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2 years ago

Which activity is a method of habitat restoration?

Physics

2 answers:

irina1246 [14]2 years ago

7 0

A. cleaning up after environmental disasters

Flura [38]2 years ago

6 0

A. Cleaning up after environmental disasters

Hope that helps

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A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu

melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

$|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)$

By taking into account the angles between the vectors r and v in each case we obtain:

v=(2,0)

r=(0,1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

r=(0,-1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

r=(-1,-1)

angle = 135°

$L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}$

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0

2 years ago

A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction

aleksley [76]

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

5 0

2 years ago

A hiker caught in a rainstorm absorbs 1.00 L of water in her clothing. If it is windy so that the water evaporates quickly at 20

masya89 [10]

Answer:

(A) Q = 2.26×10⁶J

(B) ΔT = 9°C

(C)

Explanation:

We have been given the mass of the hiker, the volume of water from which we can calculate the mass knowing that the density if water is 1000kg/m³.

Evaporation is a phase change and occurs at a constant temperature. We would use the latent heat of vaporization to calculate the amount of heat evaporated.

We would then equate this to the heat change it brings about in the hiker's body and then calculate the temperature drop.

See the attachment below for full solution.

6 0

2 years ago

This sphere is now connected by a long, thin conducting wire to another sphere of radius R2 that is several meters from the firs

alekssr [168]

Here is the full question

A metal sphere with Radius R₁ has a charge Q₁. Take the electric potential to be zero at an infinite distance from the sphere

a) What are the electric field and electric potential at the surface of the sphere?

This sphere is now connected by a long, thin conducting wire to another sphere of radius R₂ that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached:

b) what is the total charge on each sphere?

Assume that the amount of charge on the wire is much less than the charge on each sphere.

Answer:

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

Explanation:

Given that;

the radius of the sphere = R

The radius of the first sphere = $R_1$

The radius of the second sphere = $R_2$

Charge on the first sphere = $Q_1$

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b) From the question. before the part b question; we learnt that the first sphere is now connected to another sphere;

Now that the two sphere are joined . Charges flows from one to another until their potentials are equal.

As Such; We use $q_1 \ and \ q_2$ to represent their charges respectively

The potential on the surface of the first sphere;

$V_1 = \frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}$

The potential on the surface of the second sphere;

$V_2 = \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

$V_1=V_2$

∴

$\frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}= \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

Thus, we can say :

$\frac{q_1}{q_2}= \frac{R_1}{R_2}$

and $Q_1 = q_1 + q_2$

As such ;

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

3 0

2 years ago

Capillary action in trees can transport water from the roots to the tree's branches. The capillaries (Xyelem) in a certain tree

lesantik [10]

Answer:

$h=14.2857\,m$

Explanation:

Given:

radius of capillary, $r=10^{-6}\,m$

angle of contact, $\theta=0^{\circ}$

density of water, $\rho=1000\,kg.m^{-3}$

surface tension of water, $T=0.07 \,N.m^{-1}$

height, h = ?

We have the equation for the height of meniscus as:

$h=\frac{2T.cos\, \theta}{\rho.g.r}$

$h=\frac{2\times 0.07\times cos\,0^{\circ}}{1000\times 9.8\times 10^{-6}}$

$h=14.2857\,m$

No, the capillary action alone cannot be the mechanism of water transportation to the top of the trees. Transpiration also creates a suction pressure in the xylem complementary to the ascent of sap and cohesion of water being the other causes of movement of water up in the plants.

6 0

2 years ago