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Alisiya [41]
3 years ago
11

List in order from smallest to largest the grain sizes used to describe detrital rocks.

Physics
2 answers:
Masja [62]3 years ago
7 0
If there is more to the problem
I can help you
earnstyle [38]3 years ago
3 0
Where is the rest of the problem
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Two cars leave towns 680 kilometers apart at the same time and travel toward each other. one car's rate is 10 kilometers per hou
Evgen [1.6K]

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

5 0
3 years ago
This model shows an example of a fog bank formation. This can happen in the Great Lakes area as warm summer air moves across the
Vadim26 [7]

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

#SPJ1

5 0
2 years ago
Please help calculate them<br> A-f please very urganr
valentina_108 [34]
Three 40w lamp for 6 hours
8 0
3 years ago
1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it
spin [16.1K]

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. KE=\frac{1}{2}mv^2 so

   KE=\frac{1}{2}(2.5)(14)^2 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:

   380=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(380)}{2.5} } 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

   310=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(310)}{2.5} } 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:

   v=v_0+at 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 = v_0t+\frac{1}{2}at^2 and filling in the time the object was at 26 m/s:

   Δx = 0t + \frac{1}{2}(-9.8)2.7)^2 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.

4 0
2 years ago
Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi
Snowcat [4.5K]

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'.

⇒ V=20\ \Omega,\ V_1=30\ \Omega

⇒ G=1680\ \Omega,\  G_1=2930\ \Omega

Concept to be used:

Conversion of galvanometer into voltmeter.

Let G be the resistance of the galvanometer and I_g the maximum deflection in the galvanometer.

To measure maximum voltage resistance R is connected in series .

So,

⇒ V=I_g(R+G)

We have to find the value of R we know that in series circuit current are same.

For G=1680                                    For G_1=2930

⇒ I_g=\frac{V}{R+G}   equation (i)                ⇒ I_g=\frac{V_1}{R+G_1} equation (ii)

Equating both the above equations:

⇒ \frac{V}{R+G} = \frac{V_1}{R+G_1}

⇒ V(R+ G_1) = V_1 (R+G)

⇒ VR+VG_1 = V_1R+V_1G

⇒ VR-V_1R = V_1G-VG_1

⇒ R(V-V_1) = V_1G-VG_1

⇒ R =\frac{V_1G-VG_1}{(V-V_1)}

⇒ Plugging the values.

⇒ R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}

⇒ R =\frac{(50400 - 58600)}{(-10)}

⇒ R=\frac{-8200}{-10}

⇒ R=820\ \Omega

The coil resistance of the circuit is 820 Ω .

4 0
3 years ago
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