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3 years ago

Describe the motion of the box.

Physics

2 answers:

muminat3 years ago

8 0

The answer is A) The box will not move.

Since both of them are applying the same force (50 N) from both sides, the object does not move and the forces are balanced.

Andrej [43]3 years ago

6 0

I think its D but I'm not sure :/

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Please someone answer this

blondinia [14]

Answer:

you just have to draw a line from the eye reflecting from the mirror to the object shown.

Explanation:

Dont come at me if its wrong. I think thats what their asking of you.

8 0

3 years ago

A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi

Norma-Jean [14]

Answer:

Now e is due to the ring at a

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

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3 years ago

Which system works with the respiratory system for the exchange of gases to take place?

4vir4ik [10]

Yep it is. Defiantly the circulatory system

6 0

3 years ago

Describe what would happen if anything to the mass and weight when you go to jupiter

Art [367]

If you were somehow magically transported to Jupiter, your mass
would not change, but your weight would become roughly 2.5 times
your Earth weight.

6 0

3 years ago

Read 2 more answers

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)