Ask question

Ask question

All categories

3 years ago

7

A student performs an experiment and must measure the lengths of four different objects: a textbook, a pencil, a cup, and a piec

e of bread. Which would be the most appropriate unit of measurement for her to use?

2 answers:

Strike441 [17]3 years ago

7 0

Most appropriate unit of measurement should be centimetres, cm.

ehidna [41]3 years ago

6 0

Answer:

Centimeter (cm)

Explanation:

A student performs an experiment and must measure the lengths of four different objects: a textbook, a pencil, a cup, and a piece of bread.

There are so many units of measuring length of an object like centimeter, millimeter, kilometer etc. Some of units are used to measure very large distances (i.e. the distance between sun and earth etc ) are light year, astronomical units etc.

In this case, a textbook, a pencil, a cup, and a piece of bread are small object. The most appropriate unit of measurement for her to use is centimeter or cm.

You might be interested in

For a particular type of motion, the velocity is zero but the speed is a nonzero quantity. Which statement can you make about th

masha68 [24]

Answer:

because speed is the modulus of velocity which is a vector

the velocity to be zero it must be a round trip

Explanation:

This is because speed is the modulus of velocity which is a vector.

For the velocity to be zero it must be a round trip, therefore the resulting vector zero

On the other hand, the speed of the module is the same in both directions

8 0

3 years ago

Individuals who suffer from a mental disorder are most likely to commit violent crimes. true or false

Damm [24]

That is true. Some people with mental disorders envision the world differently. Some people think that people are out to get them, they are in another world, etc. They will do anything to feel safe in their mental state.

I hope this helps!
~cupcake

7 0

3 years ago

Read 2 more answers

A ________ detects the speed that a given device can handle and communicates with it at that speed

melisa1 [442]

Switch because look A switch detects the speed that given device can handle and communicates with it at that speed

4 0

3 years ago

A bicycle odometer which counts revolutions and is calibrated to report distance traveled is attached near the wheel axle and is

Bess [88]

Answer:

each rotation of the smaller wheel will show 84.382-75.3982=8.9838 inches more than the actual distance

Explanation:

d = Diameter of the wheel

The distance traveled in one rotation of the wheel is the circumference of the wheel

$\pi d=\pi\times 27\\ =84.823\ inch$

When diameter is 24 inches

$\pi d=\pi\times 24\\ =75.3982\ inch$

Therefore, each rotation of the smaller wheel will show 84.382-75.3982=8.9838 inches more than the actual distance

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:

a)

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

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

b)

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:

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

c)

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