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earnstyle [38]
3 years ago
13

What is the quotient of 4,510 - 22 ?

Mathematics
1 answer:
spayn [35]3 years ago
3 0

Answer:

The answer is 99,220!!

Step-by-step explanation:

Did you mean to type quotient?

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Suppose a man’s scalp hair grows at a rate of 0.35 mm per day. What
iVinArrow [24]
<span>1cm = 10mm
1 inch = 2.54 cm
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0.001148... ft/day 
</span>Now converting days into centuries:
1 year = 365 days
1 century = 100 year
(0.001148 ft / day) * (365 day / 1 yr) * (100 yr / 1 cent)
= <span>41.9 ft/century</span>
5 0
3 years ago
What is the result of converting 60 pounds?
Molodets [167]

Answer:

To kilos? About 132.2. To dollars? Google the exchange.

7 0
3 years ago
Find one rational and one irrational no. between 3 and 5.
Leya [2.2K]
Rational number between 3 and 5 = 3.5
irrational number between 3 and 5 = (pi) ....3.14159265359.......
4 0
3 years ago
9. Calcule el valor de la fuerza que experimenta un ascensor cuando levanta 150 kg en los siguientes casos: a) Cuando asciende c
GenaCL600 [577]

Answer:

a) La fuerza neta que experimenta un ascensor cuando asciende con una aceleración de 4 metros por segundo al cuadrado es 600 newtons hacia arriba.

b) La fuerza neta que experimenta un ascensor cuando desciende con una aceleración de 6 metros por segundo al cuadrado es 900 newtons hacia abajo.

c) Por la fuerza de tensión sobre el cable del ascensor.

Step-by-step explanation:

a) La fuerza neta experimentada por el ascensor (F), en newtons, es:

F = m\cdot a (1)

Donde:

m - Masa, en kilogramos.

a - Aceleración, en metros por segundo cuadrado.

Si sabemos que m = 150\,kg y a = 4\,\frac{m}{s^{2}}, entonces la fuerza neta del ascensor es:

F = m\cdot a

F = 600\,N

La fuerza neta que experimenta un ascensor cuando asciende con una aceleración de 4 metros por segundo al cuadrado es 600 newtons hacia arriba.

b) Si sabemos que m = 150\,kg y a = -6\,\frac{m}{s^{2}}, entonces la fuerza neta del ascensor es:

F = m\cdot a

F = -900\,N

La fuerza neta que experimenta un ascensor cuando desciende con una aceleración de 6 metros por segundo al cuadrado es 900 newtons hacia abajo.

c) De manera simplificada, el ascensor experimenta dos fuerzas que definen la fuerza y aceleración netas: (i) La fuerza de tensión sobre el cable que traslada el ascensor y el peso total del ascensor, opuesta a la anterior y en función de la aceleración gravitacional. Puesto que la masa no varía en ningún caso, se concluye que el peso es constante, entonces la diferencia de valores se debe a la fuerza por tensión del cable. En el descenso, es fuerza es menor que en el ascenso.

8 0
2 years ago
George and Paula are running around a circular track. George starts at the westernmost point of the track, and Paula starts at t
arsen [322]

Answer:

George is 43.20 ft East of his starting point.

Step-by-step explanation:

Let Paula's speed be x ft/s

George's speed = 9 ft/s

Note that speed = (distance)/(time)

Distance = (speed) × (time)

George takes 50 s to run a lap of the track at a speed of y ft/s

Meaning that the length of the circular track = y × 50 = 50y ft

George and Paula meet 14 seconds after the start of the run.

Distance covered by George in 14 seconds = 9 × 14 = 126 ft

Distance covered by Paula in 14 seconds = y × 14 = 14y ft

But the sum of the distance covered by both runners in the 14 s before they first meet each other is equal to the length of the circular track

That is,

126 + 14y = 50y

50y - 14y = 126

36y = 126

y = (126/36) = 3.5 ft/s.

Hence, Paula's speed = 3.5 ft/s

Length of the circular track = 50y = 50 × 3.5 = 175 ft

So, in 4 minutes (240 s), with George running at 9 ft/s, he would have ran a total distance of

9 × 240 = 2160 ft.

2160 ft around a circular track of length 175 ft, means that George would have ran a total number of laps (2160/175) = 12.343 laps.

Breaking this into 12 laps and 0.343 of a lap from the starting point. 0.343 of a lap = 0.343 × 175 = 60 ft

So, 60 ft along a circular track subtends an angle θ at the centre of the circle.

Length of an arc = (θ/360°) × 2πr

2πr = total length of the circular track = 175

r = (175/2π) = 27.85 ft

Length of an arc = (θ/360) × 2πr

60 = (θ/360°) × 175

(θ/360°) = (60/175) = 0.343

θ = 0.343 × 360° = 123.45°

The image of this incomplete lap is shown in the attached image,

The distance of George from his starting point along the centre of the circular track = (r + a)

But, a can be obtained using trigonometric relations.

Cos 56.55° = (a/r) = (a/27.85)

a = 27.85 cos 56.55° = 15.35 ft

r + a = 27.85 + 15.35 = 43.20 ft.

Hence, George is 43.20 ft East of his starting point.

Hope this Helps!!!

6 0
3 years ago
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