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Natalka [10]
3 years ago
15

Graphs help you see ______.

Physics
2 answers:
Anastaziya [24]3 years ago
4 0
D- if i was right please mark me as a brainliest answer and a thank you
kkurt [141]3 years ago
3 0

The answer to your question is c. this is because you would graph the numbers for the independant on one line/side of the graph, and the dependant on the other. here is an example below. Also,DarkcloudRevievived please stop insulting other users. It is not nice, and just makes <em>you </em>not sound grown up enough.

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Where could convection currents form? Check all that apply.
KatRina [158]
<h3><u>Answer;</u></h3>
  • In a freshwater lake
  • In the atmosphere
  • In Earth's mantle
<h3><u>Explanation</u>;</h3>
  • <em><u>Convection currents are types that cause the process of convection, which the transfer of heat energy that occurs in fluids.</u></em>
  • <em><u>Convection currents are circular patterns that occurs in fluids such that the less dense warm fluids rises up while denser cold fluids sinks, </u></em>it is this movement of less dense warm fluid and denser cold fluids that creates circular patterns that causes the process of convection to take place.
  • <em><u>Convection currents may occur in the atmosphere where warm air rises while cold denser air sinks or moves towards the bottom, it may also occur in the mantle of the Earth and water or water bodies such as lakes.</u></em>
8 0
4 years ago
Read 2 more answers
What type of subsurface material is able to store groundwater?
notka56 [123]
<span>Porous material has many spaces that can hold(store) groundwater.</span>
4 0
3 years ago
A skater slides across the ice with an initial velocity of 5.0 m/s. She slows 10 points
zvonat [6]

Explanation:

Given that,

The initial velocity of a skater is, u = 5 m/s

She slows to a velocity of 2 m/s over a distance of 20 m.

We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :

v^2-u^2=2as

a = acceleration

a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{(2)^2-(5)^2}{2\times 20}\\\\a=-0.525\ m/s^2

So, her acceleration is 0.525\ m/s^2 and she is deaccelerating. Also, her initial velocity is given i.e. 5 m/s.

7 0
3 years ago
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Fa
Morgarella [4.7K]

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

5 0
3 years ago
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)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

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

5 0
3 years ago
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