A child sits on a dock and notices that 8 wavelengths pass the end of the dock in 4 seconds. What is the

Daniel [21]

The red box must way more. Gravitational potential energy is the product of a an objects mass times the acceleration due to gravity (which is constant on earth) times its height. Since the objects are on the same shelf they are at the same height, and since gravitational acceleration is constant as long as we stay on planet earth, then the mass is the only possible thing that could have changed. This means that the red box must weigh more than the blue box.

3 0

3 years ago

You are driving a car with an automatic transmission; the shifter is mounted on the steering column. When you press down on the

castortr0y [4]

Answer:

From the narrative in the question, there seem to have been a break failure and the ordered step of response to this problem is to

1) Put on the hazard light to inform other road users of a problem or potential fault with your car and so they should continue their journey with caution.

2) Avoid pressing on the acceleration pedal as this might cause the car to gradually slow down due to friction and gravity

3)Try navigate the car to the service lane. This is the less busy lane where cars are sometimes parked briefly.

4) Continuously pump the breaks to try stop the car. Continuously pumping the breaks might just help you build enough pressure to stop the car because often time, there are some pressure left in the break.

5) At this point, the speed of the car should be relatively slow. So at this point, you could try apply the emergency hand break. Do not pull the emergency hand breaks if the car is on high speed. Doing this may cause the car to skid off the road.

8 0

3 years ago

You have four fixed-volume containers at STP . Container A has 0.5 mol of gas in 11.2 L. Container B has 2 mol of gas in 22.4 L.

Colt1911 [192]

Answer:

Container A and C

Explanation:

ideal gas equation gives P=nRT/V

so at constant Temperature and pressure, P=n/T

Container A and C after dividing number of moles and Volume, are found to be the same=0.0446

8 0

3 years ago

What is a radio interferometer? a device for correcting chromatic aberration in radio telescopes a device for absorbing and tran

JulsSmile [24]

Answer: a network of several radio telescopes wired together

Explanation:

A radio interferometer combines signals of several radio telescopes which are used in astronomical observations simultaneously to simulate a discretely-sampled single telescope of very large aperture

Interferometer, an instrument that uses the interference patterns formed by waves to measure certain characteristics of the waves themselves or of materials that reflect, refract, or transmit the waves. Interferometers can also be used to make precise measurements of distance.

3 0

3 years ago

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest

shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer: 3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f = 5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0

3 years ago