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ExtremeBDS [4]
2 years ago
4

think about Newton's second law. when designing a rocket ship, what could be adjusted in the design to provide more acceleration

?​
Physics
1 answer:
pashok25 [27]2 years ago
8 0

Answer:Newton's second law is not sufficient to explain rocket's design or requirements it also involves momentum

Explanation:by solving some equations involving newton's second law and momentum we get this Ma=mv M is mass of rocket and m is mass of gases so a=mv/M so as the time passes mass of rocket decreases due to burning of fuel and acceleration a increases

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The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s−1)t]. (a) at time
Natali5045456 [20]
<span>x=((12.3/100)m)cos[(1.26s^−1)t]
 v= dx/dt = -</span><span>((12.3/100)*1.26)sin[(1.26s^−1)t]
 v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
 v=<span> <span>-0.13261622 m/s
 </span></span>the object moving at  0.13 m/s <span>at time t=0.815 s</span>
6 0
2 years ago
Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield
ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

Explanation:

Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be

Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

mP = 100 kg* 0.037/0.079  =  46.835 kg

then the yield YP of the product is

YP= mP / 100 kg =  46.835 kg / 100 kg = 46.835 %

YP= 46.835 %

the concentration of solids Cs is

Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs

Cs = 100 kg* 0.128 / 46.835 kg  = 0.2733 = 27.33%

Cs = 27.33%

3 0
2 years ago
A floating leaf oscillates up and down two complete cycles in one second as a water wave passes by. The wave's wavelength is 10
postnew [5]

Answer:

C) 20 m/s

Explanation:

Wave: A wave is a disturbance that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself. Examples of wave are, water wave, sound wave, light rays, radio waves. etc.

The velocity of a moving wave is

v = λf ............................ Equation 1

Where v = speed of the wave, λ = wave length, f = frequency of the wave.

Given: f = 2 Hz (two complete cycles in one seconds), λ = 10 meters

Substituting these values into equation 1

v = 2×10

v = 20 m/s.

Thus the speed of the wave = 20 m/s

The right option is C) 20 m/s

7 0
2 years ago
) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force
Nana76 [90]
Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others. 
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
F_x=F_3+F_2cos(45^{\circ})
F_y=F_1+F_2sin(45^{\circ})
Let's figure out each component:
F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\&#10;F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}
Total force acting on the charge would be:
F=\sqrt{F_x^2+F_y^2}
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N
F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N
Now we can find the total force acting on a single charge:
F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N
As said before, intensity of the force acting on charges is the same for all of them.

5 0
2 years ago
The engine on a fighter airplane can exert a force of 105,840 N (24,000 pounds). The take-off mass of the plane is 16,875 kg. (I
FrozenT [24]

Answer:

The acceleration you can get with that engine in your car is around 70,56 (\frac{m}{s^{2} }) or 7,26 (\frac{ft}{s^{2} } ) using 1500kg of mass or 3306 pounds

Explanation:

Using the equation of the force that is:

F=m*a

So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds

a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }

a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }

<em> Note: N or Newton units are: \frac{kg * m}{s^{2} }</em>

a= 70,54 \frac{m}{s^{2} }

Also in pounds you can compared

a= \frac{2400  lf }{3 306  lf}

Note: lf in force units are: \frac{lf*ft}{s^{2} }

a=7,26 \frac{ft}{s^{2} }

8 0
2 years ago
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