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3 years ago

Types of friction in physics

Physics

1 answer:

Dennis_Churaev [7]3 years ago

3 0

-- static friction

-- kinetic friction

-- fluid friction

-- sliding friction

-- air resistance

-- drag

-- professional debate

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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–

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

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Cleavage is the tendency of a mineral to break along smooth planes parallel to zones of weak bonding. Fracture is the tendency of a mineral to break along curved surfaces without a definite shape. These minerals do not have planes of weakness and break irregularly.

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Calculate the average speed of a complete round trip in which the outgoing 300 km is covered at 93 km/h , followed by a 1.0-h lu

madreJ [45]

Answer : 63 km/h

Explanation :

Outgoing distance = 300 km.

Outgoing speed = 93 km/h.

Break time = 1 h

Return distance = 300 km

Return speed = 56 km/h

Outgoing time = outgoing distance / outgoing speed

Outgoing time = 300 km / 93 km/h = 3.225806451612 h

Return time = return distance / return speed

Return time = 300 km / 56 km/h = 5.357142867142 h

Total distance =. Outgoing distance + return distance travelled

Total distance = 300 km + 300 km = 600 km

Total time = outgoing time + break time + return time

Total time = 3.225806451612 h + 1 h+ 5. 357142867142 h = 9.582949308754 h

Average speed = total distance / total time

Average speed = 600 km / 9.582949308754 h

Average speed = 62.61120462042 km/h = 63 km/h

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3 years ago

Types of friction in physics ​

Types of friction in physics