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Elanso [62]
3 years ago
7

Suppose an object’s initial velocity is 10 m/s and its final velocity is 4 m/s. Mass is constant.

Physics
1 answer:
galina1969 [7]3 years ago
8 0

The negative work done states that the work is done by the object and not on the object.

<u>Explanation: </u>

According to work energy theorem, the work done is equal to change in kinetic energy exhibited by the body. As the mass of the object is constant, and the velocity is decreased from 10 m/s to 4 m/s, the work done will be

           W=\frac{1}{2} \times\left(u^{2}-v^{2}\right)

Here W is the work done, m is the mass and v is the final and u is the initial velocity of the object. As the initial velocity is greater than the final velocity. So

          W=\frac{1}{2} \times\left(4^{2}-10^{2}\right)=\frac{1}{2} \times(16-100)=\frac{1}{2} \times(-84)= - 42 J

So the work done is negative for the given situation. The negative work done states that the work is done by the object and not on the object.

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You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

6 0
2 years ago
A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the
tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

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The mass needed at peg 1 is a 5g mass.

The 15g should hang at peg 5.

The reason is force x distance clockwise is equal to force x distance anti-clockwise
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3 years ago
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krok68 [10]

Answer:

14. scolded

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2 years ago
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A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when
vampirchik [111]

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}

Then F = ma = 0.04 x 2500 = 100N

5 0
3 years ago
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