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OverLord2011 [107]
2 years ago
11

Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so

me have more details available when you click on them. Drag and drop each label to the corresponding area it identifies in the image.
ITEM BANK: Move to Bottom
(I'm in 8th grade and I seriously just need a answer to this)

Physics
1 answer:
Vikki [24]2 years ago
7 0
1 isla thanks for your help and I hope you are feeling 44356
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A car travels 500m in 50s, then 1,500m in 75s. Calculate its averages speed for the whole journey
SIZIF [17.4K]

Answer:

15m/s

Explanation:

500 ÷ 50 = 10m/s

1500 ÷ 75 = 20m/s

10 + 20 = 30

30 ÷ 2 = 15m/s

8 0
2 years ago
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Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh
guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

V = i R

i = \frac{V}{R}

i = \frac{120}{160}

i = \frac{3}{4} A

now the charge flow through it is given as

Q = i t

total time is t = 8 hours

Q = \frac{3}{4}(8 \times 60 \times 60)

Q = 21600 C

7 0
2 years ago
A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards
Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0
2 years ago
Some fuel cells are powered by hydrogen. Scientists are looking into the decomposition of water (H2O) to make hydrogen fuel with
mr_godi [17]

A challenge scientists face with this process is the use of ultrathin iron oxide, to pull protons off water and produce hydrogen gas, which itself is a poor electrical conductor.

5 0
2 years ago
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A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
2 years ago
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