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FinnZ [79.3K]
3 years ago
14

According to Newtons third law of motion when one object exerts a force on a second object what is true about the forces?

Physics
1 answer:
fomenos3 years ago
7 0

Answer:

D.They are equal in magnitude and opposite in direction

Explanation:

  • Newton's third law states that the action force is always associated with a reaction force.
  • When a body 'A' exerts a force on body 'B', it is called the force of action.
  • When the body 'B' in turn resist the force of 'A' is called the reaction force. It is the reactive force acted upon by the body 'B' on 'A'.
  • This reaction force is equal in magnitude of the action force.
  • If the two bodies remain in the same horizontal line, the 'A' exerts a force in the direction towards 'B' and the body 'B' exerts a reaction force in the direction towards 'A'.
  • Hence, the two forces that are exerted by the bodies are equal in magnitude and opposite in direction.
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Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M
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Answer:

sin\theta - \mu_k cos\theta = \frac{m}{M}

sin\theta - \mu_k cos\theta = 1

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

F_f = \mu Mgcos\theta

now if it is moving down the inclined plane at constant speed

so we will have

Mgsin\theta - T - \mu mgcos\theta = 0

on other side the mass "m" will go up at constant speed

so we have

T - mg = 0

so we have

Mgsin\theta = \mu Mgcos\theta + mg

so we have

sin\theta - \mu_k cos\theta = \frac{m}{M}

for special case when m = M

then we have

sin\theta - \mu_k cos\theta = 1

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3 years ago
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A bicycle on the top of the hill has the highest potential energy, and when the bike goes down, it transfers to kinetic because it is moving

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4 0
3 years ago
The nuclear equation is incomplete.<br><br><br>What particle completes the equation?​
Romashka-Z-Leto [24]

Answer:

_{6}^{12} C

Explanation:

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What is one watt ? Write the relation of watt with kilowatt , megawatt , and horsepower .​
alina1380 [7]

Answer:

See the explanation below

Explanation:

The watt (the power) is equal to the relationship between the work and the time in which that work is performed.

P = W/t

where:

W = work [J] (units of Joules)

t = time [s].

Now 1000 [W] are equal to 1 [kW]

And 1000000 [W] are equal to 1 [MW]

The horsepower is the unit of power in the imperial system of units.

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3 years ago
What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

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3 years ago
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