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sammy [17]
3 years ago
10

Why wheat flour is usually passed near a magnet before being packed?

Physics
1 answer:
densk [106]3 years ago
8 0
Becasue steel particles may contaminate the flour.
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A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?
denis23 [38]

Answer:

Workdone = 465766038 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

7 0
3 years ago
How do you calculate average speed?
Mila [183]

Divide
       (the distance covered in some period of time)
by
       (the time taken to cover the distance).

The quotient is the average speed during that period of time.
4 0
3 years ago
Read 2 more answers
How many neutrons does element X have if its atomic number is 48 and its mass number is 167?
lys-0071 [83]
If its atomic number is 48, then it has 48 protons in the nucleus
of each atom.  Any more mass than that is supplied by the neutrons
that are mixed in there with the protons.

If the mass is 167, and 48 of those are protons, then there are

             (167 - 48)  =  119 neutrons

in each nucleus.

6 0
3 years ago
Glucose can have a straight chain of ______________ atoms or, more commonly, form a _____________ structure.
vovikov84 [41]
Lipids and <span>Polysaccharides r the answers hoped i helped</span>
6 0
3 years ago
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
3 years ago
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