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2 years ago

Who watching all star draft? Luka better get picked first ong

Physics

2 answers:

Whitepunk [10]2 years ago

5 0

Na dame Dolla fo surrrreeeeee

Studentka2010 [4]2 years ago

4 0

Answer: i did

it was good

Explanation: have a nice day

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Can you help me with this?

Natasha2012 [34]

Answer:

Explanation:

so basically I am domb so I can not help you

4 0

3 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

3 years ago

Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal

NeX [460]

Answer:

366 m

Explanation:

u = 90 km/h = 25 m/s,

theta = 5 degree

acceleration, a = g Sin theta = 9.8 x Sin 5 = 0.854 m/s^2

The final velocity os zero and let the braking distance be s.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = 25 x 25 - 2 x 0.854 x s

s = 366 m

8 0

3 years ago

If you stung by a honey bee you are <br>said with baking soda why<br>?

Karolina [17]

Can you please reword your question I can’t understand

4 0

3 years ago

An 85-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical

Bas_tet [7]

Answer:

The change in gravitational potential energy of the climber-Earth system, $\Delta E=4.49\times 10^5\ J$

Explanation:

Given that,

Mass of the hiker, m = 85 kg

Time, t = 2 h

Vertical elevation of the climber, h = 540 m

We need to find the change in gravitational potential energy of the climber-Earth system. We know that due to change in position of an object, gravitational potential energy occurs. It is given by :

$\Delta E=mgh\\\\\Delta E=85\times 9.8\times 540\\\\\Delta E=4.49\times 10^5\ J$

So, the change in gravitational potential energy of the climber-Earth system is $4.49\times 10^5\ J$ . Hence, this is the required solution.

5 0

3 years ago