Losing or gaining electrons makes an atom a/an: A. Electron B. Charged atom. C. Proton. D. Ion

You are driving a metric car that has a speedometer that displays in meters per second. When you look down at your speedometer,

shepuryov [24]

Ahahhabsba babababba ab

7 0

3 years ago

Time-outs are ineffective because

maria [59]

I think the correct answer would be the child quietly sits until the timer goes off. It is ineffective because the child would be used to this process and would come to a time where he will not get something from it.Also, it would make a child feel isolated and alone which would affect hi self-esteem in a long run.

8 0

4 years ago

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When wile coyote is launched his velocity is 20 m/s. As a slightly underweight 10kg coyote, how far would the 400 N/m spring hav

Deffense [45]

Answer:

E = 1/2 m v^2 = 1/2 k x^2 equating KE of coyote and spring

x^2 = m v^2 / k = 10 kg * 20^2 m^2 / s^2 / 400 N / m

x^2 = 10 * 400 / 400 (kg m^3 / kg-m) = 10 m^2

x = 3.16 m

3 0

3 years ago

At an accident scene on a level road, investigators measure a car’s skid mark to be 88 m long. The accident occurred on a rainy

Oduvanchick [21]

Answer:

The the speed of the car is 26.91 m/s.

Explanation:

Given that,

distance d = 88 m

Kinetic friction = 0.42

We need to calculate the the speed of the car

Using the work-energy principle

work done = change in kinetic energy

$W=\Delta K.E$

$\mu\ mg\times d=\dfrac{1}{2}mv^2$

$v^2=2\mu g d$

Put the value into the formula

$v=\sqrt{2\times0.42\times9.8\times88}$

$v=26.91\ m/s$

Hence, The the speed of the car is 26.91 m/s.

3 0

3 years ago

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A mass attached to a spring oscillates in simple harmonic motion with an amplitude of 10 cm. When the mass is 5.0 cm from its eq

timama [110]

When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

<h3>Total energy of the mass</h3>

The total energy possessed by the mass under the simple harmonic motion is calculated as follows;

U = ¹/₂kA²

where;

k is the spring constant
A is the amplitude of the oscillation

<h3>Potential energy of the mass at 5 cm from equilibrium point</h3>

P.E = ¹/₂k(Δx)²

<h3>Kinetic energy of mass</h3>

K.E = U - P.E

K.E = ¹/₂kA² - ¹/₂k(Δx)²

<h3>Percentage of its energy that is kinetic</h3>

$K.E (\%) = \frac{U - P.E}{U} \times 100\%\\\\K.E (\%) =\frac{\frac{1}{2}kA^2 - \frac{1}{2}k(\Delta x)^2 }{\frac{1}{2}kA^2} \times 100\%\\\\K.E (\%) = \frac{A^2 - (\Delta x)^2}{A^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - (10-5)^2}{10^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - 5^2}{10^2} \times 100\%\\\\K.E (\%) = 75\%$

Thus, when the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

Learn more about kinetic energy here: brainly.com/question/25959744

3 0

2 years ago

Losing or gaining electrons makes an atom a/an: A. Electron B. Charged atom. C. Proton. D. Ion​

Losing or gaining electrons makes an atom a/an: A. Electron B. Charged atom. C. Proton. D. Ion