If A= +4j - 2i - 3k and C= -4i -2j - 3k , which of the following numbers is closest to the magnitude of A-C ?

Hello, I don't see a table, but I am guessing that you are referring to the one I attached (below).

Answer:

So, the correct order of events sorted chronologically is:

1. A nebula located in the Milky Way galaxy begins pulling nearby hydrogen atoms in its orbit.

2. The Nebula shrinks in its volume due to gravity, becoming denser and hotter. But, it's not hot enough for nuclear fusion.

3. The temperature in the core of the Nebula reaches 14 million Kelvin.

4. Hydrogen atoms begin shedding their electrons and colliding with one another.

5. The Sun enters the main sequence stage. The energy created as a result of its radiation begins nurturing life on planet such as Earth.

6. The Sun uses up all the hydrogen in its core.

7. The Sun expands greatly and cools. It is larger and redder.

8. The Sun completely runs out of hydrogen to fuse. Its outer layers are pushed away, and a cloud of ionized gas surrounds its core.

9. The Sun is a white dwarf with a dim glow.

Hope this helps!

Answer:

It includes earthwork, trenching, wall shafts, tunneling and underground

Explanation:

Answer:

Divide the change in distance by the change in time.

Explanation: Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).

To my best knowledge It's D

Answer:

50km/h

Explanation:

Average velocity = change in distance (or distance travelled) divided by/ the change in time (or time taken.)

The change in distance has been given as 100km.

The change in time is 3pm-1pm = 2 hours.

Therefore the average velocity was 100/2 = 50km/h (to the east).

Hope this helped!

Answer:

Explanation:

In this problem, we are required to solve for the total distance that the car travelled. and the displacement

A) the distance travelled by car

this can be gotten by summing all the distances the car has travelled.

i,e total distance= 60 miles+120 miles

total distance= 180 miles

B) the displacement of the car

the displacement can be gotten by  subtracting the final distance from the initial distance

final distance = 120 miles

initial distance= 60 miles

displacement= 120-60= 60 miles

Explanation:

Given that,

Initial length of simple pendulum, [tex]L_1=1\ m[/tex]

Initial time period, [tex]T_1=2\ s[/tex]

We need to find the length of the simple pendulum when the period is exactly 1 second.

[tex]T_2=1\ s[/tex]

We know that the time period of simple pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{L}{g}} \\\\T\propto \sqrt{L} \\\\\dfrac{T_1}{T_2}=\dfrac{L_1}{L_2}[/tex]

Put all values and find L₂

[tex]\dfrac{T_1}{T_2}=\sqrt{\dfrac{L_1}{L_2}}\\\\L_2=\dfrac{T_2^2L_1}{T_1^2}\\\\L_2=\dfrac{1^2\times 100\ cm}{2^2\ s}\\\\L_2=25\ cm[/tex]

So, the length of the pendulum with a period of exactly one second is 25 cm.

A lighted candle can increase the temperature inside a shelter by a few degrees Celsius.

A lighted candle produces heat by burning fuel. The heat from the candle is transferred to the air in the shelter by conduction, convection, and radiation. Conduction is the transfer of heat through direct contact. Convection is the transfer of heat through the movement of fluids. Radiation is the transfer of heat through electromagnetic waves.

The amount of heat that is transferred to the air in the shelter depends on the size of the candle, the number of candles, and the insulation of the shelter. The larger the candle, the more heat it will produce. The more candles, the more heat will be produced. The more insulation the shelter has, the less heat will be lost to the outside. In general, a lighted candle can increase the temperature inside a shelter by a few degrees Celsius.

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Answer:

A. The wires exert equal magnitude attractive forces on each other.

Explanation:

Magnetic field due to current i on current 2i

B₁ = 10⁻⁷ x 2 i / r where r is distance between the two wires

Force on wire II due to wire I per unit length

= magnetic field x current in wire II

= B₁ x 2 i

= [ 10⁻⁷ x 2 i / r ]  x 2i

= 4  x 10⁻⁷ i² / r

Magnetic field due to current 2i on current i

B₂ = 10⁻⁷ x 4 i / r where r is distance between the two wires

Force on wire I due to wire II per unit length

= magnetic field x current in wire I

= B₂ x  i

= [ 10⁻⁷ x 4 i / r ]  x i

= 4  x 10⁻⁷ i² / r

So final forces on each wire are same .

