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3 years ago

Q1.

Physics

1 answer:

Vesnalui [34]3 years ago

4 0

Answer:

P = 450 J

Explanation:

Given that,

Mass of a child, m = 18 kg

The vertical distance from the top to the bottom of the slide is 2.5 metres.

The Gravitational field strength = 10 N/kg

We need to find the decrease in gravitational potential energy of the child sliding from the top to the bottom of the slide.

The formula for the gravitational potential energy is given by :

P = mgh

Substituting all the values,

P = 18 kg × 10 m/s² × 2.5 m

P = 450 J

Hence, the decrease in gravitational potential energy is 450 J.

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3 years ago

A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of

Dennis_Churaev [7]

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = $85\frac{1000}{3600}=23.61\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s$

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s$

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2$

Emergency acceleration is 2.99 m/s² (magnitude)

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3 years ago

Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The ro

ehidna [41]

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

$T=2\pi\sqrt{\frac{l}{g} }$

angular frequency ω = 2π / T

= $\omega=\sqrt{\frac{g}{l} }$

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

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.5 ( ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

4 0

3 years ago