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NemiM [27]
3 years ago
14

Ingrid kicks a football with an initial velocity at 12 m/s at an angle of 45 relative to the ground

Physics
1 answer:
natali 33 [55]3 years ago
5 0
Equation of movement is

y(x) = x * tg (a)  -    \frac{g}{2 * u^{2} * cos (a)^{2} } * x^{2}

where y is y-coordinate, x - x coordinate,  u - your initial speed, a = angle above horizon.

differentiate it to get dy/dx (speed)
 
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A 30 ky child running at 7 m/s jumps onto a 10 kg sled which was initially at rest. What will be the
Ede4ka [16]
Hope this helps. If you need clarification just ask me!

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3 years ago
Which is most likely the length of a student’s textbook?
UkoKoshka [18]

For this case, what we must do is to rewrite these measurements in the same unit in order to compare them.

By writing the measurements in meters we have:

30 mm = (30) * (\frac{1}{1000}) = 0.030 m

30 cm = (30) * (\frac{1}{100}) = 0.30 m

30 dm = (30) * (\frac{1}{10}) = 3 m\\30 Hm = (30) * (100) = 3000 m

Therefore, physically the correct measure is:

0.30 m = 30 cm

Answer:

the length of a student's textbook most likely is:

30 centimeters

3 0
3 years ago
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A ball with an initial velocity of zero is dropped from a table onto the ground. After 2 s, the
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  • Initial velocity=u=0m/s
  • Acceleration=a=10m/s^2
  • Time=t=2s

Final velocity=v

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+10(2)

\\ \sf\longmapsto v=20m/s

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2 years ago
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A device used to measure a mass of an object is called​
OLga [1]

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8 0
3 years ago
A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi
frosja888 [35]

Answer:

The answer to the question

The steady state response is u₂(t) = -\frac{3\sqrt{2} }{2}cos(3t + π/4)

of the form R·cos(ωt−δ) with R = -\frac{3\sqrt{2} }{2}, ω = 3 and δ = -π/4

Explanation:

To solve the question we note that the equation of motion is given by

m·u'' + γ·u' + k·u = F(t) where

m = mass = 2.00 kg

γ = Damping coefficient = 1

k = Spring constant = 3 N·m

F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)

Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by \frac{-1+/-\sqrt{23} }{4} \sqrt{-1} =\frac{-1+/-\sqrt{23} }{4} i

This gives the general solution of the homogeneous equation as

u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which  2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t)  we have

3·B-15·A = 27

3·A +15·B = 18

Solving the above linear system of equations we have

A = -1.5, B = 1.5 and  u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4)

The general solution is then  u(t) = u₁(t) + u₂(t)

however since u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t) ⇒ 0 as t → ∞ the steady state response = u₂(t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = -\frac{3\sqrt{2} }{2}

ω = 3 and

δ = -π/4

8 0
3 years ago
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