-----------------removes nitrogen from the atmosphere.

Why is it good practice for scientists to repeat trails

It makes the data thet they collect more reliable so if they need the data again, they have already tested it a few times so therefor they know that it is right.

8 0

3 years ago

Durante uma corrida de carros, o piloto da escuderia ganhadora percorreu a primeira volta da pista com velocidade media de 310 k

lesya692 [45]

A velocidade mínima é 0. A velocidade máxima é inferior ou igual a 620 km/h.
buena suerte mi amigo

3 0

3 years ago

Calculate the change in length of a Pyrex glass dish (Coefficient of linear expansion for Pyrex is 3 * 10-6 / oC) that is 0.25 m

DIA [1.3K]

9* $10^{-5}$ m

Explanation:

Step 1:

We are given the initial length of the Pyrex glass dish at a particular temperature and need to calculate the change in the length when the temperature changes by 120° C. The coefficient of linear expansion of Pyrex is provided.

Step 2:

Change in length = Coefficient of linear expansion * Change in temperature * Initial length

Step 3:

Coefficient of linear expansion = 3* $10^{-6}$ /°C

Change in temperature = 120°C = 120 K

Initial length = 0.25 m

Step 4:

Change in length = 3* $10^{-6}$ * 120 * 0.25 = 9* $10^{-5}$ m

8 0

3 years ago

What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a

Nutka1998 [239]

The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:

Vf = Vi + at

where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time

Substituting the values gives:

30 = 20 + a(2)
a = 5 m/s^2 --> Final Answer

6 0

2 years ago

The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y,

kap26 [50]

$W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

Explanation:

The parametric representation of a line segment joining the points (a,b,c) and (l,m,n) is

r(t) = (1-t) . (a,b,c) + t . (l, m, n) where t ∈ |0, 1|

So, the parametric representation of a line segment joining the points (3,0,0) and (3,2,5) is

r(t) = (1 - t) . (3,0,0) + t . (3,2,5) where t ∈ |0, 1|

r(t) = (3(1 - t), 0, 0) + (3t, 2t, 5t) where t ∈ |0, 1|

r(t) = (3, 2t, 5t)

Given:

$F(x, y, z) = \frac{Kr}{|r|^3} \\\\F(x, y, z) = \frac{K}{(x^2 + y^2 + z^2)^3^/^2} (x, y, z)\\\\F(r(t)) = \frac{K}{(3^2 + (2t)^2 + (5t)^2)^3^/^2} (3, 2t, 5t)\\\\F(r(t)) = \frac{K}{(9 + 29t^2)^3^/^2} (3, 2t, 5t)$

dr = (0, 2, 5) dt

$Work = \int\limits^1_0 {F} \, dr$

$W = \int\limits^1_0 {\frac{K}{(9 + 29t^2)^3^/^2} } (3, 2t, 5t) . (0, 2, 5)\, dt\\ \\ = \int\limits^1_0 {\frac{K ( 4t + 25t)}{(9 + 29t^2)^3^/^2} } \, dt\\\\\\$

$W = \frac{1}{2}\int\limits^1_0 {\frac{K(29t)}{(9 + 29t^2)^3^/^2} } \, dt \\ \\$

Substitute = 9 + 29t² = u, 92tdt = du

Limit changes from 0→1 to 9 → 38

$W = \frac{K}{2} \int\limits^3_9 {\frac{du}{u^3^/^2} } \,\\\\$

On solving this, we get:

$W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

Therefore, work done is $W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

5 0

2 years ago