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nignag [31]
3 years ago
5

This causes nickel and certain other metals to be magnetic.

Physics
2 answers:
bija089 [108]3 years ago
5 0

Answer:

The answer is B

can i have brainliest

natka813 [3]3 years ago
4 0
"The pattern of electrons in their atoms" is the one among the following choices given in the question that <span>causes nickel and certain other metals to be magnetic. The correct option among all the options that are given in the question is the second option or option "B". I hope that the answer has helped you.</span>
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A phoneme is the largest unit of sound in a word.
lukranit [14]

This question is incomplete; here is the complete question:

A phoneme is the largest unit of sound in a word. Please select the best answer from the choices provided

A. T

B. F

The correct answer to this question is F (False)

Explanation:

The word "phoneme" is used to refer to the minimal unit of sound in words, and therefore in language. For example, the first phoneme in the word "man" is "m". These units of sound are essential in language because they make each word unique in meaning and sound. For example, "fan" and "man" are different due to the phonemes "m" and "f". According to this, the phone is not the largest unit of sound but the smallest unit.

7 0
3 years ago
Read 2 more answers
Which has greater value, a newton of gold on earth or a newton of gold on the moon?
jek_recluse [69]
The moon, because the acceleration due to gravity is less.
4 0
2 years ago
Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed
schepotkina [342]

Answer:

12.35m

Explanation:

Hello! To solve this problem we must consider the following:

1. The car moves with constant speed, which means that the distance traveled is equal to the multiplication of time by speed.

X = VT

we solve the equation for time

T=\frac{x}{V} =\frac{27}{17} =1.588s

2. The bolt moves with constant acceleration, with acceleration of 9.81m / s ^ 2, so we could apply the following equation.

note=remember that "a uniformly accelerated motion", means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

Y= VoT-\frac{1}{2}gt^{2}

where

Vo = Initial speed

=0

T = time

=1.588s

g =gravity=9.8m/s^2

Y =  bridge height

solving

Y= \frac{1}{2}gt^{2}\\Y=(0.5)(9.8)(1.588^2)=12.35m

4 0
3 years ago
Which part of an atom makes up most its volume
VikaD [51]

Hi! The inner most part of the atom (which is most of the atom) is occupied by electrons, therefore it’d take of most of its volume. Have a nice day!

4 0
3 years ago
A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the
Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

a = \frac{v^{2} }{r}

Where a is the centripetal acceleration

v is the speed

and r is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be v_{2}, that is

v_{2} = 2v

and without changing the length of the string means radius r is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be a_{2}.

Then we can write that

a_{2} = \frac{v_{2}^{2}  }{r}

From

a = \frac{v^{2} }{r}

v = \sqrt{ar}

Recall that

v_{2} = 2v

∴ v_{2} = 2\sqrt{ar}

Now, we will put the value of v_{2} into

a_{2} = \frac{v_{2}^{2}  }{r}

Then,

a_{2} = \frac{(2\sqrt{ar}) ^{2}  }{r}

a_{2} = \frac{4ar }{r}

a_{2} = 4a

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0
3 years ago
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