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What velocity does a 2kg mass have when its kinetic energy is 16 J

alexandr402 [8]

We can use the equation for kinetic energy, K=1/2mv².
Your given variables are already in the correct units, so we can just plug in the variables and solve for v.

K = 1/2mv²
16 = 1/2(2)v²
16 = (1)v²
√16 = v
v = 4 m/s

Therefore, the velocity of a 2 kg mass with 16 J of kinetic energy is 4 m/s.
Hope this is helpful!

7 0

2 years ago

Group of physics students are asked to demonstrate Ohm’s Law in parallel circuits. Using their knowledge of Ohm’s Law and the r

astra-53 [7]

The student’s suggestion who provides enough evidence to be able to determine the value of each resistor is student D.

<h3>What is current?</h3>

The current is the stream of charges which flow inside the conductors when connected across the end of voltage.

For the given set of parallel resistors, we need to find the resistance of each resistor.

From the Ohm's law, V =IR

R = V/I

Resistance value depends upon the voltage difference across the resistor and the current flowing through that resistance.

Thus, the student D gives enough evidence to find resistance of the circuit is

Learn more about current.

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8 0

1 year 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)