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2 years ago

6

A bus start from rest. If the acceleration of bus is 0.5 m/s square. what will be its velocity at the end of 2 minutes?what will

be the distance cover?

2 answers:

attashe74 [19]2 years ago

8 0

Answer:

$\fbox {Velocity = 60 m/s, Distance = 3600 m}$

Explanation:

<u>Part (i)</u> : Velocity at end of 2 minutes

Equation used ⇒ $\boxed {v = u + at}$

⇒ t = 2 minutes = 2 x 60 = 120 seconds

⇒ v = 0 + (0.5)(120)

⇒ v = 60 m/s

<u>Part (ii)</u> : Distance covered

Equation used ⇒ $\boxed {S = \frac{v^{2}-u^{2}}{2a}}$

⇒ S = (60)² - (0)² / 2(0.5)

⇒ S = 3600 m

Andrej [43]2 years ago

5 0

Answer:

3586

Explanation:

by using formula s=it+0.5at^2

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A cue ball has a mass of 0.5 kg. During a game of pool, the cue ball is struck and now has a velocity of 3 . When it strikes a s

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Answer: 3 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:

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The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore

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So, the final total momentum will also be

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And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:

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3 years ago

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A 50-g chunk of 80 degrees C iron is dropped into a cavity in a very large block of ice at 0 degrees C. Show that 5.5 g of ice w

Alenkasestr [34]

Answer:

5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity

Explanation:

The concept to solve this problem is given by Energy Transferred, the equation is given by,

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Where,

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Replacing the values where mass is 50g and temperature is 80°C to 0°C we have,

$Q = mc\Delta T$

$Q = 50*0.11*(80-0)$

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Then we can calculate the heat absorbed by m grams of ice at 0°C, then

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How Q_1=Q_2, so

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3 years ago

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3) 2 m/s

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3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

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2 years ago