All categories

3 years ago

PART ONE

Physics

1 answer:

GenaCL600 [577]3 years ago

7 0

Answer:

1. F = 45,458.17 N

2. P = 12,800,000 W

Explanation:

Part 1. The thrust force is the sum of the forces on the air and on the fuel.

For the air, 107 kg of air is accelerated from 281 m/s to 679 m/s in 1 second.

F = ma

F = (107 kg) (679 m/s − 281 m/s) / (1 s)

F = 42,586 N

For the fuel, 4.23 kg of fuel is accelerated from 0 m/s to 679 m/s in 1 second.

F = ma

F = (4.23 kg) (679 m/s − 0 m/s) / (1 s)

F = 2,872.17 N

So the thrust on the jet is:

F = 42,586 N + 2,872.17 N

F = 45,458.17 N

Rounded to three significant figures, the force is 45,500 N.

Part 2. Power = work / time, and work = force × distance, so:

Power = force × distance / time

Power = force × velocity

P = (45,458.17 N) (281 m/s)

P = 12,773,745.77 W

Rounded to three significant figures, the power is 12,800,000 W.

You might be interested in

Standing still at the top of the hill at point a,what kind of energy( potential and/or kinetic) would the snowboarder have ? How

Alja [10]

He would have potential, he hasn’t gone down the mountain yet so he remains potential

5 0

3 years ago

The only force acting on a 1.9 kg canister that is moving in an xy plane has a magnitude of 3.9 N. The canister initially has a

Hunter-Best [27]

Answer:

The work done on the canister is 15.34 J.

Explanation:

Given;

mass of canister, m = 1.9 kg

magnitude of force acting on x-y plane, F = 3.9 N

initial velocity of canister in positive x direction, $v_i$ = 3.9 m/s

final velocity of the canister in positive y direction, $v_j = 5.6 \ m/s$

The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.

ΔK.E = $W_{net}$

ΔK.E $= K.E_f -K.E_i$

The initial kinetic energy of the canister;

$K.E_i = \frac{1}{2} mv_i^2\\\\K.E_i = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_i = \frac{1}{2} *1.9(\sqrt{3.9^2 +0^2 + 0^2}\ )^2 = 14.45 \ J$

The final kinetic energy of the canister;

$K.E_f =\frac{1}{2} mv_j^2 \\\\K.E_f = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_f = \frac{1}{2} *1.9(\sqrt{0^2 +5.6^2 + 0^2}\ )^2 = 29.79 \ J$

ΔK.E = 29.79 J - 14.45 J

ΔK.E = $W_{net}$ = 15.34 J

Therefore, the work done on the canister is 15.34 J.

5 0

3 years ago

Middle School Physics 5+3 pts

Murljashka [212]

A) The ball on the small ball is far smaller than the force on the basketball.

B) The total momentum before and after the collision remains constant.

C) We know momentum is conserved so we do:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.1 x 5 + 0.6 x 0 = 0.1 x -4 + 0.6 x v₂
v₂ = 1.5 m/s

5 0

3 years ago

Suppose that the average U.S. household uses 15600 kWh (kilowatt‑hours) of energy in a year. If the average rate of energy consu

sweet [91]

Answer:

a) 166.4 s

b) (2.155 × 10⁷) s

Explanation:

15600 KWh for a year,

1 year consists of 365 × 24 hours = 8760 hours.

So, the power consumed in a year for an average household = (Energy/time)

= (15600/8760) = 1.781 KW = 1781 W

a) If the average rate of energy consumed by the house was instead diverted to lift a 1.80 × 10 3 kg car 16.8 m into the air, how long would it take

The power required for this lifting = (mgh/t)

m = 1800 kg

g = 9.8 m/s²

h = 16.8 m

t = ?

P = 1781 W

1781 = (1800×9.8×16.8)/t

t = (296,352/1781)

t = 166.4 s

b) how long would it take to lift a loaded Boeing 747 airplane, with a mass of 4.05 × 10 5 kg , to a cruising altitude of 9.67 km

The power required for this lifting = (mgh/t)

m = 405000 kg

g = 9.8 m/s²

h = 9.67 km = 9670 m

t = ?

P = 1781 W

1781 = (405000×9.8×9670)/t

t = (38,380,230,000/1781)

t = 21,549,820 s = (2.155 × 10⁷) s

Hope this Helps!!!

8 0

3 years ago

You’re an electrical engineer designing an alternator (the generator that charges a car’s battery). Mechanical engineers specify

makvit [3.9K]

Answer:

I will specify a value of 0.009T for the alternator’s magnetic field

Explanation:

E_peak = 14 V

d = 10cm = 0.1m, so r = 0.1/2 =0.05m

N = 250 turns

f = 1200rpm = (1200rp/m x 1m/60sec) = 20 revolutions per second

At peak performance, peak voltage is given by the equation;

E_peak = NABω

Let's make the magnetic field B the subject;

B = E_peak/(NAω)

Now we know that ω = 2πf

Thus, ω = 2π x 20 revs/s = 125.664 revs/s.

Let's convert it to the standard unit which is rad/s.

1 rev/s = 6.283 rad/s

Thus, 125.664 revs/s = 125.664 x 6.283 = 789.55 rad/s

Area (A) = πr² = π x 0.05² = 0.007854 m²

Thus, plugging in the relevant values to get;

B = 14/[(250 x 0.007854 x 789.55)] = 0.009T

8 0

3 years ago