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3 years ago

explain " you can not apply a force to an object with out that object applying the same force back to you"

Physics

2 answers:

NemiM [27]3 years ago

5 0

Newton’s third law motion

stealth61 [152]3 years ago

3 0

Answer:

Newton's Third Law Of Motion

Explanation:

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A 2.0-Coulomb bead is given 20 Joules of electric potential energy by lifting it from the "ground" to point A. If a second bead

Katen [24]

Answer:

40 J

Explanation:

$c_1$ = 4 C

$c_2$ = 2 C

$U_1$ = 20 J

The potential energy is directly proportional to the charge of the particle

$U\propto c$

$\dfrac{U_2}{U_1}=\dfrac{c_2}{c_1}\\\Rightarrow U_2=\dfrac{U_1c_2}{c_1}\\\Rightarrow U_2=\dfrac{20\times 4}{2}\\\Rightarrow U_2=40\ J$

The potential energy expected is 40 J

4 0

3 years ago

A stationary front occurs when the boundary surface between two air masses does not _____.

likoan [24]

When the two air masses meet i think

4 0

4 years ago

Read 2 more answers

a 1 kg emery stone measuring 20 cm redium, is rotating with an angular speed of 39 revolutions per minute, when the motor is tur

musickatia [10]

Answer: 13,24

Explanation:

3 0

3 years ago

A mass of 10 g of oxygen fill a weighted pistonâ€“cylinder device at 20 kPa and 110Â°C. The device is now cooled until the tempe

mezya [45]

Answer:

The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

$\Delta V=V_{2}-V_{1}$

$\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})$

Put the value into the formula

$\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)$

$\Delta V=14.297\ L$

$\Delta V=14.3\times10^{-3}\ m^3$

Hence, The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

6 0

3 years ago

A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc

nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

$m_sv_s + m_rv_r = 0$

where $m_s = 72.8, m_r = 265$ are the mass of the swimmer and raft, respectively. $v_s = 5.21 m/s, v_r$ are the velocities of the swimmer and the raft after the run, respectively. We can solve for $v_r$

$265v_r + 72.8*5.21 = 0$

$v_b = -72.8*5.21/265 = -1.43 m/s$

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0

3 years ago

explain " you can not apply a force to an object with out that object applying the same force back to you"​

explain " you can not apply a force to an object with out that object applying the same force back to you"