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finlep [7]
3 years ago
8

How do you find motion

Physics
1 answer:
saveliy_v [14]3 years ago
8 0

Answer:

idk you got me there

Explanation:

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A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity and displ
Anestetic [448]

Answer:

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

<h2>Part (a)</h2>

We are given/can infer these variables:

  • t = 4.0 s
  • a = -9.8 m/s²
  • v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

  • Δx = ?
  • v = ?

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 0 + (-9.8)(4.0)
  • v = -9.8 * 4.0
  • v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

  • Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

  • Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
  • Δx = 1/2(-9.8)(4.0)²
  • Δx = -4.9 * 16
  • Δx = -78.4 m

The displacement of the rock is -78.4 m.

<h2>Part (b)</h2>

We are given/can infer these variables:

  • v_0 = 8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 8.0 + (-9.8)(4.0)
  • v = 8.0 + -39.2
  • v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = 32 - 4.9(16)
  • Δx = -46.4 m

The displacement of the rock is -46.4 m.

<h2>Part (c)</h2>

We are given/can infer these variables:

  • v_0 = -8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = -8.0 + (-9.8)(4.0)
  • v = -8.0 - 39.2
  • v = -47.2 m/s

The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = -32 - 4.9(16)
  • Δx = -110.4 m

The displacement of the rock is -110.4 m.

8 0
3 years ago
When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th
Veseljchak [2.6K]
Given:
F = ax
where
x = distance by which the rubber band is stretched
a =  constant

The work done in stretching the rubber band from x = 0 to x = L is
W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2}  [x^{2} ]_{0}^{L} =  \frac{aL^{2}}{2}

Answer:  \frac{aL^{2}}{2}

4 0
4 years ago
Suppose the charges attracting each other in the above situation have an equal magnitude. Find the charge.
LiRa [457]

The force between the two point charge when they are separated by 18 cm is 3 N

<h3>How do I determine the force when they are 18 cm apart?</h3>

Coulomb's law states as follow:

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ => constant

F₁r₁² = F₂r₂²

Where

  • F₁ and F₂ are the initial and new force
  • r₁ and r₂ are the initial and new distance apart

With the above formula, we can obtain the force between the two point charge when they are 18 cm apart. Details below:

  • Initial distance apart (r₁) = 6 cm
  • Initial force of attraction (F₁) = 27 N
  • New distance apart (r₂) = 18 cm
  • New force of attraction (F₂) =?

F₁r₁² = F₂r₂²

27 × 6² = F₂ × 18²

972 = F₂ × 324

Divide both side by 324

F₂ = 927 / 324

F₂ = 3 N

Thus, the force when they are 18 cm apart is 3 N

Learn more about force:

brainly.com/question/28569085

#SPJ1

7 0
2 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit
Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

d_1=4.7\times 10^{10} m

v_i=9.5\times 10^4 m/s

d_2=6\times 10^{12} m

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

K_i+U_i=K_f+U_f

\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}

\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f

v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})

v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}

Using the formula

v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}

v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}

Where mass of sun=M=1.98\times 10^{30} kg

G=6.7\times 10^{-11}

v_f=58515.9 m/s

4 0
3 years ago
An object of mass 30kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.
Luba_88 [7]

That all depends on the planet toward which the mass is falling. 

If this happens to be taking place near the Earth, then the object accelerates
at the rate of about 9.8 meters per second every second.

Furthermore, if there is truly no air resistance, then it makes no difference whether
the object is a feather, a mass of 30 kg, or a school-bus.  All objects accelerate at
the same rate regardless of their mass.


7 0
3 years ago
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