All categories

4 years ago

I need help now please help me

Physics

1 answer:

alexandr1967 [171]4 years ago

8 0

The correct answer is C. Plug in x and y value for answers to see if they work. For example, 9/3 = 3. So C is the answer.

You might be interested in

Causes and effects of a social problem

aev [14]

Answer:

what is the social problem about

Explanation:

5 0

3 years ago

Read 2 more answers

A raft of mass 199 kg carries two swimmers of mass 52 kg and 70 kg. The raft is initially floating at rest. The two swimmers sim

Minchanka [31]

To solve this problem we will apply the concept related to the conservation of the Momentum. We will then start considering that the amount of initial momentum must be equal to the amount of final momentum. Considering that all the objects at the initial moment have the same initial velocity (Zero, since they start from rest) the final moment will be equivalent to the multiplication of the mass of each object by the velocity of each object, so

Initial Momentum = Final Momentum

$(m_B+m_1+m_2)v_i = m_1v_1+m_2v_2+m_Bv_B$

Here,

$m_B$ = mass of Raft

$m_1$ = Mass of swimmers 1

$m_2$ = Mass of swimmers 2

$v_i$ = Initial velocity (of the three objects)

$v_B$ = Velocity of Raft

Replacing,

$(199+52+70)*0 = (52)(4)+(70)(-4)+199v_B$

Solving for $v_B$

$vB = \frac{72}{199}$

$v_B = 0.3618m/s$

Therefore the velocity the rarft start to move is 0.3618m/s

5 0

4 years ago

How does a computer process data? *

elena55 [62]

The correct answer D: all of the above

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)