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3 years ago

How do you calculate the net force when there are multiple forces in different directions?

Physics

1 answer:

Artyom0805 [142]3 years ago

4 0

To find $F_{net}$ we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a length of 3 pointing directly east, then we use the distance formula to find the length of the net force $F_{net} = \sqrt{(3)^2+(4)^2} \\$ which gives $F_{net} = 5$ . We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using $\theta = \tan^{-1} (\frac{4}{3})$ where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.

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A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

Irina-Kira [14]

Answer:

part (a) $\alpha\ =\ -2.5\ rad/s^2$

part (b) N = 79.61 rev

part (c) $\tau\ =\ 23.54\ Nm$

Explanation:

Given,

Initial speed of the wheel = $w_o\ =\ 50.0\ rad/s$
total time taken = t = 20.0 sec

part (a)

Let $\alpha$ be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

$\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2$

part (b)

Let $\theta$ be the total angular displacement of the wheel from initial position till the rest.

$\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad$

We know, 1 revolution = $2\pi$ rad

Let N be the number of revolution covered by the wheel.

$\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev$

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

Mass of the pole = m = 4 kg
Length of the pole = L = 2.5 m
Angle of the pole with the horizontal axis = $\theta\ =\ 60^o$

Now the center of mass of the pole = $d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m$

Weight component of the pole perpendicular to the center of mass = $F\ =\ mgcos\theta$

$\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm$

3 0

3 years ago

Consider two identical objects released from rest high above the surface of Earth. (Neglect air resistance for this question.)

Nadusha1986 [10]

The question is missing its alternatives. Here is the complete question.

Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.

1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.

a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;

Answer: C) K2 = (4/3)K1

Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - $\frac{G.Me.m}{r}$ + U₀

In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.

The potencial energy of an object of mass m on the surface of the Earth is:

Usurface = - $\frac{G.Me.m}{Re}$

The potencial energy of the object in Case 1 is

U1 = - $\frac{G.Me.m}{2Re}$

For the Case 2:

U2 = - $\frac{G.Me.m}{3Re}$

The potencial change in Case 1:

ΔU1 = - G.Me.m.( $\frac{1}{Re}-\frac{1}{2Re}$ ) = - $\frac{1}{2}$ $\frac{G.Me.m}{Re}$

For Case 2:

ΔU2 = - G.Me.m( $\frac{1}{Re}-\frac{1}{3Re}$ ) = - $\frac{2}{3}$ $\frac{G.Me.m}{Re}$

Comparing ΔK1 and ΔK2 equals comparing ΔU1 and ΔU2:

Δ $\frac{U2}{U1}$ = (-2/3)(-1/2) = 4/3

So, comparing kinetic energies, K2 is 4/3 of K1.

5 0

3 years ago

A paint can is sitting on a ladder 20 m high. It has a mass of 7 kg. What is the

loris [4]

Answer:

<h3>The answer is 1400 J</h3>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 20 × 10 × 7

We have the final answer as

Hope this helps you

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3 years ago

Which statement explains the similarity between momentum and kinetic energy of an object?

brilliants [131]

The correct answer is B. Both are directly proportional to the mass and velocity of the object.

Hope this helps

-AaronWiseIsBae

8 0

3 years ago

Read 2 more answers

How do you time travel?

rosijanka [135]

You don't. Although tons of stories have been written about it,
and loads of scientific speculation about how it maybe possibly
might be done, it's never been done.

6 0

3 years ago

How do you calculate the net force when there are multiple forces in different directions?​

How do you calculate the net force when there are multiple forces in different directions?