .

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

3 years ago

What is the resistance of a 3.5 m copper wire (Rho= 1.7x10-8 Ohm·m) that 1 point

VikaD [51]

Answer:

(D)

Explanation:

Given :

l=3.5 m

$A=5.26*10^{-6}$ $m^{2}$

$p=1.7*10^{-8} ohm.m$

Resistance can be calculated as :

$R=p\frac{l}{A} \\R=1.7*10^{-8} \frac{3.5}{5.26*10^{-6} }$

$R=\frac{5.95*10^{-2} }{5.26} \\R=1.13*10^{-2}$

Resistance of the wire will be 1.1× $10^{-2}$ ohms

Option D is correct

4 0

2 years ago

A frictionless piston–cylinder device contains 7 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin

lora16 [44]

Answer: -1038.8 kJ

Explanation:

From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.

from the question we were given

Mass of nitrogen, m = 7kg

initial temperature, T1 = 250k

Final temperature, T2 = 450k

Polytropic index, n = 1.4

Specific gas constant, R = 0.2968kJ/kgK

W = [p2 * v2 - p1 * v1] / 1 - n

W = [m * R * T2 - T1] / 1 - n

W = 7*0.2968*(450 - 250)] / 1 - 1.4

W = [7*0.2968*200] / -0.4

W = 415.52 / -0.4

W = -1038.8 kJ

5 0

3 years ago