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Rom4ik [11]
3 years ago
14

A 5.00X10^5 kg rocket is accelerating straight up. Its engines produce 1.250X10^7 N of thrust, and air resistance is 4.50X10^6 N

. What is the rockets acceleration?
Physics
1 answer:
katrin [286]3 years ago
8 0
<span>6.20 m/s^2 The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be 9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N Add in the atmospheric drag and you get 4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N Now subtract that total drag from the thrust available. 1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So 3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2 Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
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2 years ago
A small airplane with a wingspan of 18.0 m is flying due north at a speed of 63.6 m/s over a region where the vertical component
choli [55]

Answer:

(a) ε = 1373.8.

(b) The wingtip which is at higher potential.

Explanation:

(a) Finding the potential difference between the airplane wingtips.

Given the parameters

wingspan of the plane is = 18.0m

speed of the plane in north direction is = 70.0m/s

magnetic field of the earth is = 1.20μT

The potential difference is given as:

ε = Blv

where ε = potential difference of wingtips

B = magnetic field of earth

l = wingspan of airplane

v = speed of airplane

ε = 1.2 x 18.0 x 63.6

ε  = 1373.8

(b) Which wingtip is at  higher potential?

The wingtip which is at higher potential.

5 0
3 years ago
Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.8 s after the eleva
Delicious77 [7]

Answer:

v= 4.0 m/s

Explanation:

  • When standing on the bathroom scale within the moving elevator, there are two forces acting on Henry's mass: Normal force and gravity.
  • Gravity is always downward, and normal force is perpendicular to the surface on which the mass is located (the bathroom scale), in upward direction.
  • Normal force, can adopt any value needed to match the acceleration of the mass, according to Newton's 2nd Law.
  • Gravity (which we call weight near the Earth's surface) can be  calculated as follows:

       F_{g} = m*g = 95 kg * 9.8 m/s2 = 930 N (1)

  • According to Newton's 2nd Law, it must be met the following condition:

       F_{net} = F_{g} -F_{n} = m*a\\  F_{net} = 930 N - 830 N = 100 N = 95 Kg * a

  • As the gravity is larger than normal force, this means that the acceleration is downward, so, we choose this direction as the positive.
  • Solving for a, we get:

       a =\frac{F_{net} }{m} =\frac{100 N}{95 kg} =  1.05 m/s2

  • We can find the speed after the first 3.8 s (assuming a is constant), applying the definition of acceleration as the rate of change of velocity:

        v_{f} = a* t = 1.05 m/s * 3.8 m/s = 4.0 m/s

  • Now, if during the next 3.8 s, normal force is 930 N (same as the weight), this means that both forces are equal each other, so net force is 0.
  • According to Newton's 2nd Law, if net force is 0, the object  is either or at rest, or moving at a constant speed.
  • As the elevator  was moving, the only choice is that it is moving at  a constant speed, the same that it had when the scale was read for the first time, i.e., 4 m/s downward.
3 0
3 years ago
Question 7(Multiple Choice Worth 4 points)
Nimfa-mama [501]
C is the correct answer, as the equation for photosynthesis is carbon dioxide +water=glucose(sugar) + oxygen
3 0
3 years ago
Read 2 more answers
ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit
horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

F = \dfrac{d^4G\delta}{8D^3N}

\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm

F = \dfrac{d^4G}{8D^3}\times 0.004

F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004

     F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

\tau = \dfrac{8FDk_s}{\pi d^3}

\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}

              = 442.6 Mpa

The tensile strength of the steel material of  ASTM A229 is equal to 1300 Mpa

now,

\tau_s \leq 0.45 S_u

\tau_s \leq 0.45 \times 1300

\tau_s \leq 585\ Mpa

since corrected stress is less than the \tau_s

hence, spring will return to its original shape.

6 0
3 years ago
Read 2 more answers
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