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Allisa [31]
3 years ago
9

Calculate the Force applied on a body of mass 100 KG Moving with the velocity 20 m/s so that it can be brought into rest in five

seconds also calculate the distance covered
Plz reply fast
Physics
1 answer:
bearhunter [10]3 years ago
7 0

Explanation:

v = u + at

0=20+a*5

a=-4

--------------

now,

f = ma

F=ma=100*-4=-400N

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Korolek [52]

Answer: C) divide: distance ÷ velocity

Explanation:

The velocity V equation is distance d divided by time t:

V=\frac{d}{t}

If we isolate t we will have:

t=\frac{d}{V}

Hence, the correct option is C: distance divided by velocity.

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HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (20pts)
anyanavicka [17]

Explanation:

potential energy =360800J

mass(m)=?

height (h)=25m

g=9.8m/s²

we have

potential energy =360800J

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If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

7 0
3 years ago
Is the following true or false? Continental glaciers fill depressions in the surface with water, where they create cirques.
Lorico [155]
False is the answer to your question
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3 years ago
A balloon is ascending at a rate of +4.00 m/s to a height of 11.0 m above the ground when a package is dropped. In the absence o
o-na [289]

Answer:

Vf = 14.7 m/s

Explanation:

Vf² = Vi² + 2 * a * Δy

given:

a = 9.81 m/s²

Δy = 11m

Vi = 0 when upon release

Vf² = 0 + 2 (9.81) 11

Vf = 14.7 m/s

6 0
3 years ago
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