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chubhunter [2.5K]

3 years ago

12

Which of the following best describes an action-reaction pair?

2 answers:

NNADVOKAT [17]3 years ago

3 0

Our weight is greater on the Earth<span> because of its stronger </span>gravity. .... track A? (b<span>) Is the speed gained by ball </span>B<span> going </span>down<span> the extra dip the .... The force of friction between your </span>back<span> foot and the </span>floor pushes<span> you forward. .... (a) Two force </span>pairs<span> act; </span>Earth's pull<span> on the apple (</span>action<span>), and the apple's </span>pull<span> on the ...</span>

Liula [17]3 years ago

3 0

Answer:

The Moon Pulls on Earth, and Earth pulls back on the moon.

Explanation:

Newton's third law of motion describes action and reaction forces. It tells that an object applies an action force on another and another object also exerts a reaction force on the first object. Both the action and reaction forces are of equal magnitude but they act in opposite direction.

Option (A) is an example of action- reaction pair i.e. "The Moon Pulls on Earth, and Earth pulls back on the moon". Earth exerts a gravitational force on the Moon. The gravitational pull of the Earth on the Moon is much more stronger than the pull of Moon on the Earth.

So, the correct option is (A).

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What creates the energy of the Sun?

Svet_ta [14]

Answer:

C

Explanation:

During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.

5 0

2 years ago

Suppose a 49-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 585 N sits on t

Andre45 [30]

Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction $,\mu_k=0.11$

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

$F=\mu_k N$

Where N= Normal=mg

$F=0.11\times 634=69.74 N$

Hence, the force is needed to pull the sled across the snow at constant speed=69.74 N

7 0

3 years ago

An airplane accelerates down a runway at 3 m/s2 for 32 s until is finally lifts off the ground. Determine the distance traveled

Alex Ar [27]

Answer:

12 or 24

Explanation:

i think it is i hope it is right

4 0

3 years ago

A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of

Alex73 [517]

Answer:

4.7 N

Explanation:

130 g = 0.13 kg

The momentum of the snowball when it's thrown at the wall is

$p = mv = 0.13*6.5 = 0.845 kgm/s$

Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s

$p = F\Delta t$

$F*0.18 = 0.845$

$F = 0.845 / 0.18 = 4.7 N$

8 0

3 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago