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3 years ago

6

The springs of a 1700-kg car compress 5.0 mm when its 66-kg driver gets into the driver seat. If the car goes over a bump, what

will be the frequency of oscillations?

1 answer:

Vadim26 [7]3 years ago

8 0

Answer:

The frequency of oscillations is 7 Hz

Explanation:

Given;

mass of car, = 1700 kg

mass of driver, = 66 kg

compression of the spring, x = 5mm = 0.005 m

The frequency of the oscillation is given as;

$F = \frac{1}{2 \pi } \sqrt{\frac{k}{m} }$

where;

k is force constant

m is the total mass of the car and the driver

m = 1700 kg + 66 kg = 1766 kg

Weight of the car and the driver;

W = mg

W = 1766 x 9.8

W = 17306.8 N

Apply hook's law, to determine the force constant;

F = kx

W = F

Thus, k = W/x

k = 17306.8 / 0.005

k = 3461360 N/m

Now, calculate the frequency

$F = \frac{1}{2\pi}\sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi}\sqrt{\frac{3461360}{1766} }\\\\F = 7 \ Hz$

Therefore, the frequency of oscillations is 7 Hz

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Answer:

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Explanation:

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λₙ= 298.5 nm

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Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How

grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object ( $\Delta s$ ), in meters, can be determined by the following expression:

$\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2}$ (1)

Where:

$v_{o}$ - Initial velocity, in meters per second.

$t$ - Time, in seconds.

$a$ - Acceleration, in meters per square second.

If we know that $v_{o} = 5\,\frac{m}{s}$ , $t = 6\,s$ and $a = 2\,\frac{m}{s^{2}}$ , then the box displacement after 6 seconds is:

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3 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

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Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

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