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2 years ago

if a cyclist traveled southwest a distance of 12,250 meters in 1.7 hrs what would the velocity of the cyclist be?

Physics

1 answer:

Paladinen [302]2 years ago

7 0

Answer:

Velocity = 7.2 km/hr SW

Explanation:

Velocity is a vector that includes both speed and direction. Speed is calculated by dividing the distance travelled by the time required for that distance.

Speed = Distance/Time

i'll convert 12,250 meters into 12.250 kilometers (km). Not needed, but it's my preference. Time is 1.7 hours:

Speed = (12.250 km)/(1.7 hr) = 7.2 km/hr (to 2 sig figs)

The direction is southwest, or SW

Velocity = 7.2 km/hr SW

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The amount of sea ice and permafrost have steadily decreased near Shishmaref, Alaska, over the last century. The loss of ice has

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Climate change, where there is an increase in the average temperature, due to continuous human activities. These activities increase greenhouse gases in the atmosphere.

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3 years ago

(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from r

Oksi-84 [34.3K]

Answer:

0.321659377 m/s²

1.383138458 m/s

0.321659377 m/s²

0.62667 m/s²

0.7044 m/s² and 27.17°

Explanation:

d = Diameter of rim = 13 in = $13\times 0.0254=0.3302\ m$

r = Radius = $\frac{d}{2}=\frac{0.3302}{2}=0.1651\ m$

$\omega_f$ = Final angular velocity = $80\times\frac{2\pi}{60}=8.37758\ rad/s$

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 4.3 s

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{8.37758-0}{4.3}\\\Rightarrow \alpha=1.94827\ rad/s^2$

Tangential acceleration is given by

$a_t=r\alpha\\\Rightarrow a_t=0.1651\times 1.94827\\\Rightarrow a_t=0.321659377\ m/s^2$

The tangential acceleration of the bug is 0.321659377 m/s²

Tangential velocity is given by

$v=r\omega\\\Rightarrow v=0.1651\times 8.37758\\\Rightarrow v=1.383138458\ m/s$

The tangential velocity of the bug is 1.383138458 m/s

The tangential acceleration is constant which is 0.321659377 m/s²

Centripetal acceleration is given by

$a_c=\frac{a_tt^2}{r}\\\Rightarrow a_c=\frac{0.321659377^2\times 1}{0.1651}\\\Rightarrow a_c=0.62667\ m/s^2$

The centripetal acceleration of the bug is 0.62667 m/s²

The resultant of the acceleration gives us total acceleration

$a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a=\sqrt{0.321659377^2+0.62667^2}\\\Rightarrow a=0.7044\ m/s^2$

Direction is given by

$\theta=cos^{-1}\frac{a_c}{a}\\\Rightarrow \theta=cos^{-1}\frac{0.62667}{0.7044}\\\Rightarrow \theta=27.17^{\circ}$

The magnitude and direction of the acceleration is 0.7044 m/s² and 27.17°

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