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hammer [34]
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
Physics
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

3 0
3 years ago
Which of the following is the same size as Pluto?
coldgirl [10]
I would say it's <span>C. the Moon
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4 0
3 years ago
Read 2 more answers
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

4 0
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.

7 0
3 years ago
The formula for average speed is
Ket [755]

Answer:

Explanation:

The most common formula for average speed is distance traveled divided by time taken. The other formula, if you have the initial and final speed, add the two together, and divide by 2

8 0
3 years ago
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