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Juliette [100K]
3 years ago
8

You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to th

e right. The snow exerts a friction force of 30 N on the sled to the left. The mass of the sled is 50 kg.
Please help me with this
Physics
2 answers:
Luden [163]3 years ago
3 0

Answer:

Explanation:

From your other post, the complete question is:

"You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to the right. The snow exerts a friction force of 30 N on the sled to the left. The mass of the sled is 50 kg.

Find the sum of the force on the sled.

Determine the acceleration of the sled .

If the sled has an initial velocity 2m/s to the right, how fast will it be traveling after 5 seconds?

"

Given the rope exerting a force of 50 N to the right and the snow exerts a friction force of 30 N to the left, the sum of forces

= 50 - 30

= 20N to the right

The mass of sled is 50 kg and force = mass * acceleration.

Acceleration = Force / mass

= 20 / 50 = 0.4 m/s^2 to the right

If the sled has an initial velocity 2m/s to the right, after 5 seconds it will be traveling at initial velocity + acceleration * time

= 2 + 0.4*5

= 4 m/s to the right

Lena [83]3 years ago
3 0

Answer:

Explanation:

net force = applied force - friction force

= 50-30

= 20N to right

acceleration = net force/mass

= 20/50

= 2/5m/s tp right

final velocity = initial velocity + acceleration*time

= 2+2/5*5

= 4m/s to right

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Answer:

Explanation:

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Technician A says that the crankshaft determines the stroke of an engine. Technician B says that the length of the connecting ro
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Answer:

Technician A is correct.

Explanation:

Stroke of the engine means how far is the piston moves during the cycle.It is determined by the crank of the crank shaft. Engine displacement volume  is calculated by multiplying length of the stroke to the piston displacement area.

                          If there are more than one cylinder in the engine, than in that case this number in multiplied by the number of the cylinders inside the engine.

Therefore technician A is correct.

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3 years ago
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block
kotegsom [21]

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  v  = 11.76  \ m/ s

Explanation:

From the question we are told that

   The  total distance traveled is  d =  1.2  \ m

    The mass of the block is  m_b  =  0.3 \ kg

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   PE  =  KE

Where  PE  is the potential energy which is mathematically represented as

      PE  =  m * g  *  h

substituting values

     PE  =   3 *  9.8  *  0.60

      PE  =  17.64 \  J

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          KE  =  \frac{1}{2}  *  m v^2

So

      \frac{1}{2}  *  m* v ^2  =  PE

substituting values  

  =>    \frac{1}{2}  *  3 * v ^2  = 17.64

=>       v  = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }

=>    v  = 11.76  \ m/ s

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A motorcycle starts at rest and accelerates at a rate of 3 meters per second squared (m/s2) over a time period of 5 seconds (s).
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Final velocity = 3 x 5 = 15 m/s
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In the diagram, 91, 92, and q3 are in a straight line. Each of these particles has a charge of -2.35 x 10-6 C. Particles q₁ and
Strike441 [17]

The net force on particle q₃ is  6.2128125 N.

<h3>What is electrostatic force?</h3>

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/d²

where k = 9 x 10⁹ N.m²/C²

Given is the diagram in which each of the particles has a charge of -2.35 x 10⁻⁶ C. Particles q₁ and q₂ are separated by 0.100 m and particles q₂ and q₃ are separated by 0.100 m.

Force acting on q₃ due to q₁

F₃₁  = 9 x 10⁹ x (-2.35 x 10⁻⁶)²/(0.1)²

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Force acting on q₃ due to q₂

F₃₂ = 9 x 10⁹ x (-2.35 x 10⁻⁶)²/(0.1+0.1)²

F₃₂ = 1.2425 N (in right direction)

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F₃ = 6.2128125 N

Thus the net force on a charged particle is  6.2128125 N to the right.

Learn more about electrostatic force.

brainly.com/question/9774180

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