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3 years ago

How do you find the density of an object?

Physics

2 answers:

Travka [436]3 years ago

8 0

Density is mass divided by volume p=m/v

Setler79 [48]3 years ago

6 0

You need to find the mass of the object aswell as its volume.

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101.32 + 20.1234 - 212.1 determine total number of significant figures

Leviafan [203]

Answer:

The total number of significant figures is twelve.

Explanation:

7 0

3 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

5 0

3 years ago

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the ot

s344n2d4d5 [400]

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

$f = \dfrac{nv}{2L}$

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

$f = \dfrac{1*343}{2*2.08}$

$f = \dfrac{343}{4.16}$

$f =82.45 \ Hz$

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = $\dfrac{nv}{2f}$

The length of the longer pipe is L = $\dfrac{1*343}{2*74.45}$

The length of the longer pipe is L = $\dfrac{343}{148.9}$

The length of the longer pipe is L = 2.30 m

6 0

3 years ago

1. To calulate the molarity of a solution, you need to know the moles of the solute and the?

Fofino [41]

1. Volume of the solution (B)

2. Celery (D)

3. Hydroxide ions in solution (A)

3 0

3 years ago

A 65 N boy sits on a sled weighing 52 N on a horizontal surface. The coefficient of friction between the sled and the snow is 0.

pogonyaev

Answer:

1.40 N

Explanation:

The magnitude of the frictional force is given by:

$F=\mu N$

where

$\mu$ is the coefficient of friction

N is the magnitude of the normal reaction

The coefficient of friction for this problem is $\mu=0.012$ . The magnitude of the normal reaction is equal to the combined weight of the boy and the sled, because the surface is horizontal, so

$N=65 N+52 N=117 N$

Therefore, the frictional force is

$F=\mu N=(0.012)(117 N)=1.40 N$

3 0

3 years ago