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marissa [1.9K]
3 years ago
8

A ball is moving with a velocity of 0.5 m/s its velocity is decreasing at the rate of 0.05 m/s² what is its velocity in 5 second

s? how much time will it take from start to stop?what is the distance to travelled by it before stopping?
​
Physics
1 answer:
Law Incorporation [45]3 years ago
4 0

Answer:

v after 5s = 0.25 m/s, it took 10s to stop, it has traveled 2.5m before stopping

Explanation

We can use the equation of motion with constant acceleration

Given: v0= 0.5 m/s a= -0.05 m/s²

v(5s) = v0 + a×t = 0.25 m/s

Stop => v=0 => v0 + a×t = 0 => t=10s

Distance at t=10s ⇒ x(10) = 0.5×10 + 0.5x(-0.05)x10² = 2.5m

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The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

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tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

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Horizontally,

u = 3.01 m / s

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s = ( 3.01 * 1.77 ) + 0

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