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Vikki [24]
3 years ago
14

Determine the length of the object shown 1.2 cm 1.3 cm 1.25 cm 1.250 cm

Physics
2 answers:
stiks02 [169]3 years ago
6 0

The correct answer is 1.25 because it is 1/2 of 1 1/2 and that is 1.25.

bogdanovich [222]3 years ago
5 0

The correct answer is C. 1.25 cm.

If you can see, the head of the nail stops right in the middle of the 2nd and 3rd lines after 1. This means it is 1 and .2+ .05 or 1.25.

Hope this helps!

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A spherical helium filled balloon (B) with a hanging passenger cage being held by a single vertical cable (C) attached to Earth
Gre4nikov [31]

Answer:

The tension is  T  = 4326.7 \  N

Explanation:

From the question we are told that

   The  total mass is  m  =  200 \  kg

    The  radius is r = 5 \  m

     The  density of air is  \rho_a  =  1.225 \ kg/m^3

Generally the upward  force acting on the balloon is mathematically represented as

        F_N  =   T  + mg

=>     (\rho_a  *  V  *  g ) =   T  + mg

=>   T  =  (\rho_a  * V  *  g   )  - mg

Here V is the volume  of the spherical helium filled balloon which is mathematically represented as

      V  =  \frac{4}{3}  * \pi r^3

=>   V  =  \frac{4}{3}  * 3.142 *(5)^3

=>   V  = 523.67\  m^3

So

    T  = (1.225 *  523.67*  9.8 ) -  200 *  9.8

   T  = 4326.7 \  N

5 0
3 years ago
Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a
Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg

After that, we compute the potential energy 1.00 m above the reference point:

U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J

Then, we compute the average kinetic energy at the specified temperature:

K=\frac{3}{2}\frac{R}{Na}T

Whereas N_A stands for the Avogadro's number for which we have:

K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0
3 years ago
What are you and the Australian Institute of Marine Biology trying to find out and 4. Why might it be important to find this inf
TEA [102]

Answer:

Established in 1972 by the McMahon Government, the institute's primary function is research for sustainable use and protection of the marine environment. The Institute investigates topics from broad-scale ecology to microbiology.

6 0
2 years ago
It is a general exercise that we do before practicing sports activities<br><br> worm-up or cool down
Montano1993 [528]

It is called a warm-up.

3 0
3 years ago
Read 2 more answers
A 650 kg steel beam is being pulled up by a crane with a force of 7020 N. What is the upwards acceleration of the beam?
zimovet [89]

Answer: 1 m/s^2

Explanation:

n= w+ma

n= Mg+Ma

7020 = (650)(9.8)+650a

7020 = 6370+ 650a

650=650a

A= 1 m/s^2

8 0
3 years ago
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