What is definition of population?

A block is suspended from a spring. The spring is stretched .2 meters. If the spring constant is 200 newtons per meters, what is

AleksandrR [38]

The spring force = kx = mg.
k is spring constant = 200 N/m
x is the spring elongation = 0.2 m
F = 200×0.2= 40 N
Now, F = mg wil be the weight of the object
Thus, weight is 40 N.
Thus, now we can also find the mass of object, m = F/g = 40 N/9.8
Because g is acceleration due to gravity = 9.8 m/sec ^2
m = 4.08 kg

4 0

3 years ago

You are given a semiconductor resistor made from silicon with an impurity concentration of resistivity 1.00×10−3ωm. the resistor

Vlad [161]

Answer: Resistance = $3.2 \Omega$ , Current = 1.56 A, Voltage =4.99 V

The resistance,

$R=\frac {\rho l}{A}$

where, $\rho$ is resistivity, A is the area and l is the length of the resistor.

It is given that:

$\rho=1.0\times10^{-3}\Omega m$

Length, l=2 mm

Area, $A= width \times height=1.25 mm\times 0.5 mm=0.625 mm^2$

Hence, $R=\frac{1.0\times10^{-3}\Omega m \times 2\times10^{-3}m}{0.625\times10^{-6}m^2}=3.2\Omega$

We know, Power, $P=I^2R$

$\Rightarrow I=\sqrt{\frac{P}{R}}$

$P=7.81 W$

$I=\sqrt{\frac {7.81 W}{3.2\Omega}}=\sqrt{2.44}A=1.56A$

We know, Voltage, $V=IR=1.56\times3.2=4.99 V$

4 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,

Leviafan [203]

Answer:

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

Explanation:

As we know that the charge per unit length of the long cylinder is given as

$\lambda = \frac{q}{L}$

here we know that the electric field between two cylinders is given by

$E = \frac{2k\lambda}{r}$

now we know that electric potential and electric field is related to each other as

$\Delta V = - \int E.dr$

$\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr$

$\Delta V = -2k \lambda ln(\frac{b}{a})$

$\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}$

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

7 0

3 years ago

Good science does not depend on interactions within the scientific community true or false

vlabodo [156]

Answer:

I would say that is false. Science can only be perfect after at least some sort of scientific communication and interaction.

6 0

3 years ago