What is the magnitude of the centripetal acceleration of an object on earth's equator owing to the rotation of earth?

Which of the following terms refers to the amount of heat needed to change 1 kg of a substance from a liquid to a gas at its boi

Nataly [62]

It's B I thinkkkkkkkk

8 0

4 years ago

Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter yo

Tems11 [23]

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

3 0

4 years ago

Two points are on a disk turning at constant angular velocity. One point is on the rim and the other halfway between the rim and

DerKrebs [107]

Answer:

The point on the rim

Explanation:

All the points on the disk travels at the same angular speed $\omega$ , since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the point from the centre of the disk

As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.

3 0

4 years ago

Which part of the following graph shows the object going at the fastest speed?

Alex73 [517]

Answer:

more point plzzzzz... eeeeeeeee

4 0

3 years ago

A fleeing student covered a distance of 210 meters in 35 seconds. what is the student's speed in meters/second? what is the stud

snow_lady [41]

Explanation:

Given that,

A student covered a distance of 210 meters in 35 seconds.

We need to find the student's speed in meters/second and also in meters/minute.

Speed, v = distance (d)/time (t)

So,

$v=\dfrac{210\ m}{35\ s}\\\\=6\ m/s$

We know that, 1 minute = 60 seconds

$6\dfrac{m}{s}=6\times \dfrac{m}{(\dfrac{1}{60})\ \text{minutes}}\\\\=360\ \text{meters/minutes}$

Hence, the student's speed is 6 m/s or 360 meters/minute.

6 0

3 years ago