Ask question

Ask question

All categories

2 years ago

5

"The energy conversions that take place from the time the plunger is pushed down to the time the explosives detonate are from...

"

1 answer:

MaRussiya [10]2 years ago

6 0

Answer:

mechanical to electrical to chemical

Explanation:

You might be interested in

A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positi

asambeis [7]

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

6 0

2 years ago

The ammonia molecule (NH3) has a dipole moment of 5.0×10?30C?m. Ammonia molecules in the gas phase are placed in a uniform elect

Neko [114]

Question (continuation)

(a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E S from parallel to perpendicular?

(b) At what absolute temperature T is the average translational kinetic energy 3/2kT of a molecule equal to the change in potential energy calculated in part (a)?

Answer:

a. 9.0 * 10^-24 Joules

b. 0.44K

Explanation:

Given

Let p = dipole moment = 5.0 * 10^-30 Cm

Let E = Magnitude = 1.8 * 10^6 N/m

a.

The charge in electric potential = Final Charge - Initial Charge

Initial Charge = Potential Energy

Initial Energy = -pE cosФ where Ф = 0

So, initial Energy = - 5.0 * 10^-30 * 1.8 * 10^6

Initial Energy = -9 * 10^-24 Joules

Final Energy = 0

Charge = 0 - (-9.0 * 10^-24)

Charge = 9.0 * 10^-24 Joules

b.

Absolute Temperature

Change in Kinetic Energy = Change in Potential Energy = 9.0 * 10^-24

Change in Kinetic Energy = 3/2kT where k is Steven-Boltzmann constant = 1.38 * 10^-23

So,

9.0 * 10^-24 = 3/2 * 1.38 * 10^-23 * T

T = (9.0 * 10^-24 * 2)/(3 * 1.38 * 10^-23)

T = (18 * 10^-24)/(4.14 * 10^-23)

T = 0.44K

6 0

3 years ago

Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s

anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of $-5.2^{\circ}$

D) The impulse exerted on C is 4.29 kg m/s at a direction of $-5.2^{\circ}$

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

$p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B$

$m_A = 0.600 kg$ is the mass of sphere A

$u_A = 4.00 m/s$ is the initial velocity of the sphere A (taking the right as positive direction)

$v_A$ is the final velocity of sphere A

$m_B = 1.80 kg$ is the mass of sphere B

$u_B = 2.00 m/s$ is the initial velocity of the sphere B

$v_B = 3.00 m/s$ is the final velocity of the sphere B

Solving for vA:

$v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s$

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

$K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J$

After the collision:

$K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J$

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

$p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s$

While the total momentum along the y-axis is zero:

$p_y = 0$

We can now write the equations of conservation of momentum along the two directions as follows:

$p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy}$ (1)

We also know the components of the momentum of B after the collision:

$p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s$

So substituting into (1), we find the components of the momentum of C after the collision:

$p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s$

So the magnitude of the momentum of C is

$p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s$

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

$v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s$

And the direction is

$\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}$

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

$p_C = 0$

While the final momentum is:

$p'_C = 4.29 kg m/s$

So the magnitude of the impulse exerted on C is

$I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s$

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore $-5.2^{\circ}$ .

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

$K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J$

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

$K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J$

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

$F=Ma$

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

#LearnwithBrainly

8 0

3 years ago

A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate

yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

$W=Pm\DeltaV$

$W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}$

$W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})$

Where,R = gas constant

T = temperature

P = pressure

$P_{atm}$ =Atmosphere pressure

m = mass

Put the value into the formula

$W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)$

$W=213\ kJ$

Hence, The work done by the steam is 213 kJ.

6 0

3 years ago

What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?

kvv77 [185]

Answer:

$W = 78.48N\\W =0.0784kN\\W = 8kgf\\$

$W= 3.6lbf\\W= 115.2 lbm*ft/s2$

Explanation:

if

$m=8kg=3.6lb\\$

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

$W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\$

$W= 3.6lbf\\W= 3.6lbm *32.16 ft/s2 =115.2 lbm*ft/s2$

5 0

3 years ago