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egoroff_w [7]
3 years ago
3

Explain why clear-cutting is a more destructive method of wood harvesting than selective cutting.

Physics
2 answers:
ddd [48]3 years ago
4 0

Answer:

the person above me is correto

Explanation:

Juliette [100K]3 years ago
3 0

Answer:

Simply, clear-cutting removes all the trees in an area. Alas, this leaves nothing behind to provide cover for delicate topsoil or homes for wildlife.

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A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total
Rudik [331]

Answer:

a) 0.41 m/s

b) 0.51 m/s

Explanation:

(a)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (15.4) + (93.1 - 0.292) V'

V' = 0.41 m/s

b)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = - 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (- 15.4) + (93.1 - 0.292) V'

V' = 0.51 m/s

6 0
4 years ago
Arcsin 0.9331 in degrees​
lisabon 2012 [21]

Answer:

68.9233231661

Explanation:

Just put it into your calculator, shift sin should do it but it will come up like this: sin^{-1} which is the same as arcsin

6 0
3 years ago
Compare the magnitude of the electromagnetic and gravitational force between two electrons separated by a distance of 2.00 m. As
Lesechka [4]
First you need to know about two laws, which are:
1) Coulomb's law
2) Newton's law of gravitation

1.
According to Coulomb's law, Electric force between TWO charges is:
F_{e} =  \frac{k*q_{1}*q_{2}}{r^{2}} -- (A)

Where, k = 1/(4*π*epsilon_not) = 9 * 10^{9} \frac{Nm^{2}}{C^{2}}

Both q_{1} and q_{2} = -1.61 x 10^{-19} C
r = Distance between the two charges = 2.00m

Plug-in the above values in (A), you would get:

F_{e} = (9 * 10^{9} ) (1.61 * 10^{-19}  * 1.61 * 10^{-19}) / (2*2)  

F_{e} = 5.83* 10^{-29} N

2.
According to Newton's law of gravitation:
F_{g} =  \frac{Gm_{1}m_{2}}{r^{2}} -- (B)

Where G = Gravitational constant = 6.674 * 10^{-11} m^{2} kg^{-1}s^{-2}

m1 = m2 = Mass of the electron = 9.11 * 10^{-31} kg
r = 2.0 m

Plug-in the above values in (B), you would get:

F_{g} = (6.674 * 10^{-11}) ( 9.11 * 10^{-31}  *  9.11 * 10^{-31}}[/tex]) / (2*2)  

F_{e} = 5.83* 10^{-29} N

F_{e} = 1.38 * 10^{-71} N

Now do Fe over Fg, you would get:
\frac{F_{e}}{F{g}}  = 4.23 * 10^{42}

Ans: So the blanks are:
1) 5.82
2) 1.38
3) 4.23

-i
6 1
3 years ago
Read 2 more answers
What is "In an isolated system two cars, each with a mass of 2,000 kg, collide. Car 1 is initially at rest , while car 2 was mov
yaroslaw [1]

Answer:

Final velocity of the combining cars = 10m/s

combined momentum = 40000kgm/s

Explanation:

Given the following :

Mass of both colliding cars = 2000kg

Initial Velocity of car 1 (U1) = 0

Initial Velocity of car 2 (U2) = 20m/s

Combuned momentum after collision =?

This is an inelastic collision which is calculated thus :

M1U1 + M2U2 = M1V + M2V

M1U1 + M2U2 = (M1 + M2) V

Where V = "final Velocity of the colliding bodies

2000 × 0 + 2000 × 20 = (2000 + 2000) V

0 + 40000 = 4000V

V = 40000 / 4000

V = 10m/s

Therefore combined momentum = (M1 + M2) V

(2000 + 2000) 10

4000(10) = 40000kgm/s

4 0
3 years ago
When you whirl a can at the end of a string in a circular path, what is the direction of the force that acts on the can
andreyandreev [35.5K]

Answer:

Toward the centre of the circular path

Explanation:

The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).

In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by

F=m\frac{v^2}{r}

where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.

3 0
3 years ago
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