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AnnZ [28]
3 years ago
13

A rocket takes off from Earth and reaches a speed of 105 m/s in 18.0 s. If the exhaust speed is 1,200 m/sand the mass of fuel bu

rned is 110 kg, what was the initial mass (in kg, including the initial fuel) of the rocket?
Physics
1 answer:
Amanda [17]3 years ago
7 0

Answer:

The initial mass of the rocket is 526.2 kg.

Explanation:

Given that,

Final speed = 105 m/s

Time = 18.0 s

Exhaust initial speed = 1200 m/s

Mass of burned fuel = 110 kg

We need to calculate the initial mass

The velocity change of rocket under gravity is defined as,

v=u\ ln(\dfrac{m_{i}}{m})-gt....(I)

We know that,

m=m_{i}-m_{bf}

Put the value of m in equation (I)

v=u\ ln(\dfrac{m_{i}}{m_{i}-m_{bf}})-gt

m_{i}=\dfrac{m_{bf}}{1-e^-{\dfrac{v+gt}{u}}}

m_{i}=\dfrac{110}{1- e^-{\frac{105+9.8\times18}{1200}}}

m_{i}=526.2 kg

Hence, The initial mass of the rocket is 526.2 kg.

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AnnZ [28]
Liquid water because if it said very high then it would be water vapor but it didn’t say that so the answer is B liquid water
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3 years ago
What happens to gravity when someone jumps up?
Alla [95]

Answer:

The direct answer to the question as written is as follows: nothing happens to gravity when someone jumps up - gravity continues exerting a force on the body of that particular someone proportional to (mass of someone) x (mass of Earth) / (distance squared). What you might be asking, however, is what is the net force acting on the body of someone jumping up. At the moment of  someone jumping up there is an upward acceleration, i.e., an upward-directed force which counteracts the gravitational force - this is the net force ( a result of the jump force minus gravity). From that moment on, only gravity acts on the body. The someone moves upward gradually decelerating to the downward gravitational acceleration until they reaches the peak of the jump (zero velocity). Then, back to Earth.


5 0
3 years ago
Kim is ice-skating going 4.6 m/s. What is her velocity after 10 seconds ?
MArishka [77]

This is a uniform rectilinear motion (MRU) exercise.

To start solving this exercise, we obtain the following data:

<h3><u>Data:</u></h3>
  • v = 4.6 m/s
  • d = ¿?
  • t = 10 sec

To calculate distance, speed is multiplied by time.

We apply the following formula: d = v * t.

We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:

\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }

\bf{d=46 \ m}

Therefore, the speed at 10 seconds is 46 meters.

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6 0
2 years ago
In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the f
zhannawk [14.2K]

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m

8 0
3 years ago
A 174 pound Jimmy Cheek is riding on a 54 ft diameter Ferris Wheel. The normal force on Jimmy Cheek is 146 pounds when Jimmy is
Veronika [31]

To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.

I will also attach a free body diagram that allows a better understanding of the problem.

For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore

F_c = W-N

m\omega^2r = W-N

Here,

m = Net mass

\omega= Angular velocity

r = Radius

W = Weight

N = Normal Force

m\omega^2r = 174-146

The net mass is equivalent to

F = mg \rightarrow m = \frac{F}{g}

Then,

m = \frac{174lb}{32.17ft/s^2}

Replacing we have then,

(\frac{174lb}{32.17ft/s^2})\omega^2 (54ft) =174lb-146lb

Solving to find the angular velocity we have,

\omega = 0.309rad/s

Therefore the angular velocity is 0.309rad/s

6 0
3 years ago
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