Ben haub hospital Is renovating

A fast teaub abd a ski train driving from city a and city b face to face the fast train has a speed of 76km/h, the slow train ha

LekaFEV [45]

The distance between the two cities is 513.24 km.

<h3>Time of motion when the two trains meet</h3>

The time spent on the journey when the two trains meet is calculated as follows;

(Va - Vb)t = d

where;

d is the distance between the trains before meeting

(76 - 65)t = 40

11t = 40

t = 40/11

t = 3.64 hr

<h3>Distance traveled by the fast train</h3>

d1 = 76 km/h x 3.64 h

d1 = 276.64 km

<h3>Distance traveled by the slow train</h3>

d2 = 65 km/h x 3.64 h

d2 = 236.6 km

The distance between the two cities = 276.64 km + 236.6 km

= 513.24 km

Learn more about relative velocity here: brainly.com/question/17228388

7 0

2 years ago

Forms when heat, pressure, or fluids act on igneous, sedimentary, or other metamorphic rock to change its form or composition, o

BARSIC [14]

I think its a metamorphic rock because i know 100% that a metamophic rocks are created by heat and pressure

4 0

3 years ago

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 5.00 T/s. Part A) What is the electric field str

Veronika [31]

Answer:

(A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is $3.75\times10^{-2}\ V/m$ .

Explanation:

Given that,

Magnetic field = 2.0 T

Diameter = 5.0 cm

Rate of decreasing in magnetic field = 5.00 T/s

(A). We need to calculate the electric field strength inside the solenoid at a point on the axis

Using formula of electric field inside the solenoid

$E=\dfrac{r}{2}|\dfrac{dB}{dt}|$

Electric field on the axis of the solenoid

Here, r = 0

$E=\dfrac{0}{2}\times5.00$

$E = 0$

The electric field strength inside the solenoid at a point on the axis is zero.

(B). We need to calculate the electric field strength inside the solenoid at a point 1.50 cm from the axis

Using formula of electric field inside the solenoid

$E=\dfrac{r}{2}|\dfrac{dB}{dt}|$

$E=\dfrac{1.50\times10^{-2}}{2}\times|5.00|$

$E=0.0375= 3.75\times10^{-2}\ V/m$

Hence, (A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is $3.75\times10^{-2}\ V/m$ .

4 0

4 years ago

The air in a room with volume 180 m3 contains 0.25% carbon dioxide initially. fresher air with only 0.05% carbon dioxide flows i

Butoxors [25]

Answer:

$p(t)=\frac{1}{20}(1+2e^{-t/90})$

Explanation:

Let $p(t)$ be % of $CO_2$ at time $t$ (time in minutes).

Amount of $CO_2$ in room=Volume of room $\times p$

Rate of change of $CO_2$ in room=Volume of room x $\frac{dp}{dt}$

$=180\times \frac{dp}{dt}$

Rate of inflow of $CO_2$ = $(\% CO_2 \ in\ Fresh\ Air \times \frac{1}{100})\times(Rate \ of Inflow\ of \ air)$

$=(0.05\times \frac{1}{100})\times 2=\frac{1}{100}$

Rate of $CO_2$ outflow

$(\% CO_2 \ in\ Fresh\ Air \times \frac{1}{100})\times(Rate \ of Outflow\ of \ air)\\=(p\times\frac{1}{100})\times2=\frac{p}{50}$

Therefore rate of change of $CO_2$ is $\frac{1}{100}-\frac{p}{50}$

% rate of change is $100\times \ Rate of \ Change=\frac{1}{10}-2p$

Therefore we have:

$180\frac{dp}{dt}=\frac{1}{10}-2p\\\frac{dp}{dt}=\frac{1-20p}{1800}\\\\\frac{dp}{20p-1}=-\frac{dt}{1800}$

Integrate both:

$\int\limits \frac{dp}{20p-1}=\int\limits-\frac{dt}{1800}$

$\frac{1}{20}In|20p-1|=-\frac{t}{1800}+In \ C$

In $|20p-1|=-\frac{t}{90}+$ In $K$

$+20In \ C=+In \ K$

Raise both sides to base $e$ ,

$20p-1=e^{-\frac{t}{90}+In \ K}\\\\p=\frac{1}{20}(1+Ke^{-t/90})$

Initially, there's only 0.25% of $CO_2$ ,

Now substitute $p=0.25$ and $t=0$

$0.25=\frac{1}{20}(1+K)\\K=4$

$p=\frac{1}{20}(1+2e^{-t/90})$

8 0

3 years ago

What's the Coulomb's law?

Ulleksa [173]

<span>
In layman's term: </span>like charges don't attract while opposite charges do<span>electrostatic forces between point A( which is charged) and point B (which is also charged) are proportional to the charge of point A and point B. </span><span>there is also something else about this law that I don't quite remember.</span>

<span />Here is the formula:

<span>What the formula means:</span>

F=force between charges

Q1 and Q2= amount of charge

d=distance between these two charges

k= Coulombs constant (proportionally constant)

________________________________________________

I think that about covers it and hopefully this helped.

4 0

3 years ago

Read 2 more answers