Which of the following hypotheses is written correctly?*

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

7 0

2 years ago

A scientist travelled 80 kilometres each day in the buggy.

7nadin3 [17]

Explanation:

In one day a scientist can travel = 80 km

In 10 days. a scientist can travel = 80* 10

= 800 km.

4 0

3 years ago

Read 2 more answers

Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to v

butalik [34]

Answer:

$u=34cm$

Explanation:

From the question we are told that:

Far point is $V=34 cm$

Near point is $u=17 cm$

Therefore

Focal Length

$f=-34cm$

Generally the equation for the Lens is mathematically given by

$\frac{1}{u}=\frac{1}{f}-\frac{1}{v}$

$\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}$

$u=34cm$

5 0

2 years ago

What does the potential energy diagram of a chemical reaction tell you

sattari [20]

By looking at the potential energies before and after the reaction, we can tell that the reaction is exothermic (final < initial) or endodermic (final > initial).

Also, the amount of activation energy gives an idea of the external energy required to initiate the reaction (for example, by heating the reactants).
Furthermore, by the same principle, we can also deduce the activation energy for the reverse reaction.

If a catalyst is available, the diagram will show a reduced activation energy, compared to a reaction without catalyst. However, it will also show that the catalyst does not alter the initial and final energies of the reaction.

4 0

3 years ago

Read 2 more answers

Modify how could you charge the electric circuit shown below to allow lightbulb a to stay lit even if lightbulb b is removed fro

shepuryov [24]

When a circuit is complete, or closed, electrons can flow from one end of a battery all the way around, through the wires, to the other end of the battery. Along its way, it will carry electrons to electrical objects that are connected to it – like the light bulb – and make them work!

5 0

2 years ago