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USPshnik [31]
3 years ago
13

The only force acting on a 3.1 kg body as it moves along the positive x axis has an x component fx = -8x n, where x is in meters

. the velocity of the body at x = 2.0 m is 11 m/s. (a) what is the velocity of the body at x = 4.5 m? (b) at what positive value of x will the body have a velocity of 5.8 m/s?
Physics
1 answer:
My name is Ann [436]3 years ago
4 0
<span>(a) -9.97 m/s
 (b) x = 2.83

   This is a simple problem in integral calculus. You've been given part of the 2nd derivative (acceleration), but not quite. You've been given the force instead. So let's setup a function for acceleration.

   f''(x) = -8x N / 3.1 kg= -8x kg*m/s^2 / 3.1 kg = -2.580645161x m/s^2

   So the acceleration of the body is now expressed as
 f''(x) = -2.580645161x m/s^2

    Let's calculate the anti-derivative from that.
 f''(x) = -2.580645161x m/s^2
 f'(x) = -1.290322581x^2 + C m/s

   Now let's use the known velocity value at x = 2.0 to calculate C
 f'(x) = -1.290322581x^2 + C 1
1 = -1.290322581*2^2 + C
 11 = -1.290322581*4 + C
 11 = -5.161290323 + C
 16.161290323 = C

   So the velocity function is
  f'(x) = -1.290322581x^2 + 16.161290323

   (a) The velocity at x = 4.5
  f'(x) = -1.290322581x^2 + 16.161290323
 f'(4.5) = -1.290322581*4.5^2 + 16.161290323
 f'(4.5) = -1.290322581*20.25 + 16.161290323
  f'(4.5) = -26.12903227 + 16.161290323
 f'(4.5) = -9.967741942

   So the velocity is -9.97 m/s

   (b) we want a velocity of 5.8 m/s
5.8 = -1.290322581x^2 + 16.161290323
 0 = -1.290322581x^2 + 10.36129032
 1.290322581x^2 = 10.36129032
 x^2 = 8.029999998
 x = 2.833725463</span>
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3 0
3 years ago
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A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.
jeka94
  • Mass of the elevator (m) = 570 Kg
  • Acceleration = 1.5 m/s^2
  • Distance (s) = 13 m
  • Let the force be F.
  • We know, F = ma,
  • Therefore, F = (570 × 1.5) N = 855 N
  • Angle between distance and force (θ) = 0°
  • We know, work done = F s Cos θ
  • Therefore, work done by the cable during this part
  • = (855 × 13 × Cos 0°) J
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<u>Answer</u><u>:</u>

<u>1</u><u>1</u><u>1</u><u>1</u><u>5</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
2 years ago
How to do this question?​
marissa [1.9K]

Answer:

i3 =11.014A

i5 = 3.15A

Explanation:

Here according to k'chofs first law

i1 =i2 + i3

i3 = i4 + i5

For determine the i1 you have to consider the resultant resistor of the system

4 , 1 and 3 resistors are in pararel

Then, Resultant is

1/4 + 1/1 + 1/3 = 1/ R

R = 12/19

For get total we have to add another remaining 3 resistor because of serious

Then Resultant is = 12/19 + 3

= 69/19

Then using V = IR

40 =i3* 69/19

i3 = 11.014 A

Other 3 resistors are parrarel because of this voltage of those resistors are same.

Then i inversely propotional to its resistor

Then ,

i5 * 2 = (i3-i5)*4/5

i 5 = 3.15 A

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v₀ = initial velocity of the baseball = - 39 m/s

v = final velocity of the baseball = 52 m/s

t = time of contact = 3 x 10⁻³ sec

F = average force between bat and ball

Using impulse-change in momentum equation

F t = m (v - v₀ )

F (3 x 10⁻³) = (0.145) (52 - (- 39))

F = 4398.33 N

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