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Vedmedyk [2.9K]
1 year ago
8

Which statement accurately describes the motion of the object in the graph above over 10 seconds? Group of answer choices The ob

ject is at rest for 3 seconds and then decreases velocity to 1 cm/s for 3 seconds. The object is at rest for 4 seconds and then moves forward for 4 seconds at 1 cm/s. The object moves forward at 1 cm/s, stops, and then increases velocity to 2 cm/s. The object moves forward at 1 cm/s, stops, and then continues at the same velocity.

Physics
1 answer:
timurjin [86]1 year ago
7 0

Answer:

Object appears to move forward at 1 cm/sec, then the velocity drops to zero for 3 sec and then moves forward at 2 cm/sec     (11 - 3) / (10 - 6) = 2 cm/sec

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Newton's third law tells us that for every force there is an equal and opposite force.  This means that if Anna exerts a force of 20 Newtons on the box, the box exerts a force of 20 Newtons on Anna.
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A metal disk weighing 1 N is resting on an index card that is balanced on top of a glass. When the index card is quickly pulled
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Sketch a position-time graph for a bear starting
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hopefully that makes sense. the position doesn't change over the 5 seconds, meaning it's stopped but time still continues. then when the slope is negative this shows the bear's position becoming negative (backing up, changing direction).

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A small block of mass 200 g starts at rest at A, slides to B where its speed is vB=8.0m/s,vB=8.0m/s, then slides along the horiz
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Answer

given,

mass of the block = 200 g = 0.2 Kg

Velocity at A = 0 m/s

Velocity at B = 8 m/s

slide to the horizontal distance = 10 m

height of the block be = 4 m

potential energy of the block

    P = m g h

    P = 0.2 x 9.8 x 4

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kinetic energy

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Work = P - KE

work = 7.84 - 6.14

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7 0
2 years ago
Consider a lawnmower of mass m which can slide across a horizontal surface with a coefficient of friction μ. In this problem the
inna [77]

Answer:

Fh = u*m*g / (cos(θ) - u*sin(θ))

Explanation:

Given:

- The mass of lawnmower = m

- The angle the handle makes with the horizontal = θ

- The force applied along the handle = Fh

- The coefficient of friction of the lawnmower with ground = u

Find:

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Solution:

- Construct a Free Body Diagram (FBD) for the lawnmower.

- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:

                                     F_net,x = m*a

- Since, the lawnmower is to move with constant speed then we have a = 0.

                                     F_net,x = 0

- The forces as follows:

                                     Fh*cos(θ) - Ff = 0

Where, Ff is the frictional force:

                                     Fh = Ff /cos(θ)

Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:

                                    - W - Fh*sin(θ) + Fn = 0

Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:

                                     Fn = W + Fh*sin(θ)

                                     Fn = m*g + Fh*sin(θ)

The Frictional force Ff is proportional to the contact force Fn by:

                                     Ff = u*Fn

                                     Ff = u*(m*g + Fh*sin(θ))

Substitute this expression in the form derived for Fh and Ff:

                                     Fh*cos(θ) = u*(m*g + Fh*sin(θ))

                                     Fh*(cos(θ) - u*sin(θ)) = u*m*g

                                     Fh = u*m*g / (cos(θ) - u*sin(θ))

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2 years ago
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