All categories

3 years ago

I DONT NEED ANYONE TO ANSWER QUESTION BUT IS THERE ANYONE THAT CAN HELP ME ON QUIZZES FOR FREE

Physics

2 answers:

olchik [2.2K]3 years ago

8 0

Answer:

Sorry

Explanation:

lol i need someone for myself to do that

pochemuha3 years ago

7 0

Answer:

Not me

Explanation:

Sorry

You might be interested in

ABCD next 3 letters??????

jasenka [17]

Answer:EFG

Explanation:AYEEE

8 0

3 years ago

Read 2 more answers

A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

valkas [14]

Op here is another problem exactly like that. Just plug in your variables instead. And remember, time is never negative.

5 0

3 years ago

How much time will it take to do 33j of work with 11 w of power

Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

$\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s$

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

$\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}$

8 0

3 years ago

2 Points

True [87]

The answer is B because the pollinators give pollen from the plant

8 0

3 years ago

Read 2 more answers

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

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

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago