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marta [7]
3 years ago
12

A 500kg wagon is pushed against a wall with a spring between the wall and the wagon. The spring has a stiffness coefficient of 3

00 N/m. If the spring is compressed a distance of .6m and the wagon continues to apply a 55N force toward the wall, then what will the acceleration of the wagon be?
Physics
1 answer:
andreyandreev [35.5K]3 years ago
5 0

Answer: 0.36 meter per second square

Explanation from hooke's law f = ke

from motion, F=ma

Ma=me

Where a= ?

K=300

e=0.6

M=500kg

500×a =300×0.6

a = 180÷500

a = 0.36 meter per second square

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dmitriy555 [2]
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How demands became the obstacles ?

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7 0
2 years ago
Two vehicles approach an intersection: a truck moving eastbound at 16.0 m/s and an SUV moving southbound at 20.0 m/s. Suppose th
mario62 [17]

Answer:25.61 m/s

Explanation:

Given

truck is moving eastbound with a velocity of 16 m/s

Velocity of truck v_t=16\hat{i}

SUV is moving south with a velocity of 20 m/s

Velocity of SUV in vector form v_s=-20\hat{j}

Velocity of truck relative to the SUV

v_{ts}=v_{t}-v_s

v_{ts}=16\hat{i}-(-20\hat{j})

Magnitude of relative velocity is

|v_{ts}|=\sqrt{16^2+20^2}

|v_{ts}|=25.61\ m/s                                                  

5 0
3 years ago
What type of electromagnetic wave can be used to transmit information?
JulijaS [17]
Radio wave only would be the answer.
5 0
3 years ago
Find the wavelength of radio waves of frequency 200kHz
Gala2k [10]
\lambda = \frac{c}{f}
λ - wavelength, c - the speed of light, f - frequency

f=200 \ kHz= 200 000 \ Hz \\ \\
\lambda=\frac{300 000 \ [\frac{km}{s}]}{200 000 \ [Hz]}=\frac{3}{2}=1.5 \ [km]

The wavelength of these waves is 1.5 km.
3 0
3 years ago
Read 2 more answers
A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c
kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately 20.3\; \rm m \cdot s^{-2}.

It would take approximately 3.9\; \rm s for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

  • Let v denote the final velocity of the car.
  • Let u denote the initial velocity of the car.
  • Let a denote the acceleration of the car.
  • Let x denote the distance that this car travelled.

v^2 - u^2 = 2\, a\cdot x.

Given:

  • u = 3\; \rm m \cdot s^{-1}.
  • a = 4.5\; \rm m \cdot s^{-2}.
  • x = 45\; \rm m.

Rearrange the equation v^2 - u^2 = 2\, a\cdot x and solve for v:

\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}.

Calculate the time required for reaching this speed from u = 3\; \rm m \cdot s^{-1} at a = 4.5\; \rm m \cdot s^{-2}:

\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}.

3 0
3 years ago
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