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3 years ago

The effects of agriculture and industry can be more harmful with the further loss of wetlands.

1 answer:

7 0

Yes, it is true that the effects of agriculture and industry can be more harmful with the further loss of wetlands since industry usually requires large amounts of land.

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A car is moving with the velocity of 90km/h. If the car come to rest after 10 seconds. Calculate the final velocity and distance

djyliett [7]

Answer:

you can learn from here

https://www.toppr.com/ask/en-bd/question/a-car-is-moving-with-a-velocity-of-10-ms-the-driver-sees-a-wall/

6 0

3 years ago

What is happening in the picture?

Veseljchak [2.6K]

The gravity is pushing rhe boat down

3 0

3 years ago

Read 2 more answers

Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu

Serggg [28]

Answer:

a) 17.086m

b) 0.1671 m

Explanation:

Given data: speed of water through the hose = 1.81 m/s

through the nozzle = 18.3 m/s

We know that maximum height of an object with upward velocity v is given by,

a) H = v^2/2g

where H is the maximum height water emerges

= 18.3^2/(2×9.8) = 17.086 m answer

b) Again,

H = v^2/2g

= 1.81^2/(2×9.8) = 0.1671 m

6 0

2 years ago

A skateboarder, starting from rest, rolls down a 10.9-m ramp. When she arrives at the bottom of the ramp her speed is 6.74 m/s.

JulijaS [17]

Answer:b

Explanation:b

4 0

3 years ago

You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio

Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

$E = \dfrac{\sigma }{2 \varepsilon_o}$

$E = \dfrac{8 \times 10^{-6} }{2 \times 8.85 \times 10^{-12}}$

$E = 4.51977 \times 10^5 \ V/m$

On the square plate; The electric force F = Eq

$F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)$

$F = 1.3559 \times 10^{-2} \ N$

The acceleration $a =\dfrac{ F}{m}{$

$a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}$

$a = 2.25988 \times 10^6 \ m/s^2$

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

$(\dfrac{1}{2}) mv^2 = \Delta Vq$

$v^2 = \dfrac{2 Eq}{dm}$

$v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }$

$v^2 = 64568142.86 \ m/s$

$v =\sqrt{ 64568142.86 \ m/s}$

$\mathbf{v = 8.035 \times 10^3 \ m/s}$

From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".

Hence, the force is an attractive force.

Similarly, there is a gradual increase exhibited by the velocity of the particle.

Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.

6 0

3 years ago