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diamong [38]
3 years ago
11

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

e one or more statements that assign to nElements the total number of elements that could be stored in the entire two-dimensional array.
Physics
1 answer:
Alex777 [14]3 years ago
3 0

Answer:

Explained

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;

}

You might be interested in
In mechanics, massless strings are often assumed. Why is that not a good assumption when discussing waves on strings?
Marizza181 [45]

In mechanics, massless strings are often assumed. but this is not a good assumption when discussing waves on strings because the speed of a wave on a massless string would be infinite.

<h3>How to explain the information?</h3>

It should be noted that waves simply means the dynamic disturbance of a quantity.

It should be noted that in mechanics, massless strings are often assumed. but this is not a good assumption when discussing waves on strings because the speed of a wave on a massless string would be infinite.

Learn more about waves in:

brainly.com/question/15663649

#SPJ4

6 0
1 year ago
A train travels southwest from point A to point B through the Arizona desert at 55 mi/h. How far will the train travel six-and-a
TiliK225 [7]

Explanation:

Distance = speed × time

d = (55 mi/hr) (6.5 hr)

d = 357.5 mi

4 0
3 years ago
He shoots the tree stump, which has a mass of (M), with a bullet of mass (m) traveling at some velocity (vbullet), and the bulle
Ymorist [56]

Answer:

Explanation:

Given

mass of tree stump is M

mass bullet is m

velocity of bullet is v

Conserving momentum for bullet and tree stump

Initial Momentum P_i=mv

Suppose v_0 is the velocity of the system

Final Momentum P_f=(M+m)v_0

Initial momentum =Final Momentum

mv=(m+M)v_0

v_0=\frac{mv}{m+M}

4 0
3 years ago
When the acceleration of a mass on a spring is zero, the velocity is at a
Sergeu [11.5K]

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

F=kx

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

F=ma

where

m is the mass

a is the acceleration

So we can write the equation as

ma=kx (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const. (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

- When acceleration decreases, velocity increases

Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a  maximum

When the velocity of a mass on a spring is zero, the acceleration is at a  maximum

6 0
3 years ago
In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an
VashaNatasha [74]

Answer:

V=3.475ft^3/s=3.48ft^3/s

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})

P= 73.7*10^{3}ft.lbf/s

we have other values such h=400ft and  \gamma= 62.42lb/ft^3 (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

V=\frac{P}{\gamma h \eta_0}

Where \eta_0 is the turbine efficiency, at which is,

\eta_0 = \frac{P}{\gamma Vh}

Replacing,

V=\frac{73.7*10^{3}}{62.42*400*0.85}

V=3.475ft^3/s

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

m= 217lbm/s

6 0
3 years ago
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