This force will be attractive in nature . The direction of force can be known from fleming's right  hand rule .

Answer:

Fe = 9 x 10⁻⁹ N

Fg = 3.97 x 10⁻⁴⁶ N

Fe = 2.26 x 10³⁷ Fg

Explanation:

First we find electric force by Coulomb's Law as follows:

Fe = kq₁q₂/r²

where,

Fe = electric force = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = q₂ = charges on electron and proton = 1.6 x 10⁻¹⁹ C

r = distance between electron and proton = 1.6 x 10⁻¹⁰ m

Therefore,

Fe = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(1.6 x 10⁻¹⁰ m)²

Fe = 9 x 10⁻⁹ N

Now we find gravitational force by Newton's Law of Gravitation as follows:

Fg = Gm₁m₂/r²

where,

Fg = gravitational force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of electron = 9.11 x 10⁻³¹ kg

m₂ = mass of proton = 1.673 x 10⁻²⁷ kg

r = distance between electron and proton = 1.6 x 10⁻¹⁰ m

Therefore,

Fg = (6.67 x 10⁻¹¹ N.m²/kg²)(9.11 x 10⁻³¹ kg)(1.673 x 10⁻²⁷ kg)/(1.6 x 10⁻¹⁰ m)²

Fg = 3.97 x 10⁻⁴⁶ N

Dividing both forces:

Fe/Fg = (9 x 10⁻⁹ N)/(3.97 x 10⁻⁴⁶ N)

Fe = 2.26 x 10³⁷ Fg

The electric force is 2.3 × 10^39 times greater than the gravitational force.

The gravitational force between the proton and the electron is obtained from;

Fg = Gmemp/r^2

Where;

me = mass of the electron

mp = mass of the proton

r = distance of separation

Fg = 6.67 × 10^−11  × 9.11 × 10^−31 × 1.673 × 10^−27/(1.6 × 10^−10)^2

Fg = 3.94 × 10^−48 N

For the electric force;

Fe= K e^2/r^2

K = electric constant

e = magnitude of charge on electron and proton

r = distance of separation

Hence;

Fe = 8.988 × 10^9 × (1.602×10^−19)^2/(1.6×10^−10)^2 =

Fe = 9 × 10^−9 N

The number of time the electric force is stronger than the gravitational force = 9 × 10^−9 N/3.94 × 10^−48 N = 2.3 × 10^39 times

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Answer:

3.15 N towards the positive x-axis

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = [tex]\frac{-kQq}{r^2}[/tex]

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = [tex]\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}[/tex] = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = [tex]\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}[/tex] = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = 3.15 N towards the positive x-axis

Answer:

About 1200J

(If we take gravity to be 10m/s^2)

Explanation:

U=mgh

m=6kg

g=10m/s^2

h=20m

U=(6)(10(20)=1200J

Explanation:

The expression is :

[tex]A=B^nC^m[/tex]

A =[LT], B=[L²T⁻¹], C=[LT²]

Using dimensional of A, B and C in above formula. So,

[tex]A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}[/tex]

Comparing the powers both sides,

2n+m=1 ...(1)

2m-n=1 ...(2)

Now, solving equation (1) and (2) we get :

[tex]n=\dfrac{1}{5}\\\\m=\dfrac{3}{5}[/tex]

Hence, the correct option is (E).

The values of the exponents n and m are [tex]\frac{1}{5}[/tex] and [tex]\frac{3}{5}[/tex] respectively.

