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3 years ago

A falling object accelerates from -10.0 m/s to -30.0 m/s. how much time does it take?

Physics

2 answers:

Zepler [3.9K]3 years ago

8 0

Answer:

2.04 s

Explanation:

v = at + v₀

(-30.0 m/s) = (-9.8 m/s²) t + (-10.0 m/s)

t = 2.04 s

KIM [24]3 years ago

8 0

Answer:

The time taken by the falling object is 2.04 sec

Solution:

The initial velocity $V = -10 m/s$

The final velocity $V_1 = -30m/s$

Let us assume that the time taken is t sec

We know that the acceleration due to gravity $g = 9.8m/sec^2$

As per the free falling object motion,

$V_1 = V -gt$

$-30 = -10 -9.8\times t$

$-30+10 = -9.8\times t$

$-20 = -9.8\times t$

$t =\frac{20}{9.8} = 2.04$

So, it takes total 2.04 sec

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For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ

slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

= 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

$180 \frac{mg}{kg} \times 71.215 kg$

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Therefore, calculate the volume which contains the amount of caffeine as follows.

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5 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$