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2 years ago

14

A sharpening wheel is traveling at 5 rad/s, it slows down to rest in 30 seconds while sharpening an axe. What is its angular acc

eleration?

1 answer:

Ratling [72]2 years ago

3 0

Answer:

Angular acceleration $= 0.167$ rad/s^2

Explanation:

Given

Initial Angular velocity (w1) $= 5$ rad/s

Final Angular velocity (w2) $= 0$ rad/s

Time taken to change velocity from w1 to w2 $= 30$ seconds

Angular acceleration is equal to the change in angular velocity to the time taken for making thing change

Hence, Angular acceleration

$\frac{w_2 -w_1}{t} \\\frac{5-0}{30}\\0.167$ rad/s^2

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cicadas produce a sound that has a frequency of 123 Hz. what is the wavelength of this sound in the air? the speed of sound in a

mojhsa [17]

Answer: 2.72 metres

Explanation:

Given that:

frequency of sound F = 123 Hz. wavelength of sound in the air = ?

speed of sound in air V = 334 m/s

Recall that wavelength is the distance covered by the wave after one complete cycle. It is measured in metres, and represented by the symbol λ.

So, apply V = F λ

λ = V /F

λ = 334m/s / 123Hz

λ = 2.72m

Thus, the wavelength of this sound in the air is 2.72 metres

4 0

2 years ago

A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration

wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

$V=X/T$

First stage:

T1=5s

$v_{f} =v_{o} - at$

But, $v_{f} =0$ (decelerates to rest)

then: $a =v_{o} /t=0.3/5=0.06m/s^{2}$

on the other hand:

$x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m$

X1=75cm

Second stage:

T2=5s

$x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m$

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

8 0

2 years ago

There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one hold 163.296 cm³ of paint

5 0

2 years ago

It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica

denis23 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is $\theta = 24.2^o$

Explanation:

From the question we are told that

The length of eire let out is $L = 250 \ m$

The emf generated is $\epsilon = 40 V$

The earth magnetic field is $B = 5.0 *10^{-5} T$

The speed of the shuttle and tether is $v = 7.80 * 10^3 \ m/s$

The emf generated is mathematically represented as

$\epsilon = L\ v\ B\ sin \ \theta$

making $\theta$ the subject of the formula

$\theta = sin ^{-1}[ \frac{\epsilon}{L * B *v} ]$

substituting values

$\theta = sin ^{-1}[ \frac{40}{250 * (5*10^{-5}) *(7.80 *10^{3})} ]$

$\theta = 24.2^o$

6 0

2 years ago

2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the

Luda [366]

Answer:

<em>12,25 km/h</em>

<em>≈ 3,4 m/s </em>

Explanation:

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>= 12,25km/h</em>

<em>or</em>

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>1h = 60m×60s = 3600s</em>

<em>= 12250m/3600s</em>

<em>≈ 3,4 m/s </em>

5 0

2 years ago