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allsm [11]
3 years ago
9

A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of i

ts engines is 2.3 x 10^5 N. In being launched from rest it moves a distance of 87 m and has a kinetic energy of 4.5 x 10^7 J at lift-off. What is the work done on the jet by the catapult?
Physics
1 answer:
kirza4 [7]3 years ago
8 0

Answer:

2.5 x 10⁷ J

Explanation:

F = thrust of the engine = 2.3 x 10⁵ N

d = distance traveled = 87  m

Work done by the engine is given as

W = F d =  (2.3 x 10⁵) (87) = 200.1 x 10⁵ J

W' = Net work done

W'' = work done by catapult

KE₀ = initial kinetic energy = 0 J

KE = final kinetic energy = 4.5 x 10⁷ J

Net work done is given as

W' = KE - KE₀

W' = 4.5 x 10⁷ J

We know that

W' = W + W''

4.5 x 10⁷ = 2.001 x 10⁷ + W''

W'' = 2.5 x 10⁷ J

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the difference between a resultant and equilibrant vector is that resultant vector is a direct quantity, one with both magnitude and direction, while the equilibrant vector is a force equal to, but opposite of, the resultant sum of vector forces, that force which balances other forces.

8 0
3 years ago
A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. x=x_{0}+vt

2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

8 0
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Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how
zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

3 0
3 years ago
A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.
Natali5045456 [20]

Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

20 x 10⁻⁹ C = ?

= 1.248 x 10¹¹ electrons

Thus,1.248 x 10¹¹ electrons were transferred

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