Choose the false statement regarding resistance.

3.How can you find the velocity of the object just before impact without directly measuring it?

e-lub [12.9K]

Ugudbis sinwajdiekbdudjwndjid

7 0

4 years ago

An atom is the most basic particle that represents what type of matter?

vagabundo [1.1K]

Answer:

b

Explanation:

8 0

3 years ago

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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago

the total mass of the sun is about 2×1030 kg, of which about 76 % was hydrogen when the sun formed. however, only about 14 % of

likoan [24]

The total amount of mass in the Sun is 2.0 x 10^30 kg, 5% of whig is hydrogen, and 13% of which becomes available for fusion. Thus, the total mass of hydrogen available for fusion over the Sun's lifetime is simply 13% of 75% of the total mass of the Sun or:

2.0 x 10^30 kg x .75 x .13

= 1.95 x 10^29 kg

Nuclear fusion occurs only in the core of the sun where temperature pressure and density are highest. The photosphere can be seen with visible light telescopes, the chromosphere with ultraviolet telescopes, and the corona most easily with X-ray telescopes.

The Sun is a typical star and also the closest star to the Earth. It is composed of 73% hydrogen, 25% helium, and 2% other elements. Since the gravitational pull of the sun on the earth is the centripetal force that causes the earth to move in a circular motion around the sun, we can use Newton's law of universal gravitation to find the mass of the sun without visiting it.

Learn more about The temperature here:- brainly.com/question/24746268

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3 0

1 year ago

A rock with a mass of 16 kilograms is put aboard an airplane in New York City and flown to Boston. How much work does the gravit

Law Incorporation [45]

The work done by the gravitational field of the earth on the rock is 9.998 x 10⁸ J.

The given parameter include:

the mass of the object, m₁ = 16 kg

Note: the mass of the earth, m₂ = 5.972 x 10²⁴ kg

The work done by the gravitational field of the earth is given as;

Work done = gravitational force (F) x radius of the earth (R)

$Work \ done = \frac{Gm_1m_2}{R^2} \times R\\\\Work \ done = \frac{Gm_1m_2}{R} \\\\where;\\\\R \ is \ the \ radius \ of \ the \ earth = 6,378 \ km = 6,378,000 \ m\\\\G \ is \ the \ universal \ gravitation \ constant = 6.674 \times 10^{-11} Nm^2/kg^2\ \\\\Work \ done = \frac{(6.674 \times 10^{-11} ) \times (5.972\times 10^{24}) \times (16)}{6,378,000 } \\\\Work \ done = 9.998 \times 10^{8} \ J$

Therefore, the work done by the gravitational field of the earth on the rock is 9.998 x 10⁸ J.

To learn more about work done by gravitational field of the earth visit: brainly.com/question/13934028

7 0

3 years ago