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4 years ago

How to Identify Newtons Three Laws in my Own Words?

Physics

1 answer:

ipn [44]4 years ago

5 0

I understand the struggle. Here's a way to do it.

Identify Newton's Three Laws, and let it be important that you understand what you're reading. If there's a passage, read the passage/text and look for any important detail you should add to your piece, and add any information you have learned from the passage/text and put it into your own words. You should make sure that the information you've written down is consistent with the passage/text and that your phrasing is not too close to the original.

-I anticipate that this would help you.

You might be interested in

One or more calcium ions (Ca+2) combine with one or more chlorine ions (Cl-1). What is the resulting formula?

N76 [4]

The formula is CaCl2 as you need two Cl ions in order to combine with one Ca ion in order to form a neutral compound i.e. You need to have the same amount of charges therefore you have two Cl ions which have -1 charges for one +2 Ca ion.
Hope this helps :).

5 0

4 years ago

Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou

SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

$\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j$

Total Force

$F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j$

The Resultant Force is given as

$|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb$

For(ii) angle

We can find the angle bu using tanα=y/x

$tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}$

7 0

4 years ago

A 55-kg mountain climber, starting from rest, climbs a vertical distance of 701 m. At the top, she is again at rest. In the proc

aksik [14]

Answer:

7.75%

Explanation:

mass of climber, m = 55 kg

height, h = 701 m

Qc = 4.5 x 10^6 J (heat exhaused by the body)

Work = m x g x h

W = 55 x 9.8 x 701

W = 377839 J

W = QH - Qc

Where, QH is the heat input

QH = 377839 + 4.5 x 10^6

QH = 4877839 J

So, the efficiency

e = W / QH

e = 377839 / 4877839

e = 0.0774 = 7.75 %

Thus, the efficiency of the body is 7.75 %.

3 0

3 years ago

An electron is released from rest in a uniform electric field of 418 N/C near a particle detector. The electron arrives at the d

Oksanka [162]

Answer:

a) 7.35 x 10¹³ m/s²

b) 5.03 x 10⁻⁸ sec

c) 9.3 cm

d) 6.23 x 10⁻¹⁸ J

Explanation:

E = magnitude of electric field = 418 N/C

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of the electron = 9.1 x 10⁻³¹ kg

acceleration of the electron is given as

$a = \frac{qE}{m}$

$a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}$

a = 7.35 x 10¹³ m/s²

v = final velocity of the electron = 3.70 x 10⁶ m/s

v₀ = initial velocity of the electron = 0 m/s

t = time taken

Using the equation

v = v₀ + at

3.70 x 10⁶ = 0 + (7.35 x 10¹³) t

t = 5.03 x 10⁻⁸ sec

d = distance traveled by the electron

using the equation

d = v₀ t + (0.5) at²

d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²

d = 0.093 m

d = 9.3 cm

Kinetic energy of the electron is given as

KE = (0.5) m v²

KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²

KE = 6.23 x 10⁻¹⁸ J

4 0

4 years ago

You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of oscillations, does

eimsori [14]

Answer:

The speed of the wave remains the same

Explanation:

Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.

We observed that the speed, v is independent of the frequency of the wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.

<u>So, If you increase the frequency of oscillations, the speed of the wave remains the same.</u>

4 0

3 years ago