A net force of 150 N is exerted on a rock. The rock has an acceleration of 20. m/s2 due to this force. What is the mass

6. Sandy adds 3 sugar packets to her tea in order to make it sweeter. She stirs in the sugar and then takes a sip. She notices t

irina1246 [14]

Answer:

C.Supersaturated

Explanation:

There are three types of solution:

SATURATED SOLUTION:

It is the solution that contains maximum amount of solute dissolved in a solution in normal conditions.

UNSATURATED SOLUTION:

It is the solution that contains less than the maximum amount of solute dissolved in a solution in normal conditions. It has space for more solute to be dissolved in it.

SUPERSATURATED SOLUTION:

It contains more than the maximum amount of solute dissolved in it. Such a solution has no capacity to dissolve any more solute under any conditions.

Since the sugar is no more dissolving in the tea and has settled down. Therefore, the solution is:

C.Supersaturated

7 0

2 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3

Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC Q₃ = 51 nC Q₂ = 30.5 nC

x₁ = - 1.715m x₃ = - 1.085m x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0

2 years ago

A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of 3 cm and a cross-sectional area of 15 cm2. T

stealth61 [152]

A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?

3 0

3 years ago

Read 2 more answers

A convex mirror has a focal length of 15cm. If the object is 35cm from the mirror and is 2cm tall, what is the magnification? Ro

Verizon [17]

Answer:

The magnification is 0.3

Explanation:

Given:

The focal length of the convex mirror, f = 15 cm

for convex mirror f = -f = -15 cm

The distance of object, u = 35 cm

Height of the object, h = 2 cm

Now, using the mirror formula, we have

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

where, v is the distance of the image

on substituting the values, we get

$\frac{1}{-15}=\frac{1}{35}+\frac{1}{v}$

or

$-\frac{1}{v}=\frac{1}{35}-\frac{1}{-15}$

or

v = -10.5 cm

now, the magnification (m) is given as:

m = $-\frac{v}{u}$

on substituting the values, we get

m = $-\frac{-10.5}{35}$

or

m = 0.3

Hence, the magnification is 0.3

6 0

2 years ago

A uniform piece of sheet steel is shaped as in the figure below. (Both axes are marked in increments of 2). Compute the x and y

Elena-2011 [213]

"increments of 8" means the major divisions are 0,8,16,24 ?

x axis, calculate the moment arms from 0 
3x4, 2x12, 1x20 
from an arbitrary C 
3(c-4) + 2(c-12) + (c-20) = 0 
3c - 12 + 2c -24 + c - 20 = 0 
6c = 56 
c = 9.33 

y axis 
3x3, 1x12, 2x20 
3(c-4) + 1(c-12) +2 (c-20) = 0 
3c - 12 + c - 12 + 2c - 40 = 0 
6c = 64 
c = 10.67 

so center is x = 9.33, y = 10.67

5 0

2 years ago

Read 2 more answers