11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

A spacecraft has a momentum of 20,000 kg-m/s, and a mass of 250 kg. What is the magnitude of its velocity?

Harlamova29_29 [7]

Answer:

Explanation:

The velocity of the spacecraft can be found by using the formula

$v = \frac{p}{m} \\$

p is the momentum

m is the mass

From the question we have

$v = \frac{20000}{250} = \frac{2000}{25} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 15

Sveta_85 [38]

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

$V_{A}$ = 1 m³

$T_{A}$ = 10°C = 283 K

$P_{A}$ = 350 kPa

$m_{B}$ = 3 kg

$T_{B}$ = 35°C = 308 K

$P_{B}$ = 150 kPa

Now, lets apply the ideal gas equation;

$P_{B}$ $V_{B}$ = $m_{B}$ R $T_{B}$

$V_{B}$ = $m_{B}$ R $T_{B}$ / $P_{B}$

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

$V_{B}$ = ( 3 × 0.287 × 308) / 150

$V_{B}$ = 265.188 / 150

$V_{B}$ = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, $m_{A}$ = $P_{A}$ $V_{A}$ / R $T_{A}$ = (350 × 1)/(0.287 × 283) = 350 / 81.221

$m_{A}$ = 4.309 kg

Total mass, $m_{f}$ = $m_{A}$ + $m_{B}$ = 4.309 + 3 = 7.309 kg

Total volume $V_{f}$ = $V_{A}$ + $V_{B}$ = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

$P_{f}$ = $m_{f}$ R $T_{f}$ / $V_{f}$

given that; final temperature $T_{f}$ = 20°C = 293 K

we substitute

$P_{f}$ = ( 7.309 × 0.287 × 293) / 2.77

$P_{f}$ = 614.6211119 / 2.77

$P_{f}$ = 221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

6 0

3 years ago

1. Consider a 1000 kg car rounding a curve on a flat road of radius 50 m at a speed of

raketka [301]

Answer:

The car will make the turn perfectly

Explanation:

Given that the centripetal force= mv^2/r

M= mass of the car

v = speed of the car

r= radius

Hence;

F = 1000 × (14)^2/50

F= 3920 N

The frictional force = μmg

μ = coefficient of static friction

m= mass

g = acceleration due to gravity

Frictional force= 0.6 × 1000× 10

Frictional force = 6000 N

The car will not skid off the curve because the frictional force is greater than the centripetal force.

6 0

3 years ago

The same strength force was exerted in the same direction on both Object A and Object B. Why did Object A go faster than Object

VashaNatasha [74]

Answer:

$m_B > m_A$

the mass of body B must be greater than the mass of body A

Explanation:

Newton's second law establishes a linear relationship between the force, the mass of the body and its acceleration

F = m a

a = F / m

Let's analyze this expression tells us that the force is of equal magnitude for the two bodies, but body A goes faster than body B, this implies that it has more relationships

a_A > a_B

$\frac{F}{m_A} > \frac{F}{m_B}$

$m_B > m_A$

Therefore, for this to happen, the mass of body B must be greater than the mass of body A

6 0

3 years ago

The second Law of Thermodynamics states that: A. spontaneous processes are characterized by the overall conversion of order to d

Georgia [21]

Answer:

Spontaneous processes are characterized by the overall conversion of order to disorder.

Explanation:

The second law of thermodynamics states that: A spontaneous process occurs only if there is an increase in entropy of a system and its surroundings.

Entropy, S, is a measure of the randomness or disorder of a system. It is measured in J/Kmol.

The change in entropy, ∆S = ∆H/T

Where ∆H = change in enthalpy, T = Temperature in Kelvin.

For,

I. An endothermic reaction, ∆S = positive (that is, ∆S is greater than zero), there is an increase in entropy, therefore, the reaction is spontaneous.

II. An exothermic reaction, ∆S = negative (that is, ∆S is less than zero) there is a decrease in entropy, so, the reaction is non-spontaneous.

III. A system at equilibrium, ∆S = 0.

Then,

The standard change in entropy of a reaction, ∆So reaction , is the difference in the standard entropies between products and reactants:

∆So reaction = n ∆Soproducts - m ∆Soreactants

Where, = sigma = sum of,

∆ = delta = change in,

n and m = stoichiometric coefficients of the products and reactants respectively.

Furthermore, the entropy of the system and surroundings is referred to as the entropy of the universe.

∆Suniverse = ∆Ssurroundings + ∆Ssystem.

Processes leading to an increase in entropy include melting, heating, vaporization, dissolving.

8 0

3 years ago