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Brums [2.3K]
3 years ago
6

What did the scientist Robert Hooke claim ?

Physics
2 answers:
trapecia [35]3 years ago
7 0
Robert Hooke<span> (1635 - 1703) The Englishman </span>Robert Hooke<span> (18th July 1635 - 3rd March 1703) was an architect, natural philosopher and brilliant </span>scientist, best known for his law of elasticity (Hooke's<span> law), his book Micrographia, published in 1665 and for first applying the word "cell" to describe the basic unit of life.</span>
alexgriva [62]3 years ago
7 0
Hooke was a brilliant scientist and his famous for his law of elasticity ( Hooke's law )

But he was also the first to apply the word "cell" to describe the basic unit of life.
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If you were stuck in the center of the bridge when it was swaying, where would be the safest place to walk back to land?
Ludmilka [50]
Straight
You already have to momentum of walking forward, and going back and forth are the same distance. If you go back then you would have to stop, turn and walk, but if you go forward you just have to walk.
4 0
3 years ago
A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an
AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0
3 years ago
What is the resistance of a voltage of 65 V and a current of 2.2 A? Include units.
kozerog [31]

Answer:

29.5 ohms

Explanation:

R= V/I

= 65 / 2.2

= 29.5 ohms

4 0
3 years ago
An equipoterntial surface that surrounds a + 3.0 pC point charge has a radius of 2.0 cm. What is the potential of this surface?​
mestny [16]

Answer:

Electric potential = 0.00054 V

Explanation:

We are given;

Charge; q = 3 pC = 3 × 10^(-12) C

Radius; r = 2 cm = 0.02 m

Formula for the electric potential of this surface will be;

V = kqr

Where;

K is a constant = 9 × 10^(9) N⋅m²/C².

Thus;

V = 9 × 10^(9) × 3 × 10^(-12) × 0.02

V = 0.00054 V

8 0
3 years ago
An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),
Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

From diagram we can write as

h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph

5 0
3 years ago
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