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2 years ago

O ln a hydraulic press , a force Of 400 N is exerted on

Physics

1 answer:

egoroff_w [7]2 years ago

7 0

Explanation:

Pressure = force / area

P = 400 N / 0.001 m²

P = 400,000 Pa

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3 years ago

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3 years ago

Newton’s first law states that if an objects velocity is changing a _____ must be acting on it.

klio [65]

<h3>Answer: </h3>

(a)

<h3>Explanation </h3>

Newtons First Law can be written as the formula

$f = ma$

since acceleration is the changing of velocity as its formula is

$\frac{v2 - v1}{t}$

Thus, a force must be acting on an accelerating/decelerating object

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2 years ago

A 240 N sphere 0.20 m in radius rolls, without slipping 6.0 m downa

Shalnov [3]

Answer:

The angular speed of the sphere at the bottom of the hill is 31.39 rad/s.

Explanation:

It is given that,

Weight of the sphere, W = 240 N

Radius of the sphere, r = 0.2 m

Angle with the horizontal, $\theta=28^{\circ}$

We need to find the angular speed of the sphere at the bottom of the hill if it starts from rest.

As per the law of conservation of energy, the total energy at the top is equal to the energy at the bottom.

Gravitational energy = translational energy + rotational energy

So,

$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

I is the moment of inertia of the sphere, $I=\dfrac{2}{5}mr^2$

Also, $v=r\omega$

h is the height of the ramp, $h=l\ sin\theta$

$mgl\ sin\theta=\dfrac{1}{2}m(r\omega)^2+\dfrac{1}{2}I\omega^2$

On solving the above equation we get :

$\omega=\sqrt{\dfrac{10gl\ sin\theta}{7r^2}}$

$\omega=\sqrt{\dfrac{10\times 9.8\times 6\ sin(28)}{7(0.2)^2}}$

$\omega=31.39\ rad/s$

So, the angular speed of the sphere at the bottom of the hill is 31.39 rad/s. Hence, this is the required solution.

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2 years ago

At atmospheric pressure, what is the characteristic boiling point of water,in degrees Celsius

Marina86 [1]

At standard atmospheric pressure, 100° C
is defined as the boiling point of water.

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3 years ago