An atom with more or less neutrons than expected is a what ?

A ring of diameter 7.70 cm is fixed in place and carries a charge of 5.00 mC uniformly spread over its circumference. (a) How mu

MatroZZZ [7]

Answer:

3.974 Joule

Explanation:

Diameter of ring = 7.7 cm

a = Distance from the center = d/2 = 3.85 cm = 0.0385 m

Q = Charge = 5 mC

q = Charge to move = 3.4 mC

k = Coulomb constant = 9×10⁹ Nm²/C²

Work done will be equal to Potential energy when mass is at center

$U=\frac{kQq}{a}\\\Rightarrow U=\frac{9\times 10^9\times 5\times 10^{-6}\times 3.4\times 10^{-6}}{0.0385}=3.974\ J$

∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule

4 0

3 years ago

Read 2 more answers

What is the relationship between net force and balanced forces

creativ13 [48]

The mass will accelerate. Balanced Forces: When forces are in balance, acceleration is zero.

5 0

3 years ago

Read 2 more answers

A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

7 0

4 years ago

Suppose a car is traveling at +25.0 m/s, and the driver sees a traffic light turn red. After 0.340 s has elapsed (the reaction t

scoundrel [369]

First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m

Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
vf^2 = vi^2 + 2*a*d where:
vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = 1/2*(625) /(7) = 44.6428 m

The total stopping distance = 8.5 + 44.6428 = 53.1428 m

3 0

3 years ago

Can someone help me figure out how to do a function formula for fx and fy

masha68 [24]

Answer: How to solve for FX and FY?

to find fx(x, y): keeping y constant, take x derivative; • to find fy(x, y): keeping x constant, take y derivative. f(x1,...,xi−1,xi + h, xi+1,...,xn) − f(x) h . ∂y2 (x, y) ≡ ∂ ∂y ( ∂f ∂y ) ≡ (fy)y ≡ f22. similar notation for functions with > 2 variables.

Explanation:

4 0

3 years ago