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2 years ago

Which scientist first proposed physical laws to mathematically describe the effect of forces on the motions of bodies?

Physics

2 answers:

Basile [38]2 years ago

6 0

Answer:

Issac Newton!

Explanation:

Hope this helps.

vodomira [7]2 years ago

3 0

Answer: Isaac Newton

Explanation:

Isaac NewtonThe three laws of motion were first compiled by Isaac Newton in his Philosophiæ Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), first published in 1687. Newtonused them to explain and investigate the motion of many physical objects and systems.Feb 21, 2017

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A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

3 years ago

Bacteria prefer a dry environment to a damp or moist environment. False True

mariarad [96]

Answer:

false

Explanation:

cause bacteria prefers moist and damp environment

7 0

2 years ago

Read 2 more answers

An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed

Tema [17]

Let the mass of 2500 kg car be $m_1$ and it's velocity be $v_1$ and the mass of 1500 kg car be $m_2$ and it's velocity be $v_2$ .

After the bumping the mass be M and it's velocity be V.

By law of conservation of momentum we have

$m_1v_1+m_2v_2 = MV$

2500 * 5 + 1500 * 1=4000 * V

V = 14000/4000 = 7/2 = 3.5 m/s

So the velocity of the two-car train = 3.5 m/s

9 0

2 years ago

a radio station broadcast a frequency 90500 Hz. these radio waves travel at speed of 30000 m/s. what is the wavelength of the ra

Feliz [49]

Answer:

0.331m

Explanation:

wave equation: v=f λ

v=30000m/s

f=90500 Hz

λ= $\frac{v}{f}$

λ= $\frac{30000}{90500}$

λ= 0.311m

5 0

1 year ago

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown

ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

$P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,$

where $h_1$ and $h_2$ are the height of the column of oil and the unkown liquid, respectively. Writing for $\gamma_{unk}$ , we have

$\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.$

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

3 0

2 years ago