Ask question

Ask question

All categories

anastassius [24]

3 years ago

9

Find the angular momentum of a hollow spinning sphere when its angular speed is 20rad/s. The rotational inertia of the sphere is

0.02kg.m2.

1 answer:

zysi [14]3 years ago

5 0

Answer:

L= 0.4 kgm²/s

Explanation:

The angular momentum of a hallow spinning sphere is

L = Iω

L = (0.02kgm²) × (20rad/s)

L= 0.4kgm²/s

You might be interested in

1. Given a list of atomic model descriptions:

laiz [17]

Answer: A: electron shells outside a central nucleus

B: hard, indivisible sphere

C: mostly empty space

Which list of atomic model descriptions represents

the order of historical development from the earliest

to most recent?

Explanation:

3

8 0

3 years ago

What is a prokaryote cell/ eukaryote cell?

Wewaii [24]

A prokaryote is a microscopic single-celled organism that has neither a distinct nucleus with a membrane nor other specialized organelles. Prokaryotes include the bacteria and cyanobacteria.

A eukaryote is an organism consisting of a cell or cells in which the genetic material is DNA in the form of chromosomes contained within a distinct nucleus. Eukaryotes include all living organisms other than the eubacteria and archaebacteria.

8 0

3 years ago

What erases the impact craters on earth and is responsible for most of the landforms that we see?

frez [133]

The big bang theory I think

6 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

A motorcycle starting from rest covers 200 metre distance in 6 second. Calculate final velocity and acceleration of the motorcyc

QveST [7]

Answer:

Explanation:I don't say you must have to mark my ans as brainliest but if you think it has really helped u plz don't forget to thank me....

3 0

3 years ago