The given parameters;

[tex]A = B^nC^m[/tex]

[tex]A = LT\\\\B = L^2T^{-1}\\\\C = LT^2[/tex]

The values of the given exponents "n" and "m" are calculated as follows;

[tex]LT = [L^2T^{-1}]^n[LT^2]^m\\\\LT= [L^{2n}T^{-n}][L^mT^{2m}]\\\\LT = [L^{2n + m} \ T^{2m-n}]\\\\L^1 = L^{2n + m} \\\\T^1 = T^{2m-n}\\\\1 = 2n \ + m \ ---(1)\\\\1 = 2m - n \ ---(2)[/tex]

from equation(2);

[tex]n = 2m - 1[/tex]

substitute the value of n into equation (1);

[tex]1 = 2(2m-1) + m\\\\1 = 4m - 2 + m\\\\1 = 5m - 2\\\\3 = 5m\\\\m = \frac{3}{5} \\\\n = 2m - 1\\\\n = 2(\frac{3}{5} ) - 1\\\\n = \frac{6}{5} - 1 \\\\n = \frac{1}{5}[/tex]

Thus, the values of the exponents n and m are [tex]\frac{1}{5}[/tex] and [tex]\frac{3}{5}[/tex] respectively.

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Answer:

732492 g

Explanation:

Given that

Room Temperature = 300 K

Cold T = 0°C = 0 + 273 =273 K

Work done = 0.375 hp

ΔH of fusion of water is 6008 J per mol

The first thing we do is to Convert the power into J/s

Given that 1 hp = 746 W (J/s), then

0.375 hp = 0.375 * 746 W/hp

0.375 = 280 J/s

Then we find the heat per unit time

Q = (cold T / hot T - cold T ) x power

Q = [273 / (300 - 273)] * 280

Q =273/27 * 280

Q = 10.1

Q = 2831 J/s

Moles of ice per second = Q/ ΔH fusion

Moles of ice per second = 2831/6008 = Moles of ice per second = 0.471 mol /s

If we convert 1 day into second, we have

1 day = 1 day * ( 24 hours/day) * ( 60 minutes/hour) * (60 seconds/minutes)

1 * 24 * 3600 = 86400 second

moles of ice /day

0.471 mol / s ) x 86400 s / day =

40694 mols of ice

Molar mass is 18 g/mol

Mass of ice

40694 * 18 = 732492 g

Mass of ice produced per day is 732492 g

Answer:

a

[tex]N  =  2.094*10^{10} \  electrons[/tex]

b

[tex]O  =  9.33*10^{-13} \  electrons[/tex]

Explanation:

From the question we are told  that

   The  mass of the lead sphere is  [tex]m  =  7.70g =  0.0077 \  kg[/tex]

   The net charge is [tex]Q_{net} =  -3.35*10^{-9} \ C[/tex]

   The  atomic number is  [tex]u  =  82[/tex]

     The  molar mass is  [tex]M  =  207 \  g/mol[/tex]

Generally the excess number of  electron on the sphere is mathematically represented as

     [tex]N  =  \frac{Q_{net}}{ e }[/tex]

Here e is the charge on the electron is [tex]e  =  -1.60 *10^{-19} \  C[/tex]  

     So

    [tex]N  =  \frac{-3.35 *10^{-19}}{ -1.60*10^{-19}}[/tex]

     [tex]N  =  2.094*10^{10} \  electrons[/tex]

Generally the number of atom present is mathematically represented as  

     [tex]n  =  N_a  *  \frac{m}{ M}[/tex]

Here [tex]N_a[/tex] is the Avogadro's number with value  [tex]N_a  =  6.0*10^{23} \  atoms[/tex]

       [tex]n  =  6.03 *10^{23} *  \frac{7.70}{ 207}[/tex]

      [tex]n  =  2.24 *10^{22} \  atoms [/tex]

Generally the electrons are there per lead atom is mathematically represented as

       [tex]O =  \frac{N}{n}[/tex]

=>   [tex]O =  \frac{2.24*10^{22}}{2.094*10^{10}}[/tex]

=>    [tex]O  =  9.33*10^{-13} \  electrons[/tex]

Answer:

solution

Explanation:

A(n) __________ is developed during the process of technological design.

