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3 years ago

6

our 3.80-kg physics book is placed next to you on the horizontal seat of your car. The coefficient of static friction between th

e book and the seat is 0.650, and the coefficient of kinetic friction is 0.550. You are traveling forward at 72.0 km/h and brake to a stop with constant acceleration over a distance of 30.0 m. Your physics book remains on the seat rather than sliding forward onto the floor. Is this situation possible

1 answer:

gtnhenbr [62]3 years ago

4 0

Answer:

Explanation:

Maximum force of friction possible = μmg

= .65 x 3.8 x 9.8

= 24.2 N

u = 72 x 1000 / 60 x 60

= 20 m /s

v² = u² - 2as

a = 20 x 20 / (2 x 30)

= 6.67 m / s²

force acting on it

= 3.8 x 6.67

= 25.346 N

Friction force possible is less .

So friction will not be able to prevent its slippage

It will slip off .

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A thin rod (length = 2.97 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.

nlexa [21]

Answer:

a) w = 2.57 rad / s , b) α = 3.3 rad / s²

Explanation:

a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground

Initial. Higher

Em₀ = U = m g h

Final. Touching the ground

$Em_{f}$ = K = ½ I w²

How energy is conserved

Em₀ = $Em_{f}$

mg h = ½ I w2

The moment of specific object inertia

I = m L²

We replace

m g h = ½ (mL²) w²

w² = 2g h / L²

The height of the object is the length of the bar

h = L

w = √ 2g / L

w = √ (2 9.8 / 2.97)

w = 2.57 rad / s

b) the angular acceleration can be found from Newton's second rotational law

τ = I α

W L = I α

mg L = (m L²) α

α = g / L

α = 9.8 / 2.97

α = 3.3 rad / s²

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Which of the following is the same size as Pluto?

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g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

$\frac{R}{2L}T= \ln \frac{5}{3.8}$

$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

= 11.45

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2 years ago

A football player kicks a ball at a 30o angle from the ground with an initial velocity of 15 m/s. What is the final velocity of

navik [9.2K]

Given that,

Angle = 30°

Initial velocity = 15 m/s

We need to calculate the time of flight

Using formula of time of flight

$T=\dfrac{2u\sin\theta}{g}$

Where, u = initial velocity

g = acceleration due to gravity

Put the value into the formula

$T=\dfrac{2\times15\sin30}{9.8}$

$T=1.5\ sec$

We need to calculate the final velocity of the ball

Using equation of motion

$v=u+gt$

$v=15+9.8\times1.5$

$v=29.7\ m/s$

Hence, The final velocity of the ball is 29.7 m/s.

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Ket [755]

Answer:

Explanation:

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