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victus00 [196]
3 years ago
13

A Porsche challenges a Honda to a 500 m race

Physics
1 answer:
zlopas [31]3 years ago
8 0
A Porsche will always win no matter what
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Besides the force due to gravity or g, what else determines the period of a pendulum?
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Length of the cord..........................................................

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Scientists have concluded that Earth is at risk of future impacts with meteoroids. What are some criteria that engineers conside
Aleksandr [31]
Answer:
Explain step by step
Explanation:
Collisions with asteroids, comets and other stuff from space have been responsible for huge landmarks in our planet’s history: global shifts in climate, the creation of our moon, the reshuffling of our deepest geology, and the extinction of species.

Asteroid threats pop up in the news every now and then, but the buzz tends to fizzle away as the projectiles pass us by. Other times, as with the 2013 Chelyabinsk meteor in Russia, we don’t know they’re here until they’re here.
Perhaps most useful to remember is that when near-Earth objects (including asteroids, comets and meteoroids) enter the atmosphere, they’re called meteors; and if there’s anything left when they hit the ground, the resulting object is called a meteorite. We tend to focus on asteroids when talking about potential collisions, because they’re more likely to hit us than other stuff like comets, but still big enough to pose a threat.
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2 years ago
Kwanzaa farts apple juice
8_murik_8 [283]

Answer:

well

Explanation:

7 0
2 years ago
What mRNA sequence would result from the following DNA sequence?
Nezavi [6.7K]

Answer:

UAC CUG AGG AUC

Explanation:

<em>The mRNA sequence from ATG GAC TCC TAG DNA sequence would be </em><em>UAC CUG AGG AUC.</em>

<u>According to Chargaff's base pairing rule, the purine bases always pair with pyrimidine bases. Specifically, Adenine base must pair with Thymine base while Guanine base must pair with Cytosine base. In RNA, Thymine base is replaced with Uracil base.</u>

Hence:

ATG   GAC   TCC   TAG will pair with

UAC   CUG   AGG   AUC

5 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
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