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Dmitrij [34]
2 years ago
11

Assume you are given an int variable named nElements and a 2-dimensional array that has been created and assigned to a2d. Write

one or more statements that assign to nElements the total number of elements that could be stored in the entire 2-dimensional array.
Physics
1 answer:
lozanna [386]2 years ago
8 0

Answer:

SEE EXPLAIN

Explanation:

public int dimension(int [][]a2d,int nElements)

{

int count = 0;

for(int i = 0;i < a2d.length ; i++)

{

count = count + a2d[i].length;

}

return count;  

}

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Which pair of graphs represent the same motion of an object
ASHA 777 [7]
To be able to identify that the object is in the same motion, we should find the graphs that has an increasing slope of displacement and with the constant velocity with varying time. Graphs on letter D satisfies these requirements.

<em>ANSWER: D</em>
6 0
2 years ago
Read 2 more answers
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
2 years ago
José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
zaharov [31]

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg


8 0
2 years ago
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
SCORPION-xisa [38]

Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

7 0
2 years ago
A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener
Bingel [31]

The ball was dropped from a height 20 meters

Explanation:

The given is

1. A ball is dropped from the top of a cliff

2. By the time it reaches the ground, all the energy in its gravitational

   potential energy store has been transferred into its kinetic energy

   store, that mean K.E = P.E

3. The ball is travelling at 20 m/s when it hits the ground

4. The gravitational field strength is 10 N/kg

We need to find the height that the ball dropped from it

The ball dropped from the top of a cliff means the initial speed is 0

→ K.E = \frac{1}{2}m(v^{2}-v_{0}^{2})

where v is the final speed, v_{0} in the initial speed and m

is the mass

→ v = 20 m/s and v_{0} = 0 m/s

→ K.E = \frac{1}{2}m(20^{2}-0^{2})

→ K.E = \frac{1}{2}m(400)

→ K.E = 200 m joules ⇒ when the ball hits the ground

→ P.E = m g h

where g is the gravitational field strength, m is the mass and h is

the height

→ g = 10 N/kg

→ P.E = m(10)(h)

→ P.E = 10 m h joules

→ P.E = K.E

→ 10 m h = 200 m

Divide both sides by 10 m

→ h = 20 meters

The ball was dropped from a height 20 meters

Learn more

You can learn more about gravitational potential energy in brainly.com/question/1198647

#LearnwithBrainly

8 0
2 years ago
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