Answer:
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Explanation:
The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.
The angle between the 2 forces and displacement is ∅ = 15°.
First we have to calculate the work done by the individual force and then we can calculate the total work.
The work done on a particle by a constant force F during a straight line displacement s is given by following formula:
W = F*s
W = F*s*cos∅
With ∅ = the angles between F and s
The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N
The total work done can be calculated as followed:
Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2
Wtotal = 2Fs*cos∅
Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°
Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>
<u />
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Answer:
3.76 m/s
Explanation:
Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.
v' = dx(t)/dt..................... Equation 1
Where v' = instantaneous velocity, x = distance, t = time.
Given the expression,
x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³
x(t) = 28 + 12.4t - 0.0450t³
Differentiating x(t) with respect to t.
dx(t)/dt = 12.4 - 0.135t²
dx(t)/dt = 12.4 - 0.135t²
When t = 8.00 s.
dx(t)/dt = 12.4 - 0.135(8)²
dx(t)/dt = 12.4 - 8.64
dx(t)/dt = 3.76 m/s.
Therefore,
v' = 3.76 m/s.
Hence, the instantaneous velocity = 3.76 m/s
Electric potential energy is defined as Ep=Q*V where Q is the magnitude of the charge and V is the potential difference. So when a charge moves between the points that have a potential difference, it's energy changes.
In our case:
Q=2e=2*(-1.6*10^-19) C
V=75 V
Ep=(-3.2*10^-19)*75
Ep=-2.4*10^-17 J
The change in potential energy of the charge is -2.4*10^-17 J
Answer:
f = 3.1 kHz
Explanation:
given,
length of human canal =2.8 cm = 0.028 m
speed of sound = 343 m/s
fundamental frequency = ?
The fundamental frequency of a tube with one open end and one closed end is,



f = 3062.5 Hz
f = 3.1 kHz
hence, the fundamental frequency is equal to f = 3.1 kHz
Answer:
Gravity provides a downward force, resulting in the diver going downward. They speed up like any falling object would, the pull of gravity is a dominant force. (There is a drag force – as a result of moving through the air.)