1. Two forces act on a box as follows: F= 100 N at 02 = 170. and F2= 75 N at

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 128 ∘C. The gas expands and, in the process, a

o-na [289]

Answer:

The final temperature of the gas is <em>114.53°C</em>.

Explanation:

Firstly, we calculate the change in internal energy, ΔU from the first law of thermodynamics:

ΔU=Q - W

ΔU = 1180 J - 2020 J = -840 J

Secondly, from the ideal gas law, we calculate the final temperature of the gas, using the change in internal energy:

$ΔU=\frac{3}{2} nRΔT$

$ΔU=\frac{3}{2} nR(T_{2} -T_{1} )$

Then we make the final temperature, T₂, subject of the formula:

$T_{2} =\frac{2ΔU}{3nR} +T_{1}$

$T_{2} =\frac{2(-840J)}{(3)(5)(8.314J/mol.K)} +128 deg.C$

$T_{2} =114.53 deg.C$

Therefore the final temperature of the gas, T₂, is 114.53°C.

7 0

3 years ago

MARKING BRAINLIST | Which situation below would have the STRONGEST gravitational force between them?

maks197457 [2]

Case d) has the strongest gravitational force

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

a) For this pair of objects:

m1 = 10 kg

m2 = 2 kg

r = 30 km = 30,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(2)}{30000^2}=1.48\cdot 10^{-18}N$

b) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 30 km = 30,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(10)}{30000^2}=7.41\cdot 10^{-18}N$

c) For this pair of objects:

m1 = 2 kg

m2 = 2 kg

r = 10 km = 10,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(2)(2)}{10000^2}=1.33\cdot 10^{-17}N$

d) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 10 km = 10,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(10)}{10000^2}=6.67\cdot 10^{-17}N$

Therefore, the strongest gravitational force is in case d).

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago

For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ

slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

= 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

$180 \frac{mg}{kg} \times 71.215 kg$

= 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

0.65% (w/w) = $\frac{0.65 g}{100 g}$

= $\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}$

= $\frac{0.65 g}{100 ml}$

Therefore, calculate the volume which contains the amount of caffeine as follows.

12818.7 mg = 12.8187 g = $\frac{12.8187 g}{\frac{0.65 g}{100 ml}}$

= 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0

3 years ago

A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn

vesna_86 [32]

Answer:

(a) Magnitude of static friction force is 109 N

(b) Minimum possible value of static friction is 0.356

Solution:

As per the question;

Horizontal force exerted by the girl, F = 109 N

Mass of the crate, m = 31.2 kg

Now,

(a) To calculate the magnitude of static friction force:

Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:

F = f = 109 N

(b) The maximum possible force of friction between the floor and the crate is given by:

$f_{m} = \mu_{s}N$

where

N = Normal reaction = mg

Thus

$f_{m} = \mu_{s}mg$

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.

$f\leq f_{m}$

$f \leq \mu_{s}mg$

$109 \leq \mu_{s}\times 31.2\times 9.8$

$\mu_{s}\geq 0.356$

7 0

3 years ago

Two metal spheres are hanging from nylon threads. When you bring the spheres close to each other, they tend to attract. Based on

Tamiku [17]

Explanation:

In the given question, the two metal spheres were hanged with the nylon thread.

When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.

The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.

During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.

It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.

4 0

3 years ago