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agasfer [191]
3 years ago
9

9. A sailor pulls a boat along a dock using a rope at an angle of 60.0° with the

Physics
1 answer:
aleksklad [387]3 years ago
6 0

Answer: (1) 3.83x10^3 J

Explanation:

(1) Fx=(255N)cos60°

   dx=30.0m

   w=Fx dx =(255)(cos60°)(30.0m)

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How much force is needed to accelerate a 2500 kg car at a rate of 3.5 m/s^2?
Sati [7]

F = ma  

F = applied force in newtons = to be determined  

m = mass of the car = 2,500 kg  

a = acceleration of the car = 3.5 m/s²  

F = (2,500 kg)(3.5 m/s²)  

F =8750

4 0
3 years ago
A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t
emmainna [20.7K]
Efficiency =  Work Output / Work Input

92%  =  Work Output / 100

0.92 =   Work Output / 100

Work Output = 0.92 * 100

Work Output  = 92 joules.
8 0
2 years ago
Which formula correctly expresses the property density?
Olenka [21]

Answer:

Option A

D = m/v

Explanation:

Density is defined as mass per unit volume of an object. Therefore, D=m/v where m is the mass of the object and v is the volume

Therefore, option A is the right option

6 0
2 years ago
A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box
forsale [732]
We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force. 

From N-Gcosα=0 we get: 

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
8 0
3 years ago
A tsunami covers a distance of 500km in 83.3 minutes. If the period of vibration of the molecules is 35.0 minutes, what is the w
marin [14]

Answer:

210079.798 m

Explanation:

pls give brainliest :)

8 0
2 years ago
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