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3 years ago

The black lines that wrap themselves around a basketball represent an example of what type of geometry?

1 answer:

5 0

In Euclidean geometry parallel lines never intersect. But in non-Euclidean geometry parallel lines can either curve away from each other, or curve towards each other. Example : the black lines that wrap themselves around the basketball.
Answer: B ) non-Euclidean

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Your roommate drops your wallet down to you from the third-floor window of your apartment, which is 11.5 m from the ground. What

Aneli [31]

Answer:

15 m/s

Explanation:

Using the law of conservation of energy, potential energy equals kinetic energy hence

$mgh=0.5mv^{2}$

Therefore

$v=\sqrt{2gh}$

where g is the acceleration due to gravity, m is the mass of the object, h is the height and v is the speed of the wallet

Taking g as 9.81 then

$v=\sqrt{2\times 9.8\times 11.5}=15.02098532 m/s\approx 15 m/s$

6 0

3 years ago

A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha

umka21 [38]

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

$R_{1} = 330\ohm$

$R_{2} = 470\ohm$

$R_{3} = 220\ohm$

Since, $R_{2} = 470\ohm$ and $R_{3} = 220\ohm$ are in parallel and this combination is in series with $R_{1} = 330\ohm$ , so,

Equivalent resistance of the circuit is given by:

$R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}$

$R_{eq} = \frac{470\times 220}{470 + 220} + 330$

$R_{eq} = 149.85 + 330 = 479.85 \ohm$

power drain on an ideal battery, P = $\frac{V^{2}}{R_{eq}}$

P = $\frac{2.9^{2}}{479.85}$

P = 0.017 W

4 0

4 years ago

Air flows into a jet engine at 70 lbm/s, and fuel also enters the engine at a steady rate. The exhaust gases, having a density o

Andrews [41]

Answer:

1387908 lbm/h

Explanation:

Air flowing into jet engine = 70 lbm/s

ρ = Exhaust gas density = 0.1 lbm/ft³

r = Radius of exit with a circular cross section = 1 ft

v = Exhaust gas velocity = 1450 ft/s

Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel

Q = (70+x) lbm/s

Area of exit with a circular cross section = π×r² = π×1²= π m²

Now from energy balance

Q = ρ×A×v

⇒70+x = 0.1×π×1450

⇒70+x = 455.53

⇒ x = 455.53-70

⇒ x = 385.53 lbm/s

∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h

8 0

3 years ago

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. it suddenly collides directly with a stationary seal of m

Sphinxa [80]

M = mass of the whale = 1000 kg

m = mass of the seal = 200 kg

V = initial velocity of whale before collision with the seal = 6.0 m/s

v = initial velocity of the seal before collision with the whale = 0 m/s

V' = final velocity of two sea creatures after collision = ?

Using conservation of momentum

M V + m v = (M + m) V'

inserting the above values in the equation

(1000 kg) (6.0 m/s) + (200 kg) (0 m/s ) = (1000 kg + 200 kg) V'

6000 kgm/s + 0 kgm/s = (1200 kg) V'

V' = (6000 kgm/s ) /(1200 kg)

V' = 5 m/s

3 0

3 years ago

Read 2 more answers

3x/y=6g/b solve for x

lora16 [44]

So looking at the problem, you are going to want to start by finding a common denominator (1) in this case: yb, and combining like terms (2). You are then going to want to multiply both sides by (yb) as the reciprocal to the fractions (3).
1) 3x 6g
---- = ---
y b

2) 3xb 6gy
------ = -----
yb yb

3) 3xb 6gy
(yb) ------ = -----
yb yb
which becomes: 3xb = 6gy

So after this, things become much more simple, as all you have to do is isolate the (x), which can be done by dividing the entire equation by (3b).

3xb 6gy
----- = -----
3b 3b

where you will then find your answer of:
2gy
x = ----- (simplified by the GCM of 3)
b

5 0

3 years ago