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2 years ago

Which question can be used to identify a physical property of an element?

Physics

2 answers:

olganol [36]2 years ago

7 0

Answer:

C. Do bubbles form when the element is mixed with an acid?

Explanation:

all the other options are incorrect because they are

BabaBlast [244]2 years ago

3 0

Answer:C

Explanation:

Do bubbles form when the element is mixed with an acid

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100 points!! word bank!

Novosadov [1.4K]

Answer:

Except in the crust, the interior of the Earth cannot be studied by drilling holes to take samples. Instead, scientists map the interior by watching how seismic waves from earthquakes are bent, reflected, sped up, or delayed by the various layers.

Explanation:

6 0

2 years ago

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl

aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0

3 years ago

ASAP ONLY CORRECT ANSWERS PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEE I RLLY NEED HELPPPPPPP

nadya68 [22]

Answer:

attached below

Explanation:

5 0

3 years ago

An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu

never [62]

Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

$R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N$ ....(I)

Where, N = number of radio active atoms

$t_{\frac{1}{2}}$ =half life

We need to calculate the number of radio active atoms

For $N_{12_{c}}$

$N_{12_{c}}=\dfrac{N_{A}}{M}$

Where, $N_{A}$ =Avogadro number

$N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}$

$N_{12_{c}}=5.02\times10^{22}\ nuclie/g$

For $N_{c_{14}}$

$N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}$

$N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}$

$N_{c_{14}}=6.526\times10^{10}\ nuclei/g$

Put the value in the equation (I)

$R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}$

$R=15.0650\ decay/min g$

100 g carbon will decay with rate

$R=100\times15.0650=1507\ decay/min$

We need to calculate the total half lives

$(\dfrac{1}{2})^{n}=\dfrac{390}{1507}$

$2^n=\dfrac{1507}{390}$

$2^n=3.86$

$n ln 2=ln 3.86$

$n=\dfrac{ln 3.86}{ln 2}$

$n =1.948$

We need to calculate the age of living tree

Using formula of age

$t=n\times t_{\frac{1}{2}}$

$t=1.948\times5700$

$t=11103.6 =11104\ years$

Hence, The age of living tree is 11104 years.

5 0

3 years ago

An object experiences a constant change in velocity in free fall.<br><br> True False

jeyben [28]

False, it experiences a constant change in ACCELERATION in free fall.

3 0

2 years ago