Lucy took 3 hours to cover 2/3 of a journey. She covered the

6. Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. Which light bulb has the greater filament resi

Nonamiya [84]

$\\ \bull\tt\longrightarrow P=\dfrac{V^2}{R}$

P is power
R is resistance

$\\ \bull\tt\longrightarrow R=\dfrac{V^2}{P}$

Hence

$\\ \bull\tt\longrightarrow R\propto V$

$\\ \bull\tt\longrightarrow R\propto \dfrac{1}{P}$

Therefore if power is low then resistance will be high.

The first bulb has less power hence it has greater filament resistance.

5 0

2 years ago

You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear

Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

$L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|$

L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?

$log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}$

$\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =15 \times 10^{\dfrac{|20-50|}{20}}$

$r_2 =15 \times 10^{-1.5}$

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0

3 years ago

1. Straight-in spaces can leave you a safer out if

Juliette [100K]

Answer:

C. you're able to reverse out of the parking spot

Explanation:

Straight-in parking is an approach of parking that allows a more flexible traffic layout where a driver can approach the spot from either direction and still safely park within the lines. It thus helps to prevent blockage of cars. Each car can move in and out freely preventing it from congestion.

This way of parking can leave you safe when you able to reverse out of the parking spot. It gives you greater control and makes it easier to maneuver out space. The benefits of Straight-in parking are,

- Allows for two-way traffic

- Drivers can line up the vehicle from multiple angles

- Saves time for drivers

3 0

2 years ago

Dense water near the poles sinks, creating a current towards the equator. What would you expect to happen to this current if tem

Rudik [331]

It's not c or d :/!!!!!

3 0

3 years ago

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0

1 year ago