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3 years ago

Find acceleration. Will give brainliest!

Physics

2 answers:

Art [367]3 years ago

8 0

Answer:

$\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}$

Explanation:

$\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}$

$\displaystyle \mathrm{a = \frac{v - u}{t}}$

$\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}$

$\displaystyle \mathrm{a = \frac{45- 0}{10}}$

$\displaystyle \mathrm{a = \frac{45}{10}}$

$\displaystyle \mathrm{a = 4.5}$

$\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }$

zalisa [80]3 years ago

7 0

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

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3 years ago

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3 years ago

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

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3 years ago

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mamaluj [8]

Answer:

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