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3 years ago

What is net force?

Physics

1 answer:

kramer3 years ago

3 0

Answer:

A. The sum of all the forces acting on an object.

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At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. Th

jekas [21]

Answer:

2.2 m/s

Explanation:

<u>solution:</u>

To calculate change in stored energy at desired extension

ΔU = 1/2*k*(δx)^2

= 1/2*3700*(0.37^2-0.180^2)

= 201 N.m

use work energy theorem

ΔU = ΔK = 1/2*m*v^2 = 201

= 2.2 m/s

calculation maybe wrong but method is correct.

8 0

3 years ago

Read 2 more answers

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of

sweet [91]

Answer:

$1.44\times 10^{-3}$ N

Explanation:

We are given that three charged particle are placed at each corner of equilateral triangle.

$q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC$

$q_1=-8.2\times 10^{-9} C$

$q_2=-16.4\times 10^{-9} C$

$q_3=8.0\times 10^{-9} C$

Side of equilateral triangle =3.3 cm= $\frac{3.3}{100}=0.033m$

We know that each angle of equilateral angle= $60^{\circ}$

Net force=F = $\sum\frac{kQq }{d^2}$

Where k= $9\times 10^9 Nm^2/C^2$

If we bisect the angle at $q_3$ then we have 30 degrees from there to either charge.

Direction of vertical force due to charge $q_1$ and $q_2$

Therefore, force will be added

Vertical force= $9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})$

Vertical net force= $9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}$

Vertical force = $9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}$

Vertical force= $-1.41\times 10^{-3}N$ (towards $q_1}$

Horizontal component are opposite in direction then will b subtracted

Horizontal force= $9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}$

Horizontal force= $0.27\times 10^-3}$ N(towards $q_2$

Net electric force acting on particle 3 due to particle = $\sqrt{F^2_x+F^2_y}$

Net force= $\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}$

Net force= $1.44\times 10^{-3}$ N

3 0

3 years ago

A hypothesis is a(n) ______________

marusya05 [52]

Answer:

It is an educated guess

Explanation:

let me know if the bottom ones need to be answered too.

6 0

3 years ago

If the angular magnification of an astronomical telescope is 42 and the diameter of the objective is 72 mm, what is the minimum

andriy [413]

Answer:

Minimum diameter=1.714

Explanation:

Angular magnification=fob/fey=42

fob=40*fey

Hence minimum diameter=diameter of objective/angular magnification

minimum diameter=72/42=1.714

4 0

2 years ago

A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the fr

vlada-n [284]

Answer:

50 m

Explanation:

F = ma

10 N = (10 kg) a

a = 1 m/s²

Given:

v₀ = 0 m/s

a = 1 m/s²

t = 10 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²

Δx = 50 m

5 0

2 years ago