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3 years ago

Worth 50 point!!!

Physics

1 answer:

olasank [31]3 years ago

3 0

Answer:

Chief Hopper

Explanation:

Mike travels at a constant speed of 3.1 m/s. To find how long it takes him to reach the school, we need to find the distance he travels. We can do this using Pythagorean theorem.

a² + b² = c²

(1000 m)² + (900 m)² = c²

c ≈ 1345 m

So the time is:

v = d / t

3.1 m/s = 1345 m / t

t ≈ 434 s

Next, Chief Hopper travels a total distance of 1900 m, starting at rest and accelerating at 0.028 m/s². So we can use constant acceleration equation to find the time.

d = v₀ t + ½ at²

1900 m = (0 m/s) t + ½ (0.028 m/s²) t²

t ≈ 368 s

So Chief Hopper reaches the school first, approximately 66 seconds before Mike does.

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A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

Tanya [424]

Answer:

$124.62\ \text{rad/s}$

$3.71\ \text{m/s}$

$1.23\ \text{km/s}^2$

$20.28\ \text{m}$

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

$\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}$

The angular speed is $124.62\ \text{rad/s}$

r = 2.98 cm

Tangential speed is given by

$v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}$

Tangential speed at the required point is $3.71\ \text{m/s}$

Radial acceleration is given by

$a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2$

The radial acceleration is $1.23\ \text{km/s}^2$ .

t = Time = 2.06 s

Distance traveled is given by

$d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}$

The total distance a point on the rim moves in the required time is $20.28\ \text{m}$ .

8 0

3 years ago

A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec

olga2289 [7]

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

$h=u't+\dfrac{1}{2}gt^2$

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

$h=\dfrac{1}{2}\times 9.8\times (4.3)^2$

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

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