Two 6 ohm resistors in parallel gives an equivalent resistance of

Carl works hard to get a grades on his report card because his mother pays him 25 dollars for each semester he earns straight as

Aleksandr [31]

He is influenced by EXTRINSIC MOTIVATION

8 0

2 years ago

Question 9 of 10<br> How many carbon (C) atoms are in a molecule of CH,O?

Juli2301 [7.4K]

Answer:

1 molecule not shure

Explanation:

6 0

3 years ago

Read 2 more answers

What is the prefix for oxygen in As2O5? tri- di- mono- penta-

ikadub [295]

The prefix for oxygen in As2O5 is PENTA.
Penta is used to show that the number of oxygen atoms present in the compound is five. Penta means five, while mono mean one, di means two, tri means three and tetra mans four and so on. The chemical name for the given compound is Arsenic pentoxide.

3 0

3 years ago

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If a crow flies west for 60 km and then south for 45 km, what is the direction of its displacement?

son4ous [18]

That's 105 km that he flew, or 65.2 miles ! I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking. But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:

The displacement is the distance and direction from the place
where the crow took off to the place where he landed.

-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.

D² = (60 km)² + (45 km)²

=    3,600 km² + 2,025 km² = 5,625 km²

   D = √(5625 km²) = 75 km .

-- It's direction is the angle whose tangent is (45 S / 60 W).

   tan⁻¹ (45/60) = tan⁻¹ (0.75) = 36.9° south of west

   = 53.1° west of south.

   = not exactly southwest but close.

7 0

3 years ago

A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

$t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085$

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m

5 0

3 years ago