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3 years ago

Which is heavier 1kg of iron or 1 kg of feathers?

1 answer:

7 0

-- On Earth, 1 kg of anything weighs 9.81 Newtons (2.205 pounds). It doesn't matter what substance comprises the kilogram.

-- Although their weights are identical, any mass of feathers occupies more volume than the same mass of iron, no matter where they happen to be located.

-- Iron is more dense than feathers, and more dense than a lot of other substances as well, too.

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state the type of force which provide the needed centripetal force in the following instance :an electron orbiting the nucleus

Vladimir [108]

i think the answer is electrostatic force hope this helps u stay safe

6 0

3 years ago

A ball is dropped off the roof of a tall building. If the ball reaches the ground in 8 seconds, how tall is the building, in met

Ghella [55]

Let b be the height of the building, and y the height of the ball at time t, given by

y = b - 1/2 gt²

where g = 9.8 m/s² is the magnitude of the acceleration due to gravity.

It takes the ball 8 s to reach the ground, at which point y = 0, so that

0 = b - 1/2 (9.8 m/s²) (8 s)²

b = 1/2 (9.8 m/s²) (8 s)²

b = 313.6 m

5 0

4 years ago

To a nighttime observer on Earth, how many degrees do the stars appear to move around Polaris in 3 hours?

STatiana [176]

The correct answer is 45 degrees. 360 degrees is 24 hours, because it takes a whole day for the earth to turn in one whole circle. So, to find how many degrees the stars seem to move per hour, simply divide the total number of degrees by the total number of hours in a day: 360/24 = 15. Since the question is asking how much the stars appear to move in 3 hours, now multiply the number of degrees per hour times 3: 15 x 3 = 45.

5 0

3 years ago

The revolving nosepiece of a compound microscope is used to: a. move the condenser up or down b. change the objective lens c. ad

Jlenok [28]

Answer:

Option B: change the objective lens

Explanation:

The revolving nosepiece is one of the parts of a microscope. Its responsibility is to hold the objective lenses.

6 0

3 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

$\alpha=\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

The angular velocity can be written as

$\omega=\frac{2\pi}{T}$

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

$\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})$

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

$T'=T+8.09\cdot 10^{-6} =0.01400809 s$

Therefore, the change in angular velocity after 1 year is

$\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s$

So, the angular acceleration of the pulsar is

$\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y$

To solve this part, we can use the following equation of motion:

$\omega'=\omega + \alpha t$

where

$\omega'$ is the final angular velocity

$\omega$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time

For the pulsar in this problem:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s$ is the initial angular velocity

$\omega'=0$ , since we want to find the time t after which the pulsar stops rotating

$\alpha = -0.259 rad/s/y$ is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

$t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y$

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

$\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y$

which means that the period of the pulsar increases by

$\Delta T=8.09\cdot 10^{-6} s$

For every year:

$\Delta t=1 y$

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

$T=T_0+\frac{\Delta T}{\Delta t}\Delta t$

where $T_0$ is the original period and

$\Delta t=869 y$

is the time that has passed; solving for T0,

$T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s$

6 0

3 years ago