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3 years ago

14

Which of the following objects will exert the greatest gravitational force on a box, if the distance between them remains the sa

me? 2 kg mass 4 kg mass 6 kg mass 8 kg mass

1 answer:

madreJ [45]3 years ago

7 0

If the surface between them is the same, then the largest force exerts the body with the largest mass, ie 8 kg

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RideAnS [48]

A group of students toured a limestone cave in northwest Georgia.
Which of these best explains how the limestone caves in Georgia were formed?
A. Plant roots split the rock.

B. Acidic water dissolved the rock.

C. Animals burrowed into the rock.

D. Ice formed and broke up the rock.

7 0

3 years ago

Tenemos un Cable de cobre de 1 km de longitud cuya sección es de 2 milímetros al cuadrado y queremos saber la resistencia que se

Nady [450]

Answer:

8.5 Ω

Explanation:

La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

La fórmula de la resistencia (R) viene dada por:

R = ρL/A

Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.

Dado que:

L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm

Sustituyendo da:

R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶

R = 8.5 Ω

7 0

2 years ago

Which star makes up Orion's knee to your right, or the end of his tunic?

lord [1]

Answer:

Betelgeuse and Rigel

Explanation:

100% correct

7 0

3 years ago

At the instant the traffic light turns green, a car that has been waiting at an intersection starts with a constant acceleration

Setler79 [48]

Answer:

(a) 300 ft

(b) 60 ft/s

Explanation:

distance $d=0.5at^{2}$ where a is acceleration and t is time

$d=0.5\times 6\times t^{2}=3t^{2}$

Also, d=vt where v is the velocity

d=30t

Therefore

$30t=3t^{2}$ hence t=10 s

Substituting t is either formula

d=30t=30*10=300 ft

Also

v=at hence $v=6\times 10=60 ft/s$

4 0

3 years ago

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C

Alex_Xolod [135]

Answer:

26.466cm³/min

Explanation:

Given:

Volume 'V'= 320cm³

P= 95kPa

dP/dt = -11 kPa/minute

pressure P and volume V are related by the equation

P $V^{1.4}$ =C

we need to find dV/dt, so we will differentiate the above equation

$V^{1.4} \frac{dP}{dt} + P\frac{d[V^{1.4} ]}{dt} = \frac{d[C]}{dt}$

$\frac{dP}{dt} V^{1.4} + P(1.4)V^{0.4} \frac{dV}{dt} =0$

lets solve for dV/dt, we will have

$\frac{dV}{dt} =\frac{-\frac{dP}{dt} V^{1.4} }{P(1.4)V^{0.4} } \\\frac{dV}{dt} =- \frac{-\frac{dP}{dt} V}{P(1.4)}$

$\frac{dV}{dt} = -\frac{(-11 ) 320}{95(1.4)}$ (plugged in all the values at the instant)

$\frac{dV}{dt}$ = 26.466

Therefore, the volume increasing at the rate of 26.466cm³/min at this instant

6 0

3 years ago

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