Ask question

Ask question

All categories

3 years ago

12

1. Is there a relationship between the volume of water displaced and the total volume of the block that has anything to do with

density? If so, what is it?

1 answer:

strojnjashka [21]3 years ago

8 0

Answer:

The volume of the block is equal to the volume of water displaced by the block.

Explanation:

Volume refers to the amount of space occupied by a given object (in this case the block). When an object such as the block is immersed in water, it displaces its own volume of water. This volume of water displaced is equal to the volume of the block. Hence we can write;

Final Volume of water - Initial Volume of water= Water Displaced = Volume of the block

Recall that the density of a body is given by;

Density= mass/volume

If we obtain the volume of the block by measuring the volume of water displaced by the block, then we weigh the block using a weighing balance, we can obtain the density of the block easily from the relationship shown above.

You might be interested in

Most energy obtained from water is converted from _____.

Snowcat [4.5K]

Potential energy behind dams

7 0

3 years ago

Read 2 more answers

A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?

Snowcat [4.5K]

Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

It makes 10 degree angle with the positive x-axis.

We need to find the magnitude of the vertical component of the force. It can be given by :

$F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N$

So, the magnitude of the vertical component of the force is 6.07 N.

5 0

3 years ago

Read 2 more answers

A 4-kg cat is resting on a bookshelf that is 2 m high. What is the cats gravitational potential energy relative to the floor if

ra1l [238]

The most exact answer is 78.4J also in this kind of options we can say answer "d"

3 0

3 years ago

Since all objects are ‘weightless’ for an astronaut in orbit, is it possible for astronauts to tell whether an object is heavy o

ivanzaharov [21]

W=gm
where g - gravitation
m - mass
w - weight
as gravitation equals to zero, multiplying by 0 gives W=0
It is not possible to tell whether and object is heavy or light

4 0

3 years ago

Read 2 more answers

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

8 0

3 years ago