The flow of ____ forms the physical basis for electric charge.

What would be the Elastic Potential Energy (EPE) stored in a spring with a constant k = 200 N/m that is pulled to stretch 0.45m?

harkovskaia [24]

Answer:

During the summer, oxygen is supplied by surface winds. M/NPT 16 to 26 0.14 to 0.45 m 1/ 77.80 $ 28 Phone: 418 81-1 Fax: 418 81-2882 SZ5-2025 SZ22-200 SZ2-250 SZ24-00 12.5 x 60 12.5 x 60 12.5 x 60 8 x 10 10 x 15 10 x 20 15 x HI-950 can take measures between 0.1 C to 999.9 C/ F. K, J & T-type: The HI-9551

(Hope this is what you're asking for)

Hope this helps Have a good day

3 0

3 years ago

A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is $v=\frac{d}{t}$ , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is $d=v\cdot t$

Now, the distance equation for the man would be:

$d_{man}=v_{man}\cdot t=1.6\cdot t$

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

$d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)$

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

$d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0$

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

$d_{man}=d_{dog}$

$1.6\cdot t=3\cdot (t-1.8)$

$1.6\cdot t=3\cdot t-5.4$

$1.4\cdot t=5.4$

$t=\frac{5.4}{1.4}$

$t=3.8571s$

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

$d_{man}=1.6\cdot t=1.6\cdot (3.8571)$

$d_{man}=6.1714m$

Finally, the dog catches up with the man 6.1714m later.

6 0

3 years ago

In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

$v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)$

The final speed of the leading car is given by the following formula:

$v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)$

4 0

3 years ago

What effect will a turning point have on an individuals life

ankoles [38]

Answer:

The turning points are those instants, moments or situations that happen in an absolutely unexpected way, as a result of which your life changes ... and nothing is the same as before.

4 0

3 years ago

Read 2 more answers

A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take f

sergejj [24]

Answer:A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take for the spitball to hit the ground?

(ignore air resistance) (include units and correct number of significant figures)

Explanation:La respuesta es porque esa es la respuesta, la respuesta al número es 9.7 1.8 Divide =53.888

3 0

3 years ago