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faust18 [17]
3 years ago
9

What is the average velocity of a rocket that travels 15 m in 0.25 seconds?​

Physics
1 answer:
Brrunno [24]3 years ago
5 0

Answer:

<em><u>60 m/s</u></em>

Explanation:

Average Speed = Distance/ Time

= 15/0.25 = 60m/s

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a pine raft (density = 373 kg/m^3) has a volume of 1.43 m^3. How much of the raft's volume is below the water line? Unit is m^3​
Slav-nsk [51]

Answer: 0.5334

Explanation:

i got it right on accellus :p

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3 years ago
Which of these is the unit of measure for work?
ryzh [129]
The unit of measurement of work is the Joules. B.
4 0
3 years ago
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete
disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) F_{net} \text {on wire }3=0

\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm

position of wire = 50 - 1.2

= 48.8cm

b)  F_{net} \text {on wire }1=0

\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A

Direction ⇒ downward

5 0
3 years ago
I SERIOUSLY can't do this type of questions so can someone solve it detailedly and putting with letters (there is a system you n
KatRina [158]

Answer:

4 Ohms

Explanation

(This is seriously not as hard as it looks :)

You only need two types of calculations:

  1. replace two resistances, say, R1 and R2, connected in a series by a single one R. In this case the new R is a sum of the two: R = R_1+R_2
  2. replace two resistances that are connected in parallel. In that case: \frac{1}{R}= \frac{1}{R_1}+\frac{1}{R_2}\\\mbox{or}\\R= \frac{R_1\cdot R_2}{R_1+R_2}

I am attaching a drawing showing the process of stepwise replacement of two resistances at a time (am using rectangles to represent a resistance). The left-most image shows the starting point, just a little bit "warped" to see it better. The two resistances (6 Ohm next to each other) are in parallel and are replaced by a single resistance (3 Ohm, see formula above) in the top middle image. Next, the two resistances (9 and 3 Ohm) are nicely in series, so they can be replaced by their sum, which is what happened going to the top right image. Finally we have two resistances in parallel and they can be replaced by a single, final, resistance as shown in the bottom right image. That (4 Ohms) is the <em>equivalent resistance</em> of the original circuit.

Using these two transformations you will be able to solve step by step any  problem like this, no matter how complex.  

5 0
3 years ago
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