3 years ago

A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg

Physics

1 answer:

Naddik [55]3 years ago

4 0

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

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A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.

Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

$F=mg$

$\dfrac{mv^2}{r}=mg$

$\dfrac{v^2}{r}=g$

$v=\sqrt{rg}$

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

$v=\sqrt{250\times9.8}$

$v=49.49\ m/s$

Hence, The speed of space station floor is 49.49 m/s.

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Explanation:

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A car travels in a straight line for 5 h at a constant speed of 72 km/h. What is it’s acceleration

Luda [366]

Answer:

$a \approx \: 0.001 \: m {s}^{ - 2}$

Explanation:

Given:

$initial \: velocity \: (u) = 0 \\ \\ Final \: Velocity \: (v) = 72 km /h \\ \\ Time \: (t) = 5 \: hours \\ \\ Acceleration \: (a) =? \\ \\ \because \: a = \frac{v - u}{t} \\ \\ \therefore \: a = \frac{72 - 0}{5} \\ \\ a = \frac{72}{5} \\ \\ a = 14.5 \: km {h}^{ - 2} \\ \\ a = \frac{14.5 \times 1000}{3600\times 3600} \: m {s}^{ - 2} \\ \\ a = 0.00111882716 \\ \\ a \approx \: 0.001 \: m {s}^{ - 2}$

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