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4 years ago

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Two forces whose resultant is 100N,are perpendicular to each other.if one of them makes an angle of 60° with the resultant, calc

ulate it's magnitude
Calculate the absolute pressure of an ocean depth of 1200atm. Assuming the density of liquid is 1200kg/m^2amd that (pa=1.01×10^6pa N/m^2

1 answer:

DIA [1.3K]4 years ago

5 0

Magnitude of the force = 50N

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<u>Explanation:</u>

Let one force be X

And the second force be Y

Both the forces are perpendicular, so α = 90°

The resultant R of the two forces = 100N

Angle between resultant and 1 force, α = 60°

We know,

R² = X² + Y² + 2XYcos 90

100² = X² + Y²

If X makes an angle of Φ = 60°

Then, tan Ф = Y sin α / X + Y cos α

Putting the values we find,

tan 60 = Y/X

Y = √3X

Putting Y = √3X in equation 1 we get,

(X)² + (√3X)² = 100 X 100

4X² = 100 X 100

X² = 2500

X = 50N

Therefore, magnitude of the force = 50N

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A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from

ivann1987 [24]

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$ (1)

Where:

$I_{T}$ - Moment of inertia of the turntable, in kilogram-square meters.

$r$ - Distance of the block regarding the center of the turntable, in meters.

$m$ - Mass of the object, in kilograms.

$\omega_{o}$ - Initial angular speed of the turntable, in radians per second.

$\omega_{f}$ - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$ (2)

Where $M$ is the mass of the turntable, in kilograms.

If we know that $\omega_{o} \approx 7.330\,\frac{rad}{s}$ , $M = 1.5\,kg$ , $m = 0.54\,kg$ and $r = 0.1\,m$ , then the angular speed of the turntable after the event is:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$

$I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}$

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$

$\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}$

$\omega_{T} = 3.004\,\frac{rad}{s}$ ( $28.687\,\frac{rev}{min}$ )

The turntable's angular speed after the event is 28.687 revolutions per minute.

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3 years ago

If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol

Mrrafil [7]

Answer:

The maximum wavelength is 492 nm.

Explanation:

Given that,

Angular separation $\theta=3.0\times10^{-5}\ rad$

Suppose a telescope with a small circular aperture of diameter 2.0 cm.

We need to calculate the maximum wavelength

Using formula of angular separation

$\sin\theta=\dfrac{1.22\lambda}{d}$

$\lambda=\dfrac{d\sin\theta}{1.22}$

Put the value into the formula

$\lambda=\dfrac{2.0\times\sin(3\times10^{-5})}{1.22}$

For small angle $\sin\theta\approx\theta$

$\lambda=\dfrac{0.02\times3\times10^{-5}}{1.22}$

$\lambda=4.92\times10^{-7}\ m$

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Hence, The maximum wavelength is 492 nm.

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3 years ago

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Gravitational force is very weaker;

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3 years ago

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hoa [83]

Answer: 0.6m

Explanation:

Given that:

force = 4.5 N

Work done = 2.7J

Distance moved by the book = ?

Since work is done when force is applied on an object over a distance, apply the formula:

work = force x distance

2.7J = 4.5N x distance

Distance = (2.7J / 4.5N)

Distance = 0.6 m

Thus, the book was moved 0.6 metres far

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3 years ago

When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where i

Gnoma [55]

Answer:

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The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.

The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.

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3 years ago