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Alenkinab [10]
3 years ago
6

Question 3 You see the coordinates 5 E longitude, 10 N latitude. You do not need to look at a map in order to deduce that this l

ocation is a. near both the equator and the prime meridian Selected: b. near the International Date Line as well as the North PoleThis answer is incorrect. c. near the International Date Line and the prime meridian d. near the equator but quite far from the prime meridian e. near both the equator and the International Date Line
Physics
2 answers:
olchik [2.2K]3 years ago
8 0

Answer:

a. near both the equator and the prime meridian

Explanation:

Lets first understand what is equator and prime meridian.

We have drawn an imaginary grid on the Earth for navigational and other purposes. This grid has two coordinates : Latitude and Longitude. Equator is the largest latitude circle that divides the Earth in two equal hemispheres. This is taken as 0° for latitude. Latitudes increase as we move towards North/South pole which lies at 90°N and 90° S respectively. Longitudes are the lines running from north pole to south pole. The longitude that crosses from Greenwich is taken as 0°. This longitude is also known as prime meridian. The value of longitudes vary from 0° to 180° East and 180° West.

Now, the given coordinates are: 5° E and 10° N. It is quite evident that this location will be close to equator and prime meridian.

san4es73 [151]3 years ago
5 0

Answer:

a. Near both the equator and the prime meridian.

Explanation:

The  equator is at 0 degrees latitude and the prime meridian is 0 degrees longitude.

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Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by
lara31 [8.8K]

Answer:

Option D

2 m/s

Explanation:

The speed is given by distance/time

The complete wave has a distance of 8 m (This is the wavelength)

Time taken for complete wave is 4 s

Therefore, velocity= 8 m/ 4 s= 2 m/s

The amplitude is the distance between crest and trough hence 9 m- 6 m= 3 m

Frequency is 1/period and in this case period is 4 s hence frequency= 1/4= 0.25 Hz

Essentially, what we wanted is velocity

3 0
4 years ago
A 50-cm wire placed in an east-west direction is moved horizontally to the north with a speed of 2.0 m/s. the horizontal compone
Strike441 [17]

When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.

Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula

EMF = v.(B x L)

given that

B = 25\mu Tj - 50\mu Tk

v = 2 m/s j

L = 0.50 m (-i)

now by using the above formula we will have

EMF = 2(j) .(25\mu j - 50\mu k) x (-0.50 i)

EMF = 2(j) .(12.5\mu k + 25\mu j)

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5 0
3 years ago
Read 2 more answers
An airplane awaiting departure is cleared to takeoff, its accelerates down a runway at 3.20m/ s2 for 32.8s until it finally lift
stich3 [128]

The final velocity before takeoff is 104.96 m / s.

<u>Explanation:</u>

The last velocity of a given object over some time defines the final velocity. The final velocity of the object is given by the product of acceleration and time and adding this product to the initial velocity.

To calculate the final velocity,

                          V = u + at

where v represents the final velocity,

           u represents the initial velocity,

           a represents the acceleration

            t represents the time taken.

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                            v   = 104.96 m / s.

7 0
3 years ago
A specimen of aluminum having a rectangular cross section 9.5 mm × 12.9 mm (0.3740 in. × 0.5079 in.) is pulled in tension with 3
ryzh [129]

Answer:

The resultant strain in the aluminum specimen is 4.14 \times {10^{ - 3}}

Explanation:

Given that,

Dimension of specimen of aluminium, 9.5 mm × 12.9 mm

Area of cross section of aluminium specimen,

A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2

Tension acting on object, T = 35000 N

The elastic modulus for aluminum is,E=69\ GPa=69\times 10^9\ Pa

The stress acting on material is proportional to the strain. Its formula is given by :

\epsilon=\dfrac{\sigma}{E}

\sigma is the stress

\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}

Thus, The resultant strain in the aluminum specimen is 4.14 \times {10^{ - 3}}

5 0
4 years ago
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Tertiary consumer is always at the top.
The producer is always at the bottom.
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