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3 years ago

Money is not a medium of exchange for acquiring goods and services. a. True b. False

Physics

2 answers:

bonufazy [111]3 years ago

7 0

False, money is a medium of exchange to acquire goods and services.

Depending on what you want to cancel, it is advisable to pay with cash or bank transfer.

Also, it is important to know that each country has its own exchange currency.

In the United States, the currency is the dollar.

In Europe, the currency is the euro.

Answer:

b. False

Ilya [14]3 years ago

6 0

False. Battering is another way to get goods and services for other goods and services.

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(30 points) Air at 500 kPa and 400 K enters an adiabatic nozzle at a velocity of 30 m/s and leaves at 300 kPa and 350 K. Using v

Sophie [7]

Answer:

Explanation:

Check attachment for solution

8 0

3 years ago

A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the

ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

3 0

3 years ago

if a car engine does 600,000 J of work over a 500m distance and the mass of the car is 250Kg then what is the final velocity of

drek231 [11]

Given that

Work = 600,000 J ,

distance(S) = 500 m ,

mass (m) = 250 Kg ,

Determine the velocity of car (v) = ?

We know that,

Work = Force × distance

=> Force = Work ÷ distance

= 600,000 ÷ 500

= 500 N .

Also Force F = m.a ; from Newtons II law

500 = 250 × a

a = 2 m/s.

<em>Final Velocity from the given formula </em>

V² = u² + 2.a.s

= 0 + 2 × 2 × 500

= \sqrt{2000}

8 0

3 years ago

At which temperature do the lattice and conduction electron contributions to the specific heat of Copper become equal.

sp2606 [1]

Answer:

At 3.86K

Explanation:

The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:

gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2

Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .

We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =

√0.7/ 0.0469 = 3.86K.

4 0

3 years ago

In regard to the Compton scattering experiment with x-rays incident upon a carbon block, as the scattering angle becomes larger,

BartSMP [9]

Answer:

Increases

Explanation:

In the Compton scattering experiment with x-rays,

The change in operation

$\Delta \lambda = \frac{h}{m_oc} [1-cos\theta]$

Now rest being constant, as \theta increases, cos\theta decreases

Hence, The change in wavelength will increase with the increase in \theta.

Hence, wavelength increases with an increase in the angle of scatttering.

6 0

3 years ago