D = distance between the cars at the start of time = 680 km
v₁ = speed of one car
v₂ = speed of other car = v₁ - 10
t = time taken to meet = 4 h
distance traveled by one car in time "t" + distance traveled by other car in time "t" = D
v₁ t + v₂ t = D
(v₁ + v₂) t = D
inserting the values
(v₁ + v₁ - 10) (4) = 680
v₁ = 90 km/h
rate of slower car is given as
v₂ = v₁ - 10
v₂ = 90 - 10 = 80 km/h
Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.
<h3>How Fog banks formed?</h3>
Fog banks form at sea where cool air moves quickly over the surface of the ocean that is warm. The cool incoming air lowers the temperature of the air just above the water surface and water vapor condenses into fog.
So we can conclude that Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.
Learn more about fog here: brainly.com/question/18943608
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Three 40w lamp for 6 hours
Answer:
Explanation:
I got everything but i. Don't know why but it's eluding me. So let's do everything but that.
a. PE = mgh so
PE = (2.5)(98)(14) and
PE = 340 J
b.
so
and
KE = 250 J
c. TE = KE + PE so
TE = 340 + 250 and
TE = 590 J
d. PE at 8.7 m:
PE = (2.5)(9.8)(8.7) and
PE = 210 J
e. The KE at the same height:
TE = KE + PE and
590 = KE + 210 so
KE = 380 J
f. The velocity at that height:
and
so
v = 17 m/s
g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:
590 = KE + PE and
PE = (2.5)(9.8)(11.6) so
PE = 280 then
590 = KE + 280 so
KE = 310 then
and
so
v = 16 m/s
h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:
and
26 = 0 + 9.8t and
26 = 9.8t so the time at 26 m/s is
t = 2.7 seconds. Now we use that in the equation for displacement:
Δx =
and filling in the time the object was at 26 m/s:
Δx = 0t +
so
Δx = 36 m
i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.
Answer:
Resistance of the circuit is 820 Ω
Explanation:
Given:
Two galvanometer resistance are given along with its voltages.
Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.
⇒ 
⇒ 
Concept to be used:
Conversion of galvanometer into voltmeter.
Let
be the resistance of the galvanometer and
the maximum deflection in the galvanometer.
To measure maximum voltage resistance
is connected in series .
So,
⇒ 
We have to find the value of
we know that in series circuit current are same.
For
For 
⇒
equation (i) ⇒
equation (ii)
Equating both the above equations:
⇒
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ Plugging the values.
⇒ 
⇒ 
⇒ 
⇒
The coil resistance of the circuit is 820 Ω .