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2 years ago

1,What were you doing when you felt most passionate and alive?

Physics

2 answers:

Luba_88 [7]2 years ago

4 0

Answer:

wen I was in the car toing home from school after a bad day n si si I have crazzyyy

yanalaym [24]2 years ago

3 0

Answer:-

1. I feel the most passionate and alive when I watch BTS, because they inspir me to do many things such as Loveing Myself and doning what I love in life and chase my dream. I also feel alive when playing soccer, dancing, playing piano, or singing/rapping, because these things are what I love to do and always loved to do.

2. Yes, I feel like I live everyday the same day, some little things change but the things I do every day are the same, never changes. Wake up, eat, homework, clean, eat, tv, eat, and sleep.

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What mass of water will fill a tank that is 100.0 cm long, 50.0 cm wide, and 30.0 cm high? Express the answer in grams.

Bond [772]

$calculate\ mass\ from\ formula\ for\ density:\\\\
density=volume*mass\\
mass=\frac{density}{volume}\\\\
volume=width*length*height\\
volume=100*50*30=150000cm^3=0,15m^3\\\\
density\ of\ water=\ 999.9720 \frac{kg}{m^3} \\\\mass=0,15*999.9720=149,9958kg=149995,8grams\\\\ Answer\ is\ :\\water\ which\ weight\ is\ equal\ to\ 149995,8grams.
$

4 0

3 years ago

A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plan

pshichka [43]

Answer:

a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

Wₓ = 1200 sin 30 = 600 N

W_y = 1200 cos 30 = 1039.23 N

Y axis

N- W_y = 0

N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

P -Wₓ -fr = 0

P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

fr = μ N

fr = 0.20 1039.23

fr = 207.85 N

we substitute

P = 600+ 207.85

P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

P + fr -Wx = 0

fr = μ N

fr = 0.20 1039.23

fr = 207.85 N

we substitute

P = Wₓ -fr

P = 600 - 207,846

P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

P - Wₓ + fr = 0

P = Wₓ - fr

fr = 0.15 1039.23

fr = 155.88 N

P = 600 - 155.88

P = 444.12 N

6 0

3 years ago

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

5 0

3 years ago

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu

Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰, force = 4.07 N

(b) When the angle, θ = 90 ⁰, force = 4.7 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;