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2 years ago

Its Human Relations

1 answer:

fomenos2 years ago

4 0

<span>The statement could be letter B: being in conflict with people isn't much of a problem. It does not also mean that you need to argue with people or make mistakes or look for other people’s mistakes all the time. A conflict means a disagreement of both parties and it is okay AS LONG AS it is a healthy one. Conflict makes people grow, it challenges them to give their opinions, to be true to their selves and to defend their stand and exercises their open-mindedness.</span>

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Select the correct answer.

chubhunter [2.5K]

I believe it’s self-referent encoding

3 0

2 years ago

Shawn and his bike have a total mass of

Trava [24]

Answer:

Shawn's kinetic energy is 61.45 J.

Explanation:

Given;

Mass of the system is, $m=52.3\ kg$

Displacement of the bike is, $d=1.6\ km=1.6\times 1000=1600\ m$

Time taken is, $t=17.4\ min=17.4\times 60=1044\ s$

Let the constant velocity be $v$ .

For constant velocity, magnitude of velocity is given as distance by time.

Therefore, $v=\frac{d}{t}=\frac{1600}{1044}=1.533\ m/s$

Now, kinetic energy of a body is given as:

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

Here, $m=52.3\ kg,\ v=1.533\ m/s$

$\therefore KE=\frac{1}{2}\times 52.3\times (1.533)^2\\KE=0.5\times 52.3\times 2.35\\KE=61.45\ J$

Therefore, Shawn's kinetic energy is 61.45 J.

3 0

3 years ago

2. With regard to the pH scale, a solution with a pH

Alexandra [31]

Answer:

ph of 10 b your welcome

Explanation:

8 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

2 years ago

A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of

Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

= 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

$P = \sqrt{P_x^2 + P_y^2}$

$P = \sqrt{4.52^2 + 2.61^2}$

P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0

2 years ago