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2 years ago

How was mount Everest made? Please i am handing out 100 points

Physics

1 answer:

Simora [160]2 years ago

7 0

Answer:

Mount Everest formed from a tectonic smashup between the Indian and Eurasian tectonic plates tens of millions of years ago.

Explanation:

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During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 31.7 rad/s. Find the angular displ

Crank

Answer:

$\Delta \theta = 3116.11\,rad$ and $\Delta \theta = 495.944\,rev$

Explanation:

The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement ( $\Delta \theta$ ), measured in radians, is:

$\Delta \theta = \omega \cdot \Delta t$

Where:

$\omega$ - Steady angular speed, measured in radians per second.

$\Delta t$ - Time, measured in seconds.

If $\omega = 31.7\,\frac{rad}{s}$ and $\Delta t = 98.3\,s$ , then:

$\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)$

$\Delta \theta = 3116.11\,rad$

The change in angular displacement, measured in revolutions, is given by the following expression:

$\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)$

$\Delta \theta = 495.944\,rev$

6 0

3 years ago

A large stick is pivoted about one end and allowed to swing back and forth with no friction as a physical pendulum. The mass of

Tamiku [17]

Answer:

$I = 94.33 kg m^2$

Explanation:

Let say the rod is slightly pulled away from its equilibrium position

So here net torque on the rod due to its weight is given as

$\tau = mg dsin\theta$

since rod is pivoted at distance of 1.4 m from centre of gravity

so its moment of inertia about pivot point is given as

$Inertia = I$

now we have

$I \alpha = mg d sin\theta$

now for small angular displacement we will have

$\alpha = \frac{mgd}{I}\theta$

so angular frequency of SHM is given as

$\omega = \sqrt{\frac{mgd}{I}}$

now we will have

$\frac{2\pi}{9} = \sqrt{\frac{4.8(9.8)1.4}{I}$

$I = 94.33 kg m^2$

5 0

3 years ago

A 5 kg bucket is lifted from the ground into the air by pulling in 10 meters of rope with linear density of 2 kg/m at a constant

iren [92.7K]

Answer:

Workdone W = 1465.1 J

Explanation:

The weight of the water = density × volume

weight of the water = 1000 kg/m³ × 100 cm³

weight of the water = 1000 kg/m³ × 0.0001 m³

weight of the water = 0.1 kg

weight of the bucket = 5 kg

weight of the rope = $2*10- 2x\\\\$

= $20- 2x$

Leakage = $\frac{0.1}{10} * x$

= $0.01 x$

Total weight = $5 + 20 - 2x - 0.01 x$

= $25 - 2.01 x$

Force = wg

Force = ( $25 - 2.01 x$ )g

Force = 9.8 ( $25 - 2.01 x$ )

Finally; the amount of work spent in lifting the bucket, rope, and water is calculated as follows:

$W = \int\limits^{10}_0 {(25-2.01 x)} \, 9.8 dx \\\\W = 9.8 (25x - \frac{2.01x^2}{2}})^{10}__0\\\\$

$W = 9.8 (25*10-2.01*50)\\\\W = 1465.1 \ \ J$

7 0

4 years ago

A 3900 kg truck is moving at 6.0 m/s what is the kinetic energy

Stolb23 [73]

Answer:

70200J

Explanation:

k.E = 1/2mv^2

K.E = 1/2(3900)(6)^2

3 0

3 years ago

Read 2 more answers

A sailboat took 25 hours to cover 1/4 of a journey. Then, it

ozzi

Answer:

Explanation:

4 0

3 years ago