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crimeas [40]
2 years ago
10

What do Newton's three laws state?

Physics
1 answer:
guapka [62]2 years ago
8 0
In the first law, we understand that an object will not change its motion unless a force acts on it. The second law states that the force on an object is equal to its mass times its acceleration. And, finally, the third law states that there is an equal and opposite reaction for every action.
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A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and
iogann1982 [59]

Answer:

p_2 = 1.76 atm

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}

Putting the values

\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}

9.733*10^{-3} = \frac{p_2 *1.68}{304}

9.733*10^{-3}*304 = p_2*1.68

\frac{9.733*10^{-3}*304}{1.68} =p_2

p_2= 1.76 atm

7 0
4 years ago
A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting
ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

T=\dfrac{38}{8}\\\\T=4.75\ s

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

k = spring stiffness constant

So,

k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m

When the cord is in air,

mg=kx

x = the extension in the cord

x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0
3 years ago
Average velocity is different than average speed because calculating average velocity involves a)Distance b)Time c)Motion d)Disp
likoan [24]

Answer:

The answer is D.

Explanation:

Average speed involve just distance and time but average velocity includes displacement and time.

(Correct me if I am wrong)

6 0
4 years ago
A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge
Kisachek [45]

Answer:

\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

5 0
3 years ago
Help needed Can a body be accelerating if it is moving in circle
belka [17]

<em>Answer:</em>

<em>When </em><em>a </em><em>body </em><em>is </em><em>moving </em><em>on </em><em>a </em><em>circle </em><em>it </em><em>is </em><em>accelerating </em><em>because </em><em>centripetal </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>it </em><em>towards </em><em>the </em><em>center.</em>

<em>Please </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em>.</em><em>.</em><em>.</em>

<em>From </em><em>the </em><em>above </em><em>diagram,</em><em>we </em><em>can </em><em>say </em><em>the </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>the </em><em>body </em><em>when </em><em>it </em><em>moves </em><em>in </em><em>a </em><em>circle.</em>

<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

5 0
3 years ago
Read 2 more answers
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