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2 years ago

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Physics

2 answers:

o-na [289]2 years ago

8 0

ice ice baby wow hello wow hello

Explanation:

yesyes

Anna007 [38]2 years ago

7 0

1. the volume of the balloon will decrease
2. the balloon will decrease in volume and then stop as it is in thermal equilibrium
3. don’t really know what it’s asking but higher thermal energy means more motion of molecules
4. the volume of the balloon is dependent on the amount of molecular motion

was kinda hard to understand the question but i hope this helps

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If i click on a mouse does that show energy transfer?

Radda [10]

Answer:

Yes

Explanation:

You are using energy to click the mouse, and the energy moves from your fingers to the mouse clicker.

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2 years ago

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the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec

vivado [14]

$Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m$

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2 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

5 0

2 years ago

A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p

levacccp [35]

Answer:

$Q = 8,345 * v$

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

$Q = 8,345 * v$

Where v is the amount of gallons that has been drained from the tub.

Have a nice day. let me know if I can help with anything else

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3 years ago

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beks73 [17]

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2 years ago