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svetlana [45]
3 years ago
15

A space traveler discovers that her weight on a new planet is 192 newtons. Her mass is 68 kilograms. What is the gravitational a

cceleration of the new planet?
A). 2.8 m/s²

В.) 9.8 m/s²

C.) 34 m/s²

D.) 68 m/s²​
Physics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

g ≈ 2.82 m/s^2

Explanation:

By W = mg,

W = weight (in newtons)

m = mass (in kg)

g = gravitational acceleration (in m/s^2)

192 = 68g

g = 2.82352941176 m/s^2

g ≈ 2.82 m/s^2

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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" ;
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            {Note the exact conversion:  " 1 cm³ = 1 mL " .}. 
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  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 " .
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5 0
3 years ago
Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?
IRISSAK [1]

Answer:

22,800 years

Explanation:

Half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.0625 = (½)^(t / 5700)

log 0.0625 = (t / 5700) log 0.5

4 = t / 5700

t = 22,800

It takes 22,800 years.

4 0
3 years ago
A 50-g chunk of 80 degrees C iron is dropped into a cavity in a very large block of ice at 0 degrees C. Show that 5.5 g of ice w
Alenkasestr [34]

Answer:

5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity

Explanation:

The concept to solve this problem is given by Energy Transferred, the equation is given by,

Q = mc\Delta T

Where,

Q= Energy transferred

m = mass of water

c = specific heat capacity

\Delta T = Temperature change (K or °C)

Replacing the values where mass is 50g and temperature is 80°C to 0°C we have,

Q = mc\Delta T

Q = 50*0.11*(80-0)

Q = 440cal

Then we can calculate the heat absorbed by m grams of ice at 0°C, then

Q_2 = mL = 80*m

How Q_1=Q_2, so

80m=440

m=\frac{440}{80}

m = 5.5g

Then 5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity

7 0
4 years ago
True or false? If two components are connected in series, the current through one component will
Molodets [167]

the answer is ( True ) .

the current is the same in series circuits .

8 0
3 years ago
Why does increasing the number of trials increase confidence in the results of the experiment?
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It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.
8 0
3 years ago
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