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1 year ago

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A 2.0 g metal cube and a 4.0 g metal cube are 6.0 cm apart, measured between their centers, on a horizontal surface. For both, t

he coefficient of static friction is 0.65. Both cubes, initially neutral, are charged at a rate of 7.0 nC/s.
a. How long after charging begins does one cube begin to slide away?
b. Which cube moves first?

1 answer:

emmasim [6.3K]1 year ago

7 0

Answer:

a) t=10.2s

b) The 2g-cube moves first

Explanation:

Since the electric force is the same on both cubes and so is the coefficient of static friction, the first one to move will be the one with less mass.

So, on the 2g-cube the sum of forces are:

$\left \{ {{Ff-Fe=0} \atop {N-m*g=0}} \right.$

Replacing the friction on the first equation:

$\mu*m*g-Fe=0$ Thus $Fe=\mu*m*g=12.74*10^{-3}N$

The electric force is:

$Fe = \frac{K*q^2}{d^2}$ Solving for q:

q=71.44nC

This amount divided by the rate at which they are being charged:

t = 71.44nC / 7nC/s = 10.2s

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A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

6 0

2 years ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

2 years ago

Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are

ehidna [41]

Answer

The Value of r = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

$R_T = R_1 +R_2$

And from Ohm law

$I_s =\frac{ V}{R_T}$

$I_s = \frac{V_0}{R_1 +R_2} ---(1)$

The mathematical representation of the two resistors connected in parallel is

$R_T = \frac{1}{R_1} +\frac{1}{R_2}$

$= \frac{R_1 R_2}{R_1 +R_2}$

From the question $I_p =10I_s$

=> $I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} } = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)$

Dividing equation 2 with equation 1

=> $\frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}$

$10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)$

We are told that $r = \frac{R_1}{R_2} \ \ \ \ \ = > R_1 = rR_2$

From equation 3

$10 = \frac{(1-r)^2}{r}$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1+r^2 + 2r = 10r$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0$

Using the quadratic formula

$r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}$

a = 1 b = -8 c =1

$= \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}$

$r= \frac{8+ \sqrt{60} }{2} \ or \ r = \frac{8 - \sqrt{60} }{2}$

$r = \ 7.87\ or \ r \ = \ 0.127$

Now r = 0.127 because it is the least value among the obtained values

4 0

2 years ago

Bradley gets an x-ray at a radiology clinic that employs its own technologists and radiologists. Would the coder at the clinic r

KengaRu [80]

Answer:

Explanation:

If Bradley examination was done and interpreted in the same facility, the radiologist code is used example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.

if the X-ray was taken by Dr X but Dr X does not read or interpret the image but forward it to the radiologist for initial report, then a 26- modifier is used. E.g A reports by the technologist would be, procedure code 72050-Radiologic examination, spine, cervical, 2 or 3

views or 72050- TC in certain situations and the consulting radiologist would report 72050-26.

if Bradley’s x-ray were sent to an independent radiologist for interpretation, then the procedure code 76140 is used in reporting.

8 0

1 year ago

(1 point) Which of the following statements are true?A.The equation Ax=b is referred to as a vector equation.B.If the augmented

Anvisha [2.4K]

Answer:

A. False

B. False

C. True

D. True

E. True

F. True

Explanation:

A. The equation Ax=b is referred to as a matrix equation and not vector equation.

B. If the augmented matrix [ A b ] has a pivot position in every row then equation Ax=b may or may not be consistent. It is inconsistent if [A b] has a pivot in the last column b and it is consistent if the matrix A has a pivot in every row.

C. In the product of Ax also called the dot product the first entry is a sum of products. For example the the product of Ax where A has [a11 a12 a13] in the first entry of each column and the corresponding entries in x are [x1 x2 x3] then the first entry in the product is the sum of products i.e. a11x1 + a12x2 +a13x3

D. If the columns of mxn matrix A span R^m, this states that every possible vector b in R^m is a linear combination of the columns which makes the equation consistent. So the equation Ax=b has at least one solution for each b in R^m.

E. It is stated that a vector equation x1a1 + x2a2 + x3a3 + ... + xnan = b has the same solution set as that of the linear system with augmented matrix [a1 a2 ... an b]. So the solution set of linear system whose augmented matrix is [a1 a2 a3 b] is the same as solution set of Ax=b if A=[a1 a2 a3] and b can be produced by linear combination of a1 a2 a3 iff the solution of linear system corresponding to [a1 a2 a3 b] takes place.

F. It is true because lets say b is a vector in R^m which is not in the span of the columns. b cannot be obtained for some x which belongs to R^m as b = Ax. So Ax=b is inconsistent for some b in R^m and has no solution.

8 0

2 years ago