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2 years ago

Look at the two question marks between zinc (Zn) and arsenic (As). At the time, no elements were known

Physics

1 answer:

marissa [1.9K]2 years ago

6 0

Answer:

Mendeleev predicted the atomic mass of each element along with compounds they each should form.

Explanation:

Based on other elements in the same group he predicted the existence of eka-aluminum and eka-silicon, later to be named gallium (Ga) and germanium (Ge).

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the steel bed of a suspension bridge is 200m long at 20 C. If the extremes of temperature to which it might be exposed are -30 C

raketka [301]

Answer:

The steel bed will contract by 0.13 m, and expand by 0.052 m

Explanation:

For contraction,

α = ΔL/(LΔθ)..................... Equation 1

Where α = Linear expansivity of steel, ΔL = decrease in length/ Increase in length, L = Original length, Δθ = Change in temperature

make ΔL the subject of the equation

ΔL = α(LΔθ)................. Equation 2

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = -30-20 = -50 °C

Substitute into equation 2

ΔL = 13×10⁻⁶(200)(-50)

ΔL = -0.13 m

Similarly, For expansion,

Using equation 2

ΔL = α(LΔθ)

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = 40-20 = 20 °C

Substitute into equation 2

ΔL = 13×10⁻⁶(200)(20)

ΔL = 0.052 m.

Hence the steel bed will contract by 0.13 m, and expand by 0.052 m

3 0

2 years ago

Given equal time periods, which statement is correct?

Mekhanik [1.2K]

The correct option is B.
Nuclear fission and fusion are two different types of nuclear reactions, through which energy may be obtained. Nuclear fission involves the splitting of a molecule into two different part in order to generate energy while nuclear fusion reaction involves the joining together of two elements in other to form one product. Nuclear fission generate much radioactive waste than nuclear fusion. The radioactive waste that is obtainable during nuclear fusion is less than 1% of that produce by nuclear fission.

3 0

2 years ago

Read 2 more answers

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat

andrezito [222]

Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

$B(t)=5t\ T$ , t is time in seconds

$\dfrac{dB}{dt}=5\ T$

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

$E(2\pi x)=\dfrac{d\phi}{dt}$

x is the distance from the axis of the solenoid

$E(2\pi x)=\pi r^2\dfrac{dB}{dt}$ , r is the radius of the solenoid

$r^2=\dfrac{2xE}{(dE/dt)}$

$r^2=\dfrac{2\times 2\times 1.1}{(5)}$

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

4 0

2 years ago

Read 2 more answers

Calculate the average speed and average velocity of a complete round-trip in which the outgoing 250 km is covered at 95 km/h fol

ser-zykov [4K]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.

PART A)

The time taken to travel a distance of 250km with a speed of 95km/h is

$t = \frac{d}{v}$

$t = \frac{250km}{95km/h}$

$t = 2.63h$

Time taken for the lunch is

$t = 1h$

The time taken travel a distance of 250km with a speed of 55km/h

$t = \frac{d}{v}$

$t = \frac{250}{55}$

$t = 4.54h$

The total time taken is

$t = t_{outgoing}+t_{lunch}}t_{return}$

$t = 2.63+1+4.54$

$t = 8.17h$

The average speed is the ratio of total distance and total time

$v = \frac{250+250}{8.17}$

$v = 61.15km/h$

PART B)

As the displacement is zero the average velocity is zero.

5 0

2 years ago

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago