Answer:<br><br> A. 2y(x³ + 9x - 5x² - 45)<br><br> B. 2y(x² + 9)(x

A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the

Shkiper50 [21]

Answer:

ω = 0.467 rad/s

Explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round = 144 Kg

angle = 33.2°

moment of inertia

$I = \dfrac{1}{2} m R^2$

$I = \dfrac{1}{2}\times 144 \times 2.75^2$

I = 544.5 kg.m²

torque = force x radius

τ = 37.7 x 2.75

τ = 103.675 N.m

angular acceleration

$\alpha= \dfrac{\tau}{I}$

$\alpha= \dfrac{103.675}{544.5}$

α = 0.190 rad/s²

now ,

distance = $33.2\times \dfrca{2\pi}{360}$

d = 0.579 rad

we know,

using equation of rotational motion

$d = \omega t + \dfrac{1}{2}\alpha t^2$

$0.579 = \dfrac{1}{2}\times 0.190\times t^2$

t = 2.46 s

angular speed

ω = α x t

ω = 0.19 x 2.46

ω = 0.467 rad/s

7 0

2 years ago

Single Replacement: HCI + Zn

Kay [80]

During a single replacement, the Zn and H will switch places. That means the product formed will be ZnCl and H2.

4 0

3 years ago

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,

Andre45 [30]

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules$

Power

$P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W$

Converting to hp

$1\ W=\frac{1}{745.7}\ hp$

$\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp$

The power developed by the cheetah is 9241.6 W or 12.39318 hp

7 0

3 years ago

A machine gun fires 50-g bullets at the rate of 4 bullets per second. The bullets leave the gun at a speed of 1000 m/s. What is

TEA [102]

Answer:

Average recoil force experienced by machine will be 200 N

Explanation:

We have give mass of each bullet m = 50 gram = 0.05 kg

There are 4 bullets

So mass of 4 bullets = 4×0.05 = 0.2 kg

Initial speed of the bullet u = 0 m/sec

And final speed of the bullet v = 1000 m/sec

So change in momentum $P=m(v-u)=0.2\times (1000-0)=200kgm/sec$

Time is given per second so t = 1 sec

We know that force is equal to rate of change of momentum

So force will be equal to $F=\frac{200}{1}=200N$

So average recoil force experienced by machine will be 200 N

5 0

3 years ago

Which statement describes the redox reaction involved in photosynthesis?

NeTakaya

Answer:

D. O2 is removed from the atmosphere, and CO2 is released

Explanation:

Hope it helps you

3 0

2 years ago

Read 2 more answers

Answer: A. 2y(x³ + 9x - 5x² - 45) B. 2y(x² + 9)(x - 5)