The force of gravity is less between two objects that

Check all that are true about waterfalls

Lerok [7]

Waterfalls are created when a river flows following a descending rapid slope. The waterfall, then, flows from the source (where it starts) to the mouth (where it ends).
Waterfalls are created when the erosion of the rocks at the bottom of the slope is more powerful than the erosion of the rocks on the top.
After many years the water is able to erode the rocks on the top as well, and the waterfall slowly disappears.

Therefore the options that apply are:
b) waterfalls move towards their mouth;
c) the top or cap rock is resistant to erosion;
f) waterfalls indicate a youthful river

4 0

3 years ago

Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$

In words, the value of q₃ must be 5.3nC.

7 0

2 years ago

In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a

timofeeve [1]

Answer:

a) F₁ = 1.48 x 10³ N

b) P = 1.88*10⁵ Pa

c) The work is equal in both pistons

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F₁) on a small area piston (A₁), then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F₂) can be exerted that is proportional to the area(A₂) of the piston.

Pressure is defined as the force per unit area:

$P=\frac{F}{A}$ Formula (1)

P₁=P₂

$\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }$ Formula (2)

Data

r₁= 5 cm = 0.05 m

r₂= 15 cm = 0.15 m

F₂= 13300N

Area of the pistons (A₁,A₂)

A=π*r² : Area of the circle

A₁ = π*(0.05)²=7.85*10⁻³ m²

A₂= π*(0.15)²= 70.69*10⁻³ m²

a) Force that compressed air must exert to lift a car weighing 13300 N

We replace data in the formula (2)

$\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }$

$F_{1} = \frac{13300*7.85*10^{-3} }{70.69*10^{-3} }$

F₁ = 1.48 x 10³ N

b) Air pressure produced by F₁

We replace data in the formula (1)

$P=\frac{F}{A}$

F₁ = 1.48 x 10³ N , A₁ = 7.85*10⁻³ m²

$P=\frac{1.48*10^{3} }{7.85*10^{-3} }$

P= 1.88*10⁵ Pa

c)The volume of liquid displaced by the small piston is distributed in a thin layer on the large piston, so that the product of the force by the displacement (the work) is equal in both pistons.

3 0

3 years ago

Why do metals have similar properties?

ra1l [238]

Core electrons are also referred as non-valence electrons. Two different elements have similar chemical properties when they have the same number of valence electrons in their outermost energy level. Elements in the same column of the Periodic Table have similar chemical propertie

5 0

3 years ago

Mercury is a liquid metal having a density of 13.6 g/mL. What is the

sergij07 [2.7K]

Answer:

$\boxed {\boxed {\sf v \approx 33.088 \ mL}}$