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3 years ago

What two factors exert the greatest influence over a terrestrial biome?

1 answer:

Elena-2011 [213]3 years ago

5 0

I would presume the factors would be soil and the sunlight. These two, I observe, are the needed factors for survival of the terrestrial organisms. And help them thrive.

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two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f

Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

$E = 1.44 \times 10^7 N/C$

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

$E = 2\frac{kq}{r^2}$

now we have

$q = 2\times 10^{-6} C$

$r = 5 cm = 0.05 m$

now we have

$E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}$

$E = 1.44 \times 10^7 N/C$

6 0

2 years ago

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in

olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N$

7 0

3 years ago

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

$T = \frac{(10+26.2)(9.8)}{cos(16.2)}$

$T = 369.42N$

6 0

3 years ago

You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on

just olya [345]

Answer:

a) $v_{f}$ = 0.25 m / s b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

m₁ = 0.40 kg

v₁₀ = 9.0 m / s

m₂ = 14 kg

v₂₀ = 0

Initial

po = m₁ v₁₀

Final

$p_{f}$ = (m₁ + m₂) vf

po = pf

m₁ v₁₀ = (m₁ + m₂) $v_{f}$

$v_{f}$ = v₁₀ m₁ / (m₁ + m₂)

$v_{f}$ = 9.0 (0.40 / (0.40 +14)

$v_{f}$ = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

K₀ = ½ m₁ v₁₀² +0

K₀ = ½ 0.40 9 2

K₀ = 16.2 J

Final

$K_{f}$ = ½ (m₁ + m₂) $v_{f}$ 2

$K_{f}$ = ½ (0.4 +14) 0.25 2

$K_{f}$ = 0.45 J

ΔK = K₀ - $K_{f}$

ΔK = 16.2-0.445

ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

v₁₀’= v₁₀ -u

v₁₀’= 9 -.025

v₁₀‘= 8.75 m / s

v₂₀ ‘= v₂₀ -u

v₂₀‘= - 0.25 m / s

$v_{f}$ ‘= $v_{f}$ - u

$v_{f}$ = 0

Initial

K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

Ko = 15.31 + 0.4375

K o = 15.75 J

Final

$k_{f}$ = ½ (m₁ + m₂) vf’²

$k_{f}$ = 0

All initial kinetic energy is transformed into internal energy in this reference system

3 0

2 years ago

Which vector is the sum of the vectors shown below?

qaws [65]

Where is the picture??

4 0

2 years ago