Ask question

Ask question

All categories

3 years ago

6

More buffer zones would be needed around a large protected area vs. several patches of protected area of equal square footage.

2 answers:

Fudgin [204]3 years ago

6 0

Your answer is it is True. Please no mean comments I'm new to brainly

Paha777 [63]3 years ago

6 0

False you need more buffer zones in smaller areas

You might be interested in

An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's a

devlian [24]

Answer:

a

$D = 1162.7 \ m$

b

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + \frac{1}{2}(-g)t^2$

=> $-532 = 71.83 t + \frac{1}{2}(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^{-1}[\frac{v_y}{v_x } ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

So

$\beta = tan ^{-1}[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

6 0

3 years ago

What term describes the rate at which charge passes a point in a circuit

docker41 [41]

That's electric "current", usually described in 'Amperes'.

1 Ampere = 1 Coulomb per second ... "the rate at which
charge passes a point in the circuit".

8 0

3 years ago

Read 2 more answers

As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.15 mm apart and position

Artist 52 [7]

Answer:

The distance of the first bright fringe is given as $Y_C = 1.22 *10^{-3}m$

The distance of the second dark fringe from the central bright fringe is given as $Y_D = 0.00192 \ m$

Explanation:

From the question we are told that

The slit separation distance is $d = 1.15 mm = \frac{1.15}{1000} =0.00115 m$

The distance of the slit from the screen is $D = 3.23 m$

The wavelength is $\lambda = 633 nm$

For constructive interference to occur the distance between the two slit is mathematically represented as

$Y_C =\frac{m \lambda D}{d}$

Where m is the order of the fringe which has a value of 1 for first bright fringe

Substituting values

$Y_C = \frac{1 * 633 *0^{-9} * 3.23}{0.00115}$

$Y_C = 1.22 *10^{-3}m$

For destructive interference to occur the distance between the two slit is mathematically represented as

$Y_D = [n + \frac{1}{2} ] \frac{\lambda D}{d}$

m = 2

so the formula to get the dark fringe is $n = \frac{1}{2} * 1$

$n=1$

Now substituting values

$Y_D = [ 1 + \frac{1}{2} ] * \frac{633 *10^{-9} * 3.23 }{0.00115}$

$Y_D =1.5 * \frac{633 *10^{-9} * 3.23 }{0.00115}$

$Y_D = 0.00192 \ m$

4 0

3 years ago

During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electr

WINSTONCH [101]

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

$E = \frac{\Delta V}{d}$

<u>Where:</u>

E: is electric field

ΔV: is the potential difference = 3.95 mV

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

$E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m$

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

7 0

2 years ago

A driverless car collides head on with a stationary sign. The collision brings the car to a stop, but does not move the sign.

neonofarm [45]

Answer:

I'm sorry hindi ko po alam ung sagot sorry po

4 0

2 years ago

Read 2 more answers