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WITCHER [35]
2 years ago
5

A rocket in deep space has an exhaust-gas speed of 2000 m/s. When the rocket is fully loaded, the mass of the fuel is five times

the mass of the empty rocket. Part A What is the rocket's speed when half the fuel has been burned?
Physics
1 answer:
notka56 [123]2 years ago
3 0

Answer:

 v_{f} = 1,386 m / s

Explanation:

Rocket propulsion is a moment process that described by the expression

       v_{f} - v₀ =  v_{r} ln (M₀ / Mf)

Where v are the velocities, final, initial and relative and M the masses

The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)

We consider that the rocket starts from rest (v₀ = 0)

At the time of burning half of the fuel the mass ratio is that the current mass is    

       M = 2.5 Mf

       v_{f} - 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2

       v_{f} = 1,386 m / s

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Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many
Vlada [557]

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

8 0
2 years ago
If you add 700 kJ of heat to 700 g of water at 70 degrees C, how much water is left in the container? The latent heat of vaporiz
makkiz [27]

Answer:A

Explanation:Find attached picture file for details

3 0
2 years ago
1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra
UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.  

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

F_r = mgsin\theta

F_r = \mu N

F_r = \mu mg cos\theta

Matching the two equation we have,

\mu N = \mu mg cos\theta

\mu = tan\theta

Applying energy conservation,

\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1

h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})

Frictionless surface

\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2

\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2

h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between h_1 and h_2 is given by

\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}

\frac{h_1}{h_2} = \frac{1}{2}

8 0
2 years ago
The length of the side of a cube having a density of 12.6 g/ml and a mass of 7.65 g is __________ cm.
qaws [65]

Density is the characteristic property of a substance.  It is the measure of mass of the substance divided by its volume (density= mass/volume). Manipulate the given formula to come up with the formula for the volume. Therefore, volume is equals to mass of a substance divided by its density (Vol= mass/density). Given 12.6 g/ml as density and 7.65 g mass, volume is equals to 0.60714 ml, since 1 ml = 1cm^3, volume is equals to 0.60714 cm^3 then extract the cube root of the volume to get the length of the cube in cm which is equal to 0.84677 cm.




4 0
2 years ago
In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce
Mandarinka [93]

Answer:

0.22 m

Explanation:

We are told that the driver can survive an acceleration of 50g only if the collision lasts no longer than 30 ms. So,

t = 30 ms = 0.030 s

The acceleration is

a=-50g = -50(9.8)=-490 m/s^2

where the negative sign is due to the fact that this is a deceleration, since the driver comes to a stop in the collision.

First of all, we can find what the initial velocity of the car should be in this conditions by using the equation:

v=u+at

And since the final velocity is zero, v=0, and solving for u,

u=-at=-490(0.030)=14.7 m/s

And now we can find the corresponding distance travelled using the equation:

d=ut+\frac{1}{2}at^2 = (14.7)(0.030)+\frac{1}{2}(-490)(0.030)^2=0.22 m

8 0
2 years ago
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