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Eduardwww [97]
3 years ago
5

I will give brainliest) According to Newton's second law of motion, when an object is acted on by an unbalanced force, how will

that object respond?
Physics
2 answers:
Dafna11 [192]3 years ago
5 0
The object that is acted on by an unbalanced force will accelerate.

evablogger [386]3 years ago
4 0
Newton's<span> first </span>law of motion<span> has been frequently stated throughout this lesson. An</span>object<span> at rest stays at rest and an </span>object<span> in </span>motion<span> stays in </span>motion<span> with the same speed and in the same direction unless </span>acted<span> upon by an </span>unbalanced force<span>.</span>
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Explanation:

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4 years ago
V=I/R correctly expresses the relationship between voltage, current, and resistance.
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The correct answer is true
5 0
2 years ago
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Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
I need HELP Please !!
aliya0001 [1]
You have to figure it out
5 0
3 years ago
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23
Burka [1]

Answer:

a) v(2) = 3.9\,\frac{m}{s}, b) v(4) = -15.7\,\frac{m}{s}

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

v(t) = 23.5 -9.8\cdot t

The velocities at given instants are, respectivelly:

v(2) = 3.9\,\frac{m}{s}

v(4) = -15.7\,\frac{m}{s}

8 0
3 years ago
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