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2 years ago

11

A bike that is coasting down a steep hill increases its speed from 8.0 m/s to 14 m/s. The length of the hill is 55 meters. How m

uch time passes during the descent?

2 answers:

masya89 [10]2 years ago

8 0

t=5s

it was correct on my do-now

so I hope it was useful for you

e-lub [12.9K]2 years ago

5 0

Answer:

Time, t = 5 seconds

Explanation:

It is given that,

Initial speed of the bike, u = 8 m/s

Final speed of the bike, v = 14 m/s

Length of the hill, d = 55 m

Firstly, we can find the acceleration of the bike. Let it is equal to a. It can be calculated using third equation of motion as :

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(14)^2-(8)^2}{2\times 55}$

$a=1.2\ m/s^2$

Let t is the time passes during the descent. It can be calculated using the first equation of motion as :

$t=\dfrac{v-u}{a}$

$t=\dfrac{14-8}{1.2}$

t = 5 seconds

So, the time passes during the descent is 5 seconds. Hence, this is the required solution.

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A. Formula: F=ma or F/m=a

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B. Formula: a= $\frac{V-V_{0} }{t}$ and s=d/t

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s=26.3m/s

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D. The force that most likely caused this difference is friction forces

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1 year ago

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Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

$\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}$

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

$\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)$

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

$\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T$

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2 years ago

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

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2 years ago

The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20

NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

= 3.1536 x 10⁷ +0.840

= 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

= 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

= 86400.0023 s

The time period of rotation is given by the formula,

<em>Tₐ = 2π /ωₐ</em>

ωₐ = 2π / Tₐ

Substituting the values,

ωₐ = 2π / 365.046306

= 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

= 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

= 2π /86400

= 7.272205217 x 10⁻⁵ rad/s

The average angular acceleration

α = (ωₓ - ωₐ) / T₁

= (7.272205217 x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

α = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²

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2 years ago

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position. Why?

spin [16.1K]

Answer:

The body's rotational inertia is greater in layout position than in tucked position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in layout position to complete the backflip.

Explanation:

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

When the body is straight , its moment of rotational inertia is more than the case when he folds his body round. Hence rotational inertia ( moment of inertia x angular velocity ) is also greater. To achieve that inertia , there is need of greater imput of energy in the form of kinetic energy which requires greater effort.

So a gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

6 0

2 years ago