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krek1111 [17]
3 years ago
10

Hi i need help. thNKS

Physics
1 answer:
Ede4ka [16]3 years ago
3 0

Answer:

8

Explanation:

Groups go down, periods go across :)

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What would be the mass of an object that has a force of 88N and an acceleration of 12 m/s2?
MA_775_DIABLO [31]

Answer:

7.33 kg

Explanation:

F = ma

88 N = m(12 m/s^2)

m = 7.33 kg

6 0
3 years ago
The density of aluminum is 2.7 g/cm3. A metal sample has a mass of 52.0 grams and a volume of 17.1 cubic centimeters. Could the
Fudgin [204]
 Answer:  
__________________________________________________
            No;  the sample could not be aluminum;
since the density of aluminum, " 2.7 g/cm³ " , is NOT close enough to the density of the sample, " 3.04 g/cm³ " .
________________________________________________
Explanation:
________________________________________________
Density is expressed as "mass per unit volume" ;

  in which:
     "mass, "m", is expressed in units of "g" (grams);  and:
     "Volume, "V", is expressed in units of "cm³ " (such as in this problem); or                                                   in units of "mL" ;
__________________________________________________
            {Note the exact conversion:  " 1 cm³ = 1 mL " .}. 
__________________________________________________
  The formula for density:  D = m/V ;

Given:  The density of aluminum is:  2.7 g/cm³.

Given:  A sample has a mass of 52.0 g ; and Volume of 17.1 cm³ ; could it be aluminum?
_________________________________________________________
Let us divide the mass of the sample by the volume of the sample;
by using the formula:
___________________________________________
            D = m / V ;  

     and see if the value is at, or very close to "2.7 g/cm³ ".  

If it is, then it could be aluminum.
____________________________________________________
The density for the sample:

  D = (52.0 / 17.1)   g/cm³ = 3.0409356725146199 g/cm³ ;
                                              →round to "3 significant figures" ;
                                          = 3.04 g/cm³ .
_______________________________________________
No; the sample could not be aluminum; since the density of aluminum, 
   "2.7 g/cm³ "   is NOT close enough to the density of the sample,
                        "3.04 g/cm³ " .
____________________________________________________
5 0
3 years ago
Question 7 (2 points)
DiKsa [7]
This has to do with independent and dependent variables. The independent variable is what affects the dependant variable, so the question is, does a students mark on a test affect their hours of sleep or does their hours of sleep affect their test results? Which one makes more sense? I would say the latter.
6 0
3 years ago
Describe how Rutherford's experiments changed the accepted scientific model of the atom.
lorasvet [3.4K]

Rutherford's model of the atom (ESAAQ) Rutherford carried out some experiments which led to a change in ideas around the atom. His new model described the atom as a tiny, dense, positively charged core called a nucleus surrounded by lighter, negatively charged electrons.

4 0
3 years ago
A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0
3 years ago
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