You must always include a direction when recording an amount of energy. Is this true?

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

4 0

4 years ago

An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N

algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

K.E = 15.57 x 10⁻¹⁷ J

4 0

3 years ago

.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum accelerat

AlexFokin [52]

Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg

assume the greatest scale reading = 78.09 kg

Determine the maximum acceleration for these elevators

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N

At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

3 0

3 years ago

2. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this wee

Alex17521 [72]

Answer:

62 seconds
no

Explanation:

The total travel time Suzette experiences is the sum of the times in each hallway. Using

time = distance/speed

we can add the times.

(35.0 m)/(3.50 m/s) +(48.0 m)/(1.20 m/s) +(60 m)/(5.0 m/s)

= 10 s + 40 s + 12 s

= 62 s

It takes Suzette 62 seconds to get to class. She does not beat the bell.

3 0

4 years ago

a body is dropped from a height of 240m. (i)How long does it take to reach the ground. (ii) What is the velocity with which it w

kramer

Answer:

7.0 s, 69 m/s

Explanation:

If we take down to be positive, then the time to reach the ground is:

x = x₀ + v₀ t + ½ at²

240 m = (0 m) + (0 m/s) t + ½ (9.8 m/s²) t²

t = 7.0 seconds

The final velocity is:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²) (240 m - 0 m)

v = 69 m/s

5 0

3 years ago