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Sauron [17]
3 years ago
15

A trip is taken that passes through the following points in order

Physics
1 answer:
IgorLugansk [536]3 years ago
8 0

Answer:

The displacement from point B to point E is 25.0 m left

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A person is pulling a freight cart with a force of 58 pounds. how much work is done in moving the cart 70 feet if the cart's han
Kobotan [32]

<span>The person is dragging with a force of 58 lbs at an angle of 27 degrees relating to the ground. We want to use cosine function to look for the horizontal force component. And then we can compute for W = (Horizontal Force) x (Distance). We want the horizontal force component since that is the component that is parallel to the direction the cart is moving. </span><span>

(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal force component.) 
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>

6 0
3 years ago
A 2.55 kg block starts from rest on a rough inclined plane that makes an angle of 60 degree with the horizontal. the coefficient
lorasvet [3.4K]

Answer:

Change in mechanical energy = work done by friction

so it is equal to

W = -8.16 J

Explanation:

As we know that change in mechanical energy must be equal to the work done by non conservative forces only

So here when block moves down the inclined plane then the work done by friction force is given as

W = F.d

here we have

F = \mu F_n

here we know that

F_n = mg cos\theta

so we have

F_n = 2.55(9.81)(cos60)

F_n = 12.5 N

Now the friction force on the block is given as

F_f = \mu F_n

F_f = 0.25 \times 12.5

F_f = 3.13 N

now work done by the friction is given as

W = -(3.13)(2.61)

W = -8.16 J

7 0
3 years ago
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
2 years ago
What determines the velocity of an object?
BARSIC [14]
I think it’s A not 100% sure
4 0
2 years ago
The revolving nosepiece of a compound microscope is used to: a. move the condenser up or down b. change the objective lens c. ad
Jlenok [28]

Answer:

Option B: change the objective lens

Explanation:

The revolving nosepiece is one of the parts of a microscope. Its responsibility is to hold the objective lenses.

6 0
3 years ago
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