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kodGreya [7K]
3 years ago
13

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an

d the other at an angle 13.0 east of north, as they pull the tanker a distance 0.700 km toward the north.
What is the total work done by the two tugboats on the supertanker?

Express your answer in joules, to three significant figures.

Physics
1 answer:
Juliette [100K]3 years ago
3 0

Answer:W=1.93\times 10^9 J      

Explanation:

Given

Force F=1.6\times 10^{6} N

one at an angle of 13^{\circ} East of North and another at 13^{\circ} West of North

Net Force is in North Direction

F_{net}=2F\cos 13

Forces in horizontal direction will cancel out each other

thus Work done will be by north direction forces  

W=2F\cdot \cos 30\cdot s

here s=0.7 km

W=2\times 1.6\times 10^{6}\cdot \cos 30\cdot 700

W=1.93\times 10^9 J                  

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A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0
3 years ago
How does an electromagnet become permanent
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What is the primary cause of diffusion?
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What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in ai
olga nikolaevna [1]

Answer:

The magnetic flux density is 2.11\times10^{-6}\ T

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

B =\dfrac{\mu_{0}I}{2r}

Where,

r = radius

I = current

Put the value into the formula

B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}

B=2.11\times10^{-6}\ T

Hence, The magnetic flux density is 2.11\times10^{-6}\ T

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julsineya [31]

Answer:

5N

Explanation:

(25 N - 20 N = 5 N)

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