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Rus_ich [418]
3 years ago
14

Please help I'll give brainliest!!

Physics
2 answers:
photoshop1234 [79]3 years ago
4 0

Answer:

this may be wrong but when i looked it up it said “Seafloor Mapper”

Explanation:

Llana [10]3 years ago
3 0

Answer:

the technology and research for your career is the tech used is sonar tracker or map as well as the technology and research used is to track the habitants of the ocean floor  

Explanation:

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If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser
Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em></em>

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0
3 years ago
The term pressure most nearly refers to which of the following?
Arte-miy333 [17]

The term pressure refers to the force per area created by the weight
of anything whose weight is distributed over an area, such as the
Earth's atmosphere, a lake, a gas inside a sealed jar, or a pointy
high heel.

5 0
3 years ago
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What tool is used to measure air pressure?
alukav5142 [94]
A<span> barometer is used to measure air pressure. </span>
8 0
3 years ago
1) A car is moving to the right with the constant speed of 1.2 m/s. If the car starts
Marina CMI [18]
(100, 108)
Due to
1.2x90=108

100, 108
7 0
3 years ago
A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
3 years ago
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