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3 years ago

8a.The mass of a girl is 40 kg. Calculate her weight. (g = 9.8 m/s)

Physics

2 answers:

Archy [21]3 years ago

8 0

Answer:

The right answer of 8 a is 392 Newton

The right answer of 8b is 10 kg

Look at the attached picture

Hope it will help you

Good luck on your assignment

yaroslaw [1]3 years ago

3 0

Explanation:

Here,

Given,

Mass(m)=40 kg

Gram=9.8m/s

Now,

Weight=m x g

or, weight= 40x9.8

=392.0

Hope you have understood

Mark me as brainliest plz

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What is the function of eye lens of the human eye

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Answer:

Lens of the human Eye is a important and one of that most complex sense organ.

Explanation:

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2 years ago

A water balloon is thrown at 20 m/s from the top of a 20 m high building, what is its speed when it hits the ground? Does the an

Oksana_A [137]

Answer:

The final velocity is 28.14 m/s

Yes the angle of projection matters

Explanation:

Given;

initial velocity of the water balloon, u = 20 m/s

height of the building, h = 20 m

let the final speed of the ball when it hits the ground = v

The final speed is calculated as follows;

v² = u² + 2gh

v² = (20)² + 2(9.8)(20)

v² = 400 + 392

v² = 792

v = √792

v = 28.14 m/s

Yes the angle matters, if the balloon had been dropped at a certain angle, the final velocity would have been estimated using the following formula;

$v_y^2 = u_y^2 sin^2(\theta) + 2gh_y$

where;

θ is the angle of projection, which accounts for the vertical component of the velocity.

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Halogen is most likely to react to what

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Halogens therefore react most vigorously with Group 1 and Group 2 metals of all main group elements.

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2 years ago

A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc

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Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

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K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

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V² = 7.84×10^-13 / 3.32×10^-27

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V=√2.36×10^14

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F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

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