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Blababa [14]
3 years ago
8

In a closed system, _____ energy is equal to potential energy plus kinetic energy.

Physics
1 answer:
OlgaM077 [116]3 years ago
7 0
The answer would be "mechanical". Hope I helped. :)
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What group is Ne in? What is its period?
Gemiola [76]

Answer:

Neon (Ne), chemical element, inert gas of Group 18 (noble gases) of the periodic table, used in electric signs and fluorescent lamps. Colourless, odourless, tasteless, and lighter than air, neon gas occurs in minute quantities in Earth's atmosphere and trapped within the rocks of Earth's crust.

Explanation:

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A vector is 253m long and points in a 55.8degree direction. Find the y-component of the vector.
kati45 [8]

Answer:

209 m

Explanation:

The y-component of a vector is the magnitude times the sine of the angle.

y = 253 sin 55.8°

y = 209

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3 years ago
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An important thing to consider when responding to a driver in front of you that stops suddenly is:a. the mental state of the oth
Gre4nikov [31]
The correct answer is option A. i.e. An important thing to consider when responding to a driver in front of you that stops suddenly is: the mental state of the other driver.

Our talk or discussion can disrturb the balance of the driver or he can get distracted. So, we must try not to speak much while the driver is driving because by doing this we are putting the life of ourselves in danger. Any distrcaction of driver can cause accident.
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3 years ago
The least penetrating form of radiation is ____. a. beta radiation c. alpha radiation b. gamma radiation d. X rays
svet-max [94.6K]
C) alpha radiation seems to be the answer

8 0
3 years ago
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Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
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