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tia_tia [17]
2 years ago
7

Two tougboats are toeing a ship each exerts a force of 6000N and the angle between the two ropes is 60 calculate the resultant f

orce on the ship
Physics
1 answer:
Arlecino [84]2 years ago
7 0

Answer:

10392.30N

Explanation:

We proceed by computing the individual force exerted by the boats

For the first boat

The angle is 30 degree to the vertical

Hence

Force = F cos θ

F=6000 cos 30

F=6000*0.866

F=5196.15 N

Since the boats are two and also at the same angle and also exerting the same force

The Net force = 2*5196.15

Net force=10392.30N

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The potential energy of a negative charge moved from point A to point B will increase.A negative charge moved from point A to po
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Answer:

<em>The K.E from A to B won't increase...</em>

Explanation:

That's because the P.E from A to B is increasing. The K.E will increase if charge moves from a higher potential to a lower potential i.e., from B to A.

That is the reason there is no effect on net K.E when moving from a potential to same potential over and over (A to C).

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3 years ago
If earth's axis was tilted 45 degrees at what latitudes
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23.5 DEGREES is the answer. 
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2 years ago
Use this table of a school bus during morning pickups to calculate its average speed between 0 h and 2.340 h.
Gelneren [198K]

The average speed between 0 h and 2.340 h is 6.97 Km/h

Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.

Average \: speed =  \frac{total \: distance}{total \: time}  \\  \\

With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:

  • Total time = 2.340 – 0 = 2.340 h
  • Total distance = 16.3 – 0 = 16.3 Km
  • Average speed =?

Average \: speed =  \frac{total \: distance}{total \: time}  \\  \\Average \: speed =  \frac{16.3}{2.340}  \\  \\ Average \: speed = 6.97 \: Km/h \\  \\

Learn more about average speed: brainly.com/question/24884027

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What kind of rock is this?
Katyanochek1 [597]
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Read 2 more answers
A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s
3241004551 [841]

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer.  The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\  340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

8 0
2 years ago
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