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4 years ago

10

Can someone help me with this. I'm not really sure if the right answer is c.

1 answer:

taurus [48]4 years ago

6 0

It might be c I'm not sure either. Hope I helped

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Can someone please help me ASAP?

Wittaler [7]

Answer:

B

Explanation:

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3 years ago

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The brakes of an automobile are suddenly applied at the instant when its velocity is 20 m/s. If the automobile comes to a stop a

Gre4nikov [31]

Data:

Time (t) = 5 seconds
Final Velocity (v) = 0 [as the car comes to rest]
<span>Initial velocity (u) = 20 m/s
</span>Acceleration = ? <span>(Negative because it is braking)</span>

$Acceleration = \frac{v-u}{t}$
$Acceleration = \frac{0-20}{5}$
$\boxed{Acceleration = - 4m/s^2}$

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3 years ago

What is the mass of a ball that is moving at 25 m/s and has 93.75 kg*m/s of momentum ?

melisa1 [442]

Answer:

m = 3.75 [kg]

Explanation:

We must remember that momentum is defined as the product of mass by Velocity, therefore it can be represented by means of the following equation.

$P=m*v$

where:

P = momentum = 93.75 [kg*m/s]

m = mass [kg]

v = velocity = 25 [m/s]

Now replacing, we can clear the mass:

$P=m*v\\m=P/v\\m=93.75/25\\m=3.75 [kg]$

7 0

3 years ago

When a compound is broken down in the chemical reaction, its atoms are

Makovka662 [10]

A. used to form new compound

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3 years ago

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2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after

sweet [91]

Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

$A(t)=A_oe^{\frac{-bt}{2m}}$

or amplitude for unforce oscillation

$\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}$

or

$\frac{6bt}{2m}=1$

or

$b=\frac{m}{3T}$

Now, provided in the question Amplitude of the driven oscillation

$A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}$

the value of the maximum amplitude is obtained $(k=m\omega_d^2)$

thus,

$A=\frac{F_{max}}{(b\omega_d}$

Now, for n people on the bridge

Fmax = nF

thus,

max amplitude

$0.075=\frac{nF}{((\frac{m}{3T})2\pi}$

or

n = 1810

hence, there were 1810 people on the bridge

b) $A=\frac{F_{max}}{(b\omega_d}$

since the effect of damping in the millenium bridge is 3 times

thus,

b=3b

therefore,

$A=\frac{F_{max}}{(3b\omega_d}$

or

$A=\frac{1}{(3}A_o$

or

$A=\frac{1}{(3}0.075$

or

A = 0.025 m = 25 mm

6 0

3 years ago