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3 years ago

Which are causes of desertification?

Physics

2 answers:

Alex_Xolod [135]3 years ago

8 0

Answer:

a,c& d , global climate change ,deforestation, overgrazing

on edg

;)

rodikova [14]3 years ago

7 0

Deforestation because it gets rid of the nature and soon the land will be empty and dry

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A 121 turn 121 turn circular coil of radius 2.85 cm 2.85 cm is immersed in a uniform magnetic field that is perpendicular to the

sashaice [31]

Answer:

0.074 V

Explanation:

Parameters given:

Number of turns, N = 121

Radius of coil, r = 2.85 cm = 0.0285 m

Time interval, dt = 0.179 s

Initial magnetic field strength, Bin = 55.1 mT = 0.0551 T

Final magnetic field strength, Bfin = 97.9 mT = 0.0979 T

Change in magnetic field strength,

dB = Bfin - Bin

= 0.0979 - 0.0551

dB = 0.0428 T

The magnitude of the average induced EMF in the coil is given as:

|Eavg| = |-N * A * dB/dt|

Where A is the area of the coil = pi * r² = 3.142 * 0.0285² = 0.00255 m²

Therefore:

|Eavg| = |-121 * 0.00255 * (0.0428/0.179)|

|Eavg| = |-0.074| V

|Eavg| = 0.074 V

5 0

3 years ago

Read 2 more answers

What element is used as a lining in aprons to protect people in x rays

strojnjashka [21]

It's lead. That's why the "apron" is so heavy.

4 0

3 years ago

What is the elevation of Y and Z? Y=3250, Z=2950 Y=3200, Z=2900 Y=3250, Z=2900 Y=3000, Z=2850

TiliK225 [7]

I think the elevation of Y and Z are the following:
<span>Y=3200,
Z=2900 </span>

8 0

3 years ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

4 years ago

Which wave has a disturbance that is parallel to the wave motion?

Leto [7]

The answer is D.<span>longitudinal</span>

6 0

3 years ago