Myoncedyret

Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?

shtirl [24]

Answer:

$L=6.21m$

Explanation:

For the simple pendulum problem we need to remember that:

$\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0$ ,

where $\theta$ is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

$\omega^{2}=\frac{g}{L}$ ,

where $\omega$ is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

$\omega=\frac{2\pi}{T}$ ,

where T is the oscillation period. Now, we can easily solve for L:

$(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m$

3 0

3 years ago

Which of the following is a chemical change?

klemol [59]

a). Water is still H₂O after it freezes.

b). Ice is still H₂O after it melts.

c). Wire is still Cu when it's bent.

d). Paper combines with the O₂ in the air, and turns into
a lot of new compounds when it burns.

3 0

3 years ago

A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t

andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = $\frac{r}{2}$

f = $\frac{0.40}{2} = 0.20 m$

Now,

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

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

$\frac{v}{u}$ = 2 (2)

Now, by lens maker formula:

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

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

v = $\frac{uf}{u - f}$ (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = $\frac{uf}{u - f}$

2 = $\frac{f}{u - f}$

2(u - 0.20) = 0.20

u = 0.30 m

6 0

3 years ago

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

2 years ago

DETERMINE THE BCD EQUIVALENT OF 1100111110101001

Solnce55 [7]

Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.

6 0

2 years ago