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Irina-Kira [14]

3 years ago

7

What book did Galileo write that would influence sir issac newton

2 answers:

zimovet [89]3 years ago

4 0

The book Two New Sciences is the book that would later influence Sir Isaac Newton

Romashka [77]3 years ago

3 0

Galileo Galilei wrote The Discourses and Mathematical Demonstrations Relating to Two New Sciences, published in 1638, which was his final book and a scientific testament covering much of his work in physics over the preceding thirty years. A book that would later influence sir Issac Newton.

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Three charges are arranged on the circle as shown in the figure Q1 = Q2 = 20μC. and Q3 = -10 μC on the y-axis. The electric fiel

Gnoma [55]

Explanation:

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don't know but please mark me as brainliest please

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3 years ago

Physical science!!!!!!helpppp

ipn [44]

Answer:

Option B

10.36 m/s

Explanation:

Using the first given equation, then velocity=distance/time

Since distance is provided as 200 m and time, t is 19.3 seconds then substituting these figures yields

v=200/19.3=10.3626943

Rounding off to 2 decimal places, then

v=10.36 m/s

6 0

3 years ago

According to Newton's third law, which describes what happens when a 4.5 kg pumpkin is launched from a catapult with a force of

max2010maxim [7]

3. Law: Every action has a reaction equal in magnitude and opposite in direction.

6 0

3 years ago

A black spider jumps with a velocity of 10 m/s at an angle of 30 degrees. At what time is the spider moving the slowest?

Tpy6a [65]

Answer:

right after jumping so 0s and as it falls its speed will keep increasing

Explanation:

5 0

3 years ago

A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang

Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]$

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]$

$E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]$

Since $\lambda = \frac{Q}{L}$

where;

L = length

Q = charge

λ = density of the charge;

then substituting $\frac{Q}{L}$ for λ, we have :

$E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]$

$E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}$

substituting our given parameter; we have:

$E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}$

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0

3 years ago