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2 years ago

Vector A has a magnitude of 623. What are its west and north components?

Physics

1 answer:

vovangra [49]2 years ago

3 0

Answer:

D) 623 N

Explanation:

The vectors' direction is in the North And when we try to take Y and X components Cos(90°) is Zero ... So the X (West) component will be zero

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Write some interesting facts about isaac newton

Hatshy [7]

Sir Isaac Newton was an English Mathematician,physicist,astronomer,theologian and author who is widely recognized as one of the most influential scientist of all time because of his discovery of gravity(force that attracts a body toward center of earth).

8 0

3 years ago

Estimate the number of photons emitted per second from 1.0 cm2 of a person's skin if a typical emitted photon has a wavelength o

Diano4ka-milaya [45]

Answer:

2.63 x 10^18

Explanation:

A = 1 cm^2 = 1 x 10^-4 m^2

λ = 10,000 nm = 10,000 x 10^-9 m = 10^-5 m

T = 37 degree C = 37 + 273 = 310 k

Energy of each photon = h c / λ

where, h is the Plank's constant and c be the velocity of light

Energy of each photon = (6.63 x 10^-34 x 3 x 10^8) / 10^-5 = 1.989 x 10^-20 J

Energy radiated per unit time = σ A T^4

Where, σ is Stefan's constant

Energy radiated per unit time = 5.67 x 10^-8 x 10^-4 x 310^4 = 0.05236 J

Number of photons per second = Energy radiated per unit time / Energy of

each photon

Number of photons per second = 0.05236 / (1.989 x 10^-20) = 2.63 x 10^18

4 0

3 years ago

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.

Lelechka [254]

Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

Kinetic energy.
Potential energy.

Kinetic energy $=\frac{1}{2} mv^2$

Potential energy = $\frac{Kq_1q_2}{d}$

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

$=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}$

=(1280-337.5)J

=942.5 J

Total energy of a system remains constant.

Therefore,

E $=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}$

$\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750$

$\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750$

$\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}$

$\Rightarrow v= 1961.19$ m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

5 0

3 years ago

When an object is lifted 12 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 2

alexandr402 [8]

I think the answer is twice as much

8 0

2 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

8 0

3 years ago