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3 years ago

What's the momentum of a 3.5-kg boulder rolling down hill at 5 m/s

Physics

1 answer:

ICE Princess25 [194]3 years ago

6 0

P = mv
p = 3.5 × 5
p = 17.5 kg .m/s

Hope this helps!

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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

2 years ago

Two Forces are exerted on an Object In a Vertical Direction: A 20N force downwards, and 10 N force upwards. The mass of the obje

SVEN [57.7K]

When you add up (20N down) and (10N up),
you get a sum of (10N down).

The mass of the object has no effect on the forces.

Now ... I see 5 points for the answer.
Where are the other 25 coming from ?

6 0

3 years ago

An instrument used to detect the current in a<br> circuit is called

Tanzania [10]

Answer:

Ammeter

Explanation:

An ammeter is used to measure current.

4 0

2 years ago

Read 2 more answers

An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown

stiv31 [10]

The constant acceleration of the object is $0.414\ m/s^2$ .

<h3>Acceleration of the object</h3>

The acceleration of the object is constant, and the values at the three different positions is the same.

<h3> For the first position and time</h3>

$s = v_0 t + \frac{1}{2} at^2\\\\37.7 = v_o(9.6) +\frac{1}{2} a(9.6)^2\\\\ 37.7 = 9.6v_0 + 46.08a$

<h3>For the second position and time</h3>

$39.3 = v_o(18.048) +\frac{1}{2} a(18.048)^2\\\\ 39.3 = 18.048v_0 + 162.87a$

<h3>Solve the first and second equation together</h3>

$37.7 = 9.6v_0 + 46.08a\\\\39.3 = 18.048v_0 + 162.87a\\\\18.048: \ \ 680.41 = 173.261v_0 + 831.651a\\\\9.6: \ \ \ \ -(377.28 = 173.261 v_0+ 1563.552a)\\------------------\\303.13 = -731.901a\\\\-a = \frac{303.13}{731.901} \\\\-a = 0.414 \ m/s^2\\\\|a| = 0.414 \ m/s^2$

Thus, the constant acceleration of the object is $0.414\ m/s^2$ .

Learn more about acceleration here: brainly.com/question/14344386

4 0

1 year ago

ASE (National Institute for Automotive Service Excellence) certified professionals must be retested every _______ years to keep

Bess [88]

Hi,

The correct answer is C. five

I hope I helped :)

4 0

3 years ago