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dybincka [34]
2 years ago
12

The force of attraction between two objects, which is related to the object's mass is the definition of

Physics
2 answers:
ICE Princess25 [194]2 years ago
8 0

Answer:

Explanation:

The gravitation force of attraction between the two bodies is directly proportional tp the product of masses of two bodies and inversely proportional to the square of distance between them.

Andrews [41]2 years ago
7 0
This would be the definition of gravitational force, I believe. The force of attraction between 2 objects based on the mass of the objects.
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3 years ago
A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at
Vaselesa [24]

Answer:24.70 ^{\circ}C

Explanation:

Given

mass of lead piece m_l=234 gm\approx 0.234 kg

mass of water in calorimeter m_w=611 gm\approx 0.611 kg

Initial temperature of water T_w=24^{\circ}C

Initial temperature of lead piece T_l=24^{\circ}C

we know heat capacity of lead and water are 125.604 J/kg-k and 4.184 kJ/kg-k respectively

Let us take T ^{\circ}C be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

m_lc_l(T_l-T)=m_wc_w(T-T_w)

0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)

86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)

86-T=86.97T-2087.49

T=\frac{2173.491}{87.97}=24.70^{\circ}C

3 0
3 years ago
If you help me with the correct answer I will give you the brainlist
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Answer:

1. A force is a push or pull upon an object resulting from the object's interaction with another object.  

Explanation:

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3 years ago
A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed
GrogVix [38]

Answer:

  • a. \tau =  2.1161 s
  • b. V(18.8 \ s) = 0.0256 \ V

Explanation:

<h3>a.</h3>

The equation for the voltage V of  discharging capacitor in an RC circuit at time t is:

V(t) = V_0 e^{(- \frac{t}{\tau}) }

where V_0 is the initial voltage, and \tau is the time constant.

For our problem, we know

V_0 = 185 \ V

and

V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

So

185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }

ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })

- \frac{10 \ s}{\tau}  = ln (\frac{1.64 \ V}{ 185 \ V })

\tau =  \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }

This gives us

\tau =  2.1161 s

and this is the time constant.

<h3>b.</h3>

At t = 18.8 s we got:

V(18.8 \ s) = 185 \ V  \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 0.0256 \ V

4 0
3 years ago
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