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kodGreya [7K]
4 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]4 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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3 years ago
A 10 kg box is pushed with a force of 20N. What is the box's acceleration?
Rasek [7]
The following formula gives the connection between the force, the mass and the acceleration:
F=ma
In our case we know the force and the mass. F=20N and m=10kg
a=F/m
a=20m/s^2/10=2<span>m/s^2</span>
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A mechanical wave is created when a medium ____, or moves back and forth or up and down
Zielflug [23.3K]
I think it is traverse wave
6 0
4 years ago
Read 2 more answers
The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the elect
RideAnS [48]

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

E=\dfrac{\lambda}{2\pi \epsilon_o r}

It is clear that the electric field is inversely proportional to the distance. So,

\dfrac{E}{E'}=\dfrac{r'}{r}

E'=\dfrac{Er}{r'}

E'=\dfrac{125\times 3.5}{1.5}  

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

5 0
4 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
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