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daser333 [38]
3 years ago
10

Newton's third law is applicable only to objects at rest. True False

Physics
2 answers:
Wittaler [7]3 years ago
8 0
Newton's third law applies even when objects do not move. Here a gymnast pushes downward on the bars. The bars push back on the gymnast with an equal force. an action-reaction pair is because one of the objects is often much more massive and appears to remain motionless when a force acts on it.
DiKsa [7]3 years ago
5 0

Answer:

False

Explanation:

Newton's third law - For every action there's an equal and opposite reaction, action and reaction act on different objects

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Thermometer for measuring air and sea surface temperature

Barometer for measuring atmospheric pressure

Hygrometer for measuring humidity

Anemometer for measuring wind speed

Pyranometer for measuring solar radiation

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3 years ago
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A car moving with a velocity 15m/s and accelerating by 5m/s² attempts to reach a car moving with 30 m/s velocity.What distance s
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I think A is correct
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. There are fewer organisms at the highest trophic levels, because
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Answer:

Because there is less energy at higher trophic levels

Explanation:

  • In a food chain that shows the feeding levels of organisms in an ecosystem, energy decreases as you move from the lowest trophic level to the highest trophic level.
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  • Therefore, organisms at the highest trophic level are fewer in number but large in size.
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Which of the following is the best definition of viscosity? A. the tendency of a fluid to support a floating object B. the abili
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C. example. honey has a higher viscosity than water.
6 0
3 years ago
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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a
pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

120^2=130^2+50^2-2\times130\times50\cos\alpha

\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}

\alpha=\cos^{-1}(0.3846)

\alpha=67.38^{\circ}

We need to calculate the angle β

Using cosine law

50^2=130^2+120^2-2\times130\times120\cos\beta

\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}

\beta=\cos^{-1}(0.923)

\beta=22.63^{\circ}

We need to calculate the force on 130 cm side

Using formula of force

F_{130}=ILB\sin\theta

F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0

F_{130}=0

We need to calculate the force on 120 cm side

Using formula of force

F_{120}=ILB\sin\beta

F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63

F_{120}=0.1385\ N

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

F_{50}=ILB\sin\alpha

F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38

F_{50}=0.1385\ N

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0
3 years ago
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