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weqwewe [10]
3 years ago
5

A box of unknown mass is sliding with an initial speed vi = 4.70 m/s across a horizontal frictionless warehouse floor when it en

counters a rough section of flooring d = 4.50 m long. the coefficient of kinetic friction between the rough section of flooring and the box is 0.100. using energy considerations, determine the final speed of the box after sliding across the rough section of flooring.
Physics
1 answer:
malfutka [58]3 years ago
8 0

Here Change in Kinetic Energy = Work Done by Friction

Therefore, substituting the given values to the equation, we get

0.5 * m * (vFinal^2 - vInitial^2) = µ m g * d

Therefore

0.5*( 5.90^2 - Vfinal^2 ) = 0.100*9.8*2.10

Therefore

vfinal = 5.54 m/sec

<span> </span>

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Flauer [41]
Stars having less mass collapses early than those with more mass. This can be explained by Einstein's equation E=mc².
According to this equation, mass of stars is converted into light due to thermonuclear reactions occuring in the core of star which acts as engine of the stars. This thermonuclear reactions keeps star alive. Thermonuclear reactions occurs slowly in massive stars hence massive stars live more than light stars.
6 0
3 years ago
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 38 m/s when i
inysia [295]

Explanation:

look !

speed= 38m/s

start from rest= 0

5 0
2 years ago
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

8 0
3 years ago
A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t
Talja [164]

Answer:

P = 133.13 Watt

Explanation:

Initial angular speed of the ferris wheel is given as

\omega_i = 2\pi f

\omega_i = 2\pi(8.5/3600)

\omega_i = 0.015 rad/s

final angular speed after friction is given as

\omega_f = 2\pi f

\omega_f = 2\pi(7.5/3600)

\omega_f = 0.013 rad/s

now angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{0.015 - 0.013}{15}

\alpha = 1.27 \times 10^{-4} rad/s^2

now torque due to friction on the wheel is given as

\tau = I \alpha

\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})

\tau = 8875.3 N m

Now the power required to rotate it with initial given speed is

P = \tau \omega

P = 8875.3 \times 0.015

P = 133.13 Watt

8 0
3 years ago
A box has a momentum of
Kipish [7]

Answer:

67.9 kg*m/s

Explanation:

Pi = 38 kgm/s

F = 88.3N and ∆t = 0.338s

Final momentum Pf = Pi + F∆t = 38 + (88.3)(0.338) = 38 + 29.8454

=) Pf = 67.8454 kgm/s = 67.85kg*m/s

Your answer is 67.9kg*m/s with three significant figures

hope this helps your troubles!

6 0
3 years ago
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