Do you wont to be friends

A car with a total mass of 1800 kg (including passengers) is driving down a washboard road with bumps spaced 4.9 m apart. The ri

Drupady [299]

Answer:

k = 9.6 x 10^5 N/m or 9.6 kN/m

Explanation:

First, we need to use the expression to calculate the spring constant which is:

w² = k/m

Solving for k:

k = w²*m

To get the angular velocity:

w = 2πf

The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:

f = V/x

f = 5.7 / 4.9 = 1.16 Hz

Now the angular velocity:

w = 2π*1.16

w = 7.29 rad/s

Finally, solving for k:

k = (7.29)² * 1800

k = 95,659.38 N/m

In two significant figures it'll ve 9.6 kN/m

5 0

2 years ago

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What is the name of the compound that goes with the chemical formula MgCl2

frez [133]

Magnesium chloride is your answer

8 0

2 years ago

ILL GIVE BRAINLYEST

lawyer [7]

Answer:

3rd picture straight line going up right

Explanation:

3rd picture

5 0

3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

4 0

2 years ago

A car moves with an average speed of 75 kmh^-1 from town P to town Q in 2 hours. By using information, you may calculate the dis

-Dominant- [34]

Answer:

patron

Explanation:

The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.

Explanation:

patron

Explanation:

The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.

7 0

1 year ago

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