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S_A_V [24]
2 years ago
7

what would the net force be on the box in the problems shown below.( both force and direction).​ for all four diagrams. please e

xplain answers.​

Physics
1 answer:
DIA [1.3K]2 years ago
8 0

Answer:
A) object moves 20 N [West] or -20 N [East]
B) object moves 6 N [South] or -6 N [North]
C) object moves 90 N [West] or -90 N [East]
D) object does not move and is at rest*

*Rest means 0


Why:

A)both forces from north and south that are pushing against the object neutralize each other. Assume that north is positive and south is negative: 20 [N] + (-20) [S] = 0
On West and east, you can see that west has a greater force. Assume that west is negative and east is positive: 50 [E] + (-70) [W] = -20 [E]
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PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
Anastaziya [24]

Answer:

As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.

This will mean that the denser objects will always go to the bottom.

This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.

There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.

The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.

6 0
2 years ago
The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m
BARSIC [14]

A) The resultant force is 30.4 N at 25.3^{\circ}

B) The resultant force is 18.7 N at 43.9^{\circ}

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

F_1 = 20 N at 0^{\circ} above x-axis

F_2 = 15 N at 60^{\circ} above y-axis

Resolving each force:

F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N

F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N

So, the components of the resultant are:

F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude of the resultant is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

\theta=180^{\circ}-60^{\circ}=120^{\circ}

So we have:

F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N

So, the components of the resultant this time are:

F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
3 years ago
Can someone please help me with science.
kherson [118]

Answer:

Yes

Explanation:

5 0
3 years ago
Read 2 more answers
Are colavent bonds more likely to be found in compounds contaning both materals and nonmetals or compounds contaning only nonmet
atroni [7]

Answer:

bonds between 2 alike atoms or most non-metals get a covalent bond when reacting to the ... It tells what kinds of atomas are in a compound and how many.

Hope this helped

-Gavin

7 0
2 years ago
Read 2 more answers
Devise an exponential decay function that fits the given​ data, then answer the accompanying questions. Be sure to identify the
7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

P=P_0e^{-kh}

where,

P_0 =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

P=\dfrac{1}{3}P_0

\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}

Now

P=\dfrac{1}{2}P_0

ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft

The altitude will be 22145.27733 ft

P=0.02P_0

0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft

The elevation is 124984.76055 ft

6 0
2 years ago
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