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3 years ago

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What is the weight of a 435 kg object on earth? 957.98 N 957.98 N None of these answers are correct. None of these answers are c

orrect. 197.53 N 197.53 N 7.20 N 7.20 N 7203 N

1 answer:

mixer [17]3 years ago

7 0

Answer:

none of these option are correct....

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a An object is tarown up with a velocity v = 6.02 +7.0j. Calculate the (1) time taken reach the maximum height (ii) the horizont

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the update is coming down 55th 55 cm long

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The idea that our bodies help our minds think is called Embodied Cognition.

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The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel

aniked [119]

Answer:

= 2.52 x 10^ 6 m/s

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv) ÷ d

Here, d = distance between the two plates

q = charge of the charged particle

Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a = (q) (Δv) ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv) ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv) ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 = ut + 1/2 at^2 [proton}

put d/2 = ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron) ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

Note, e = 1.60 x 10^-19

m(electron) = 9.11 X 10^-31

m(proton) = 1.67 X 10^-27

Input these values into the formula above, initial speed, UI =

= 2.52 x 10^ 6 m/s

7 0

3 years ago

A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by

lana66690 [7]

Answer:

Angle: $48.19^o$

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

$v_b\ cos\alpha$

where $v_b$ is the speed of the boy in still water and $\alpha$ is the angle respect to the shoreline. If the river flows at speed $v_s$ , we now set

$v_b\ cos\alpha=v_s$

$\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}$

$\alpha=48.19^o$

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3 years ago

The __________-second rule applies to any speed in ideal weather and road conditions.

SOVA2 [1]

The two-second rule.

It is a common guideline to follow while driving.

It means that any given driver should be AT LEAST two seconds behind any vehicle that is driving in front of his vehicle. It might apply for any kind of vehicle.

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3 years ago