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3 years ago

Give 2 examples of situations in which environmental values could come into conflict with other values

Physics

1 answer:

gtnhenbr [62]3 years ago

6 0

Answer:

Example of situations in which environmental values could come into conflict with other values, such as economy values of family values, include deciding whether or not a businessman should fly across the country for a meeting or whether a family should buy a gas car or an electric car.

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When a rattlesnake strikes, its head accelerates from rest to a speed of 32 m/s in 0.63 seconds. Assume for simplicity that the

irina [24]

Answer:

Power, P = 162.53 Watts

Explanation:

Given that,

Mass, m = 200 g = 0.2 kg

Initial speed of the snake, u = 0

Final speed of the snake, v = 32 m/s

Time, t = 0.63 s

Power of an object is given by :

$P=\dfrac{W}{t}$

$P=\dfrac{\Delta K}{t}$

$P=\dfrac{mv^2}{2t}$

$P=\dfrac{0.2\times (32)^2}{2\times 0.63}$

P = 162.53 Watts

So, the power of the rattlesnake is 162.53 Watts. Hence, this is the required solution.

4 0

4 years ago

A 35.0-kg child swings in a swing supported by two chains, each 3.02 m long. The tension in each chain at the lowest point is 41

n200080 [17]

Answer:

Explanation:

Total tension must support the weight of the child and provide the necessary centripetal force.

2T = m(g + v²/R)

2(414) = 35.0(9.81 + v²/3.02)

828 = 35.0(9.81 + v²/3.02)

23.65 = 9.81 + v²/3.02

13.847 = v²/3.02

41.81 = v²

v = 6.47 m/s

F = 2(414) = 828 N (ignoring the weight of the chains and seat)

3 0

3 years ago

Coulomb's law is similar to newton's law of gravitation in several ways. which one of the statements is not a similarity between

PIT_PIT [208]

The only dissimilarity between Coulomb's law and Newton's law of gravitation other than the type of force considered is Newton's law only considers negative force while Coulomb's law considers both forces.

In these contexts, it is considered that repulsive forces are positive and attractive forces are negative. So, Newton's law of gravitation considers gravitational force which only attracts. So, Newton's law of gravitation takes only negative force into account while Coulomb's law considers both electric charges.

According to Newton's law of gravitation,

F = G $m_{1} m_{2}$ / $r^{2}$

According to Coulomb's law

F = k $q_{1} q_{2}$ / $r^{2}$

Therefore, the only dissimilarity between Coulomb's law and Newton's law of gravitation other than the type of force considered is Newton's law only considers negative force while Coulomb's law considers both positive and negative forces.

To learn more about Coulomb's law

brainly.com/question/506926

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4 0

2 years ago

A NFL linebacker runs the 100m sprint in 12s. What is his final velocity?

zaharov [31]

Answer:

Final velocity of NFL line backer is 16.67 m/s.

Explanation:

From the question, we have following data about the NFL line backer:

Initial Speed of line backer = Vi = 0 m/s (Since, he starts from rest)

Distance covered by NFL line backer = s = 100 m

Time taken by the NFL line backer to complete 100 m sprint = t = 12 s

Acceleration of NFL line backer during sprint = a

Final Velocity of NFL line backer = Vf = ?

First we need to find the acceleration of the NFL line backer. For that purpose we will use 2nd equation of motion:

s = (Vi)(t) + (0.5)at²

using values:

100 m = (0 m/s)(12 s) + (0.5)(a)(12 s)²

100 m/72 s² = a

a = 1.39 m/s²

Now, we use 1st equation of motion to find Vf:

Vf = Vi + at

Vf = 0 m/s + (1.39 m/s²)(12 s)

7 0

4 years ago

A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s

loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

$m*V_i = m_1V_1+m_2V_2$

Here,

m = Total mass (8Kg at this case)

$m_1=m_2 =$ Mass each part

$V_i =$ Initial velocity

$V_2 =$ Final velocity particle 2

$V_1 =$ Final velocity particle 1

The initial kinetic energy would be given by,

$KE_i=\frac{1}{2}mv^2$

$KE_i = \frac{1}{2}8*5^2$

$KE_i = 100J$

In the end the energy increased 100J, that is,

$KE_f = KE_i KE_{increased}$

$KE_f = 100+100 = 200J$

By conservation of the moment then,

$m*V_i = m_1V_1+m_2V_2$

Replacing we have,

$(8)*5 = 4*V_1+4*V_2$

$40 = 4(V_1+V_2)$

$V_1+V_2 = 10$

$V_2 = 10-V_1$ (1)

In the final point the cinematic energy of EACH particle would be given by

$KE_f = \frac{1}{2}mv^2$

$KE_f = \frac{1}{2}4*(V_1^2+V_2^2)$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$ (2)

So we have a system of 2x2 equations

$V_2 = 10-V_1$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: $V_1 = 10m/s$

Then replacing in (1) we have that

PART B: $V_2 = 0m/s$

8 0

3 years ago