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mote1985 [20]
3 years ago
11

The acceleration due to gravity on the surface of the moon is 1.62m/s^2. The moon's radius is Rm+1738km. A) What is the weight i

n newtons on the surface of the moon of an object that has a mass of 10kg? B) Determine the force extend on the object by the gravity of the moon if the object is located 1738km above the moon's surface.
Engineering
1 answer:
Anastaziya [24]3 years ago
4 0

Answer:

weight is 12.6 N

force is 4.05 N

Explanation:

given data

acceleration = 1.62 m/s²

radius = 1738 km

mass = 10 kg

distance = 1738 km

to find out

weight and force

solution

we apply here weight formula that is

weight =  mass × acceleration    ...................1

put here value

weight =  10 × 1.26

weight = 12.6 N

and

force = mass × An

force = 10 × 1.62 (1738/ 1738+1738)² = 4.05 N

so force is 4.05 N

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If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the cond
Dennis_Churaev [7]

Answer:

The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²

Explanation:

The electric field intensity due to a thin conducting sheet is given by the following formula:

Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)

From this formula:

Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)

We have the following data:

Electric Field Intensity = 1.5 N/C

Permittivity of free space = 8.85 x 10^-12 C²/N.m²

Therefore,

Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)

<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>

Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².

7 0
3 years ago
The spring has a stiffness k = 200 N&gt;m and an unstretched length of 0.5 m. If it is attached to the 3-kg smooth collar and th
ruslelena [56]

Answer:

15.467 m/s

Explanation:

See attached picture.

4 0
3 years ago
Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for-end loop the program exa
ELEN [110]

Answer:

%Program prompts user to input vector

v = input('Enter the input vector: ');

%Program shows the value that user entered

fprintf('The input vector:\n ')

disp(v)

%Loop for checking all array elements

for i = 1 : length(v)

   %check if the element is a positive number

   if v(i) > 0

       %double the element

       v(i) = v(i) * 2;

   %else the element is negative number.

   else

       %triple the element

       v(i) = v(i) * 3;

   end

end

%display the modified vector

fprintf('The modified vector:\n ')

disp(v)

4 0
3 years ago
Convert 25 mm into in.
mihalych1998 [28]

Answer:

25 mm = 0.984252 inches

Explanation:

Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:

<u>1 mm = 1/25.4 inches</u>

From the question, we have to convert 25 mm into inches

Thus,

<u>25 mm = (1/25.4)*25 inches</u>

So,

25 mm=\frac{25}{25.4} inches

Thus, solving we get:

<u>25 mm = 0.984252 inches</u>

4 0
3 years ago
The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg
kiruha [24]

Answer:

percentage change in volume  = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × 10^{11}  N/m²

bulk stress = 10^{7}  N/m²

solution

we will apply here bulk modulus formula that is

bulk modulus = \frac{bulk\ stress}{bulk\ strain}   ...............1

put here value and we get

3.5 × 10^{11} = \frac{10^7}{bulk\ strain}  

solve it we get

bulk strain = 2.8571 × 10^{-5}

and

bulk strain = \frac{change\ volume}{original\ volume}  

so that percentage change in volume is = 2.8571 × 10^{-5}  × 100

percentage change in volume  = 0.00285 %

6 0
3 years ago
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