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3 years ago

Scientists have found that the most destructive and deadly tornadoes occur from rotating thunderstorms called

Physics

2 answers:

Julli [10]3 years ago

8 0

Answer:

Supercells

Explanation:

supercells are rotating thunderstorms that has a well-defined radar circulation called a mesocyclone. They can sometimes produce destructive hail, severe winds, frequent lightning, and flash floods.

dezoksy [38]3 years ago

4 0

Answer:

Supercells

Explanation:

edge2020

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2. Find the time taken by the bus to reach the stop. need only group B, 2 answer

Molodets [167]

Answer:

t = 2 seconds

Explanation:

In 2nd question, the question is given the attached figure.

Initial speed of the bus, u = 0

Acceleration of the bus, a = 8 m/s²

Final speed, v = 16 m/s

We need to find the time taken by the car to reach the stop. Acceleration of an object is given by :

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

t is time taken

$t=\dfrac{v-u}{a}\\\\t=\dfrac{16-0}{8}\\\\t=2\ s$

The bus will take 2 seconds to reach the stop.

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3 years ago

sunlight, wind, and running water are essentially "free." Yet renewable energy sources are a very small part of our energy consu

dimulka [17.4K]

The heat and energy of coal burning is more easier and more efficient to turn into electricity than any method using renewable resources.

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3 years ago

Read 2 more answers

For phyics again , pls !!!

CaHeK987 [17]

Answer: same

Explanation: They both weigh a kilogram and there is no friction

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2 years ago

If a pendulum has a period of 4 seconds at the north pole with gravity force 9.83 ms^-2 and a 3.97m pendulum length, how could t

sladkih [1.3K]

Answer:

The same pendulum could be adjusted to have the same period, in the equator must have a length of 3.949m.

Explanation:

Tnp= 4 sec

gnp= 9.83 m/sec²

Lnp= 3.97m

Tequ= 4 sec

gequ= 9.78 m/sec²

Lequ=?

Lequ= (Lnp* gequ) / gnp

Lequ= 3.949 m

5 0

2 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