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2 years ago

The speed limit on some interstate highways is roughly 90 km/h. (a) What is this in meters per second

Physics

1 answer:

attashe74 [19]2 years ago

7 0

Answer:

25 m/s

Explanation:

⠀⠀⠀⇒ Speed = 90 km/h

⠀⠀⠀⇒ Speed = 90 km/60 min [As 1 hr = 60 min]

⠀⠀⠀⇒ Speed = (90 × 1000) m/(60 × 60)sec

⠀⠀⠀⇒ Speed = 90000 m/3600 sec

⠀⠀⠀⇒ Speed = 900 m/36 sec

⠀⠀⠀⇒ Speed = 300 m/12 sec

⠀⠀⠀⇒ Speed = 100 m/4 sec

⠀⠀⠀⇒ Speed = 25 m/s

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A box slides down a frictionless ramp.if it starts at rest, what is it’s speed at the bottom?

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8.854 m/s is the speed of the box after it reaches bottom of the ramp.

<u>Explanation</u>:

From the figure we came to know that height of the block is 4 m.

We know that,

Total "initial energy of an object" = Total "final energy of an object "

Total "initial energy of an object" is = "sum of potential energy" and "kinetic energy" of an object at its initial position.

$\text { "g" acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}$

$\text { Total initial energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{i}}+\frac{1}{2} \mathrm{m} v_{i}^{2}$

Initial velocity is “0” as the object does not have starting speed

$\text { Height of the block where the object is placed initially }\left(h_{i}\right) \text { is } 4 \mathrm{m} \text { . }$

$\text { Total initial energy }=\mathrm{m} \times 9.8 \times 4+\frac{1}{2} \mathrm{m} 0^{2}$

Total initial energy = 39.2 × m

$\text { Total final energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{f}}+\frac{1}{2} m v_{f}^{2}$

$\text { We need to find final velocity } v_f$

$\text { Height of the block where the object is travelled to bottom (h_) is } 0 \mathrm{m} \text { . }$

$\text { Total final energy }=\mathrm{m} \times 9.8 \times 0+\frac{1}{2} m v_{f}^{2}$

Now, Total initial energy of an object = Total final energy of an object

$39.2 \times \mathrm{m}=0.5 \mathrm{m} v_{f}^{2}$

$\frac{39.2}{0.5}=v_{f}^{2}$

$v_{f}^{2}=78.4$

$v_{f}=\sqrt{78.4}$

$v_{f}=8.854 \mathrm{m} / \mathrm{s}$

Final speed is 8.854 m/s.

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3 years ago

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

nevsk [136]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

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