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Alina [70]
3 years ago
5

Why is the fuse used in a circuit called safety fuse?​

Physics
2 answers:
Mazyrski [523]3 years ago
6 0

Answer:

The guy above me is right

Explanation:

Sunny_sXe [5.5K]3 years ago
4 0

Answer:

The maximum current which can flow through a fuse without melting it, is called its rating. ... If current higher than 8 A flows through the fuse, it would melt and circuit gets broken. ... Hence, fuse acts as a safety device

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There isn’t a best anime , but in my opinion any anime on Netflix is good other than DBZ too many episodes and storyline gets messed up.
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Improving your cardiovascular endurance, such as aerobic activity, can get your heart pumping, make you sweaty, and quicken your
Thepotemich [5.8K]

Answer:

The correct answer is - oxygen.

Explanation:

Cardiovascular endurance is the ability to do physical activity or exercise without affecting or being tired for a longer time period. Aerobic exercise or activity can help to increase cardiovascular endurance.

Improving the cardiovascular system helps in delivering oxygen as per the demands with the help of the heart that pumps the oxygen-rich blood to different organs of an individual and lungs by inhaling enough oxygen.

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The newton is defined as the:
irakobra [83]

The Newton is defined as the:

C. Force that can give a 1-kilogram mass an acceleration of 1m/sec squared.

<h3>Hope it helps..</h3>

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4 0
2 years ago
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Which choice most accurately defines resistance and voltage of a circuit?
devlian [24]

Answer:

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7 0
3 years ago
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
3 years ago
Read 2 more answers
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