What unit of electric current,the ampere,is equivalent to

A monochromatic light falls on two narrow slits that are 4.50 um apart. The third destructive fringes which are 35° apart were f

STatiana [176]

Answer:

y = 1.19 m and λ = 8.6036 10⁻⁷ m

Explanation:

This is a slit interference problem, the expression for destructive interference is

d sin θ = m λ

indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m

λ = d sin θ / m

let's calculate

λ = 4.50 10⁻⁶ sin 35 /3

λ = 8.6036 10⁻⁷ m

for the separation distance from the central stripe, we use trigonometry

tan θ= y / L

y = L tan θ

the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits

L = 2 -0.30

L = 1.70 m

let's calculate

y = 1.70 tan 35

y = 1.19 m

3 0

3 years ago

You drop a rock off the top of a building. It takes 3.4 s to hit the ground. What is the velocity at impact (final velocity)? I

SCORPION-xisa [38]

-- The acceleration of gravity (on Earth) is 9.8 m/s².

-- That means that during every second an object falls,
it adds 9.8 m/s of speed.

Now ! If it adds 9.8 m/s of speed every second, then
how fast is it falling at the end of 3.4 seconds ?

This is as close as I can bring you to the answer
without dropping it at your feet, or handing it to you
on a golden tray.

3 0

3 years ago

What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force i

CaHeK987 [17]

Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

The net friction force's magnitude will be:

⇒ $F_{net}=ma$

$=5\times 1.7$

$=8.5 \ N$

(b)

For m₃,

⇒ $ma=\mu m_3 g$

Or,

⇒ $\mu=\frac{a}{g}$

$=\frac{1.7}{9.8}$

$=0.173$

5 0

3 years ago

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until

Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

Initial angular velocity, $\omega_0 = 30.0 \text{ rad/s}$

Angular acceleration, $\alpha =35.0 \text{ rad/s}^2$

(A)

At t = 2.00 s, the angular displacement, θ, is given by

$\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}$

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

$\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}$

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

$\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t$