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3 years ago

Clarissa thought that learning to play the banjo was a waste of time. She didn't bother to practice after her lessons

Physics

2 answers:

emmasim [6.3K]3 years ago

7 0

Answer:

The answer ro this question is fear of failure

Alenkasestr [34]3 years ago

6 0

Answer:

i believe the answer is low expectations

Explanation:

hope this helped :)

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In covalent bonds what is being shared

gulaghasi [49]

Answer:

Image result for In covalent bonds what is being shared

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

8 0

2 years ago

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before

Darina [25.2K]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

$\\$

☯ As we know that,

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

$\\$

$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

$\\$

$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

$\\$

$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

$\\$

$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

$\\$

$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

$\\$

$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

$\\$

$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

4 0

2 years ago

Please give an answer that is coherent

Semmy [17]

Answer:

480

Explanation:

resistance equals to potential difference divide by electric current

120÷0.25

=480

6 0

3 years ago

Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. de

atroni [7]

Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c

5 0

3 years ago

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed

beks73 [17]

Answer:

v = 5.9 x 10⁷ m/s

Explanation:

The kinetic energy of the electron in terms of potential difference is given as:

$K.E = eV$ --------------- equation (1)

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

V = Potential Difference = 9.9 KV = 9900 Volts

The kinetic energy in general is given as:

$K.E = \frac{1}{2}mv^{2}\\$ --------- equation (2)

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

Therefore, comparing equation (1) and equation (2), we get:

$\\\frac{1}{2}mv^{2} = eV\\\\\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)v^{2} = (1.6\ x\ 10^{-19}\ C)(9900\ volts)\\\\v = \sqrt{34.81\ x\ 10^{14}} \\$

8 0

3 years ago