A carbon atom can form 4 covalent bonds how many valence electrons does a carbon atom have?

Which of the following is equivalent to 140 centiliters?

stiks02 [169]

<span>The answer 0.00140 kiloliters
</span><span>ecause one litre=100 centiliters, and one kilolitre=one litre which means that one kilolitre =100000 centiliters So 140 centiliters are 140 divided by 100000 kiloliters=0.00140 kiloliters</span>

7 0

3 years ago

Read 2 more answers

19. The current in a hair dryer is 12 A. The hair dryer is plugged into a 120-V outlet. How

patriot [66]

Answer:

10 ohms

Explanation:

R = V / I

3 0

3 years ago

I’M DESPERATE PLZ HELP!! 40 POINTS!!

Ierofanga [76]

Answer:

a. The acceleration of the hockey puck is -0.125 m/s².

b. The kinetic frictional force needed is 0.0625 N

c. The coefficient of friction between the ice and puck, is approximately 0.012755

d. The acceleration is -0.125 m/s²

The frictional force is 0.125 N

The coefficient of friction is approximately 0.012755

Explanation:

a. The given parameters are;

The mass of the hockey puck, m = 0.5 kg

The starting velocity of the hockey puck, u = 5 m/s

The distance the puck slides and slows for, s = 100 meters

The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration

The velocity of the hockey puck after motion, v = 0 m/s

The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;

v² = u² + 2·a·s

Therefore, by substituting the known values, we have;

0² = 5² + 2 × a × 100

-(5²) = 2 × a × 100

-25 = 200·a

a = -25/200 = -0.125

The constant acceleration of the hockey puck, a = -0.125 m/s².

b. The kinetic frictional force, $F_k$ , required is given by the formula, F = m × a,

From which we have;

$F_k$ = 0.5 × 0.125 = 0.0625 N

The kinetic frictional force required, $F_k$ = 0.0625 N

c. The coefficient of friction between the ice and puck, $\mu_k$ , is given from the equation for the kinetic friction force as follows;

$F_k = \mu_k \times Normal \ force \ of \ hockey \ puck = \mu_k \times 0.5 \times 9.8$

$\mu_k = \dfrac{0.0625}{0.5 \times 9.8} \approx 0.012755$

The coefficient of friction between the ice and puck, $\mu_k$ ≈ 0.012755

d. When the mass of the hockey puck is 1 kg, we have;

Given that the coefficient of friction is constant, we have;

The frictional force $F_k = 0.012755 \times 1 \times 9.8 = 0.125 \ N$

The acceleration, a = $F_k$ /m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a

The frictional force as calculated here, $F_k = 0.125 \ N$

The coefficient of friction $\mu_k$ ≈ 0.012755 is constant

5 0

3 years ago

Consider two laboratory carts of different masses but identical kinetic energies and the three following statements. I. The one

kolbaska11 [484]

Answer:

d) I and III only.

Explanation:

Let be $m_{1}$ and $m_{2}$ the masses of the two laboratory carts and let suppose that $m_{1} > m_{2}$ . The expressions for each kinetic energy are, respectively: