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gavmur [86]
2 years ago
6

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 15.0 ∘ west of north,

and the other at an angle 15.0 ∘ east of north, as they pull the tanker a distance 0.810 km toward the north.
What is the total work done by the two tugboats on the supertanker?
Physics
1 answer:
Darina [25.2K]2 years ago
5 0

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

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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time
Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

 

4 0
3 years ago
A rifle bullet with a mass of 11.5 g traveling toward the right at 251 m/s strikes a large bag of sand and penetrates it to a de
BabaBlast [244]

To look for the acceleration, it will come from:

vf^2=v0^2+2ad 
where:
vf = final velocity = 0 
v0 = initial velocity =251 m/s 
a = acceleration 
d= distance traveled = 0.237 m 

0=251^2+2a(0.237 ) 
a= -251 ^2 / (2*0.237) =-132 913.502 m/s/s 

we find the force from: 

F = ma = 0.0115kg*(-1.32x10^5m/s/s) = -1518 N 

the negative sign shows that the force is in the direction contradictory the bullet's motion

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3 years ago
A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. Th
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Answer: 0.0180701 s

Explanation:

Given the following :

Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

= √0.0003265

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5 0
3 years ago
A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity
Crank

Answer:

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Explanation:

Using the law of conservation of momentum

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m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

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u2 = 3.2m/s

Required

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Substitute the given values into the formula

64(4.2)+25(3.2) = (65+25)v

268.8+80 = 90v

348.8 = 90v

v = 348.8/90

v = 3.88m/s

Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s

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