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2 years ago

If the mass of an ideal pendulum is doubled how will the oscillation period change?

1 answer:

5 0

Answer: Doubling the mass of the bob will half the period. b) If the length is doubled, the period will increase by a factor of √2 . The mass of the bob will not effect the period

Explanation:

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How did Archimedes measure the mass of the block of gold?

MA_775_DIABLO [31]

Archimedes found a piece of gold and a piece of silver with exactly the same mass. He dropped the gold into a bowl filled to the brim with water and measured the volume of water that spilled out. Then he did the same thing with the piece of solver. Although both metals had the same mass, the silver gad a larger volume; therefore, it displaced more water than the gold did. That's because the silver was less dense than gold. Afterwards he applied the same method to the crown for the king he served who had got a new crown from a jeweler who gave it to him. Archimedes found a piece of pure gold that had the same mass as the crown. He placed the pure gold chuck and the crown in water, one at a time. The crown displaced more water the piece of gold. Therefore, its density was less than pure gold.

5 0

3 years ago

What is the momentum of a 200 kg football player running west at a speed of 8m/s

lutik1710 [3]

Answer:

<h3>The answer is 1600 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 200 kg

velocity / speed = 8m/s

We have

momentum = 200 × 8

We have the final answer as

Hope this helps you

7 0

3 years ago

The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t>0 , where is measured in seconds and

8090 [49]

The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at $0$ .

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>

A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

$v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s$

The velocity is positive at when the time of motion is as follows;

$0$ .

The total distance traveled in the first 10 seconds is calculated as follows;

$2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m$

Learn more about motion of particles here: brainly.com/question/11066673

4 0

2 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

How are theories supported in the scientific community?

mafiozo [28]

They are unproven but accepted as fact.
Many experiments support them but they can be disproven by the results of a single experiment. Until then, they stand.
The third statement is correct.

7 0

3 years ago