Answer:

A. Using

Pgauge= Pmanometer

And we know that

Pgauge= deta(hpg)

So deta h = Pgauge/density x g

So

= 7.5(6894.76/1psig)/ 13.6*10^3*9.8m/s²)

= 387.9mm

So to find height of pipe connected to the pipe we say

= h -deta h

= 900-387.97mm

=512.02mm

B. We use manometry principle

Pgas+density xg(25*10^3)-PX density{h-(H-0.25)=0

So

Finally Pgas= 6.54psig

Answer:

With this analysis the correct answer is

Metal bar. The answer is uniform charge  distribute

Non-metallic bar. As we see the charge of the ball, at the ends closest to the ball a charge of the same sign as the ball and at the far end the charge of equal magnitude and opposite sign

Explanation:

When the ball touches the bar, a charge transfer process occurs, leaving half in the bar and the other half in the ball, when the ball bounces and clears it, two dependent processes can occur, the bar is metallic or non-metallic , let's analyze the two cases

* metal bar, the load that is on the bar is distributed evenly throughout the bar, with each bounce of the ball the load on the bar increases and decreases in the balls until the two loads are equal and appear from this rebound no there is more load transfer

* Non-conductive insulating bar, when the ball touches the bar for the first time, it creates a load equal to q / 2, when the ball moves away from the bar, for this non-conductive, the transferred load remains in the same place. Therefore this load induces a load of equal magnitude, but opposite sign at the other end of the bar.

When the ball approaches for the second time, the opening already has a load, therefore the amount of load deposited is less, with between processes in each bounce the bar of the bar sounds, the process continues until the load of the ball and the bar have been equal.

With this analysis the correct answer is

Metal bar. The answer is uniform charge  distribute

Non-metallic bar. As we see the charge of the ball, at the ends closest to the ball a charge of the same sign as the ball and at the far end the charge of equal magnitude and opposite sign

Answer:

[tex] \boxed{ \sf{see \: below}}[/tex]

Explanation:

The unit of speed is metre per second or m / s. It clearly depends on two fundamental units ; unit of length ( metre ) and unit of time ( second ). Hence , m/s is a derived unit.

Hope I helped!

Best regards! :D

The unit of speed is m/s. The two terms metre and second are fundamental quantities.

As,m/s is derived from these two fundamental quantities. Therefore, Unit of speed is a derived unit.

Answer:

An virus

Explanation:

An computer virus can come from websites that try to steal your Information or apps that's not necessarily scanned and protected so that yor computer can be safe from viruses.

A computer may encounter problems or get "ill" for a variety of reasons. The system can become infected with malware and viruses, which can cause performance issues and data loss.

A failing hard drive or defective RAM are examples of hardware issues that can lead to instability and crashes. Overheating can lead to hardware damage or performance deterioration and is frequently brought on by insufficient cooling or clogged air vents.

Systems may become unstable as a result of software conflicts between out-of-date or incompatible programmes.

Electrical problems or power surges can harm components and interfere with proper operation. Computer issues can also be caused by human error, such as inadvertent file deletion or poor configuration.

Thus, a computer can be kept in good condition with routine maintenance, security precautions, and appropriate usage.

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Explanation:

It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

The gravitational force is given by :

[tex]F=\dfrac{Gm_em_m}{r^2}[/tex]

It means [tex]F\propto \dfrac{1}{r^2}[/tex]

So,

[tex]\dfrac{F_1}{F_2}=\dfrac{r_2}{r_1}[/tex]

r₁ = 3.9x10⁸ km

r₂= 1.5x10⁸ km

So,

[tex]\dfrac{F_1}{F_2}=\dfrac{1.5\times 10^8}{3.9\times 10^8}\\\\\dfrac{F_1}{F_2}=\dfrac{5}{13}[/tex]

Hence, the ratio of the gravitational forces exerted by Earth and the Sun on the Moon is 5:13.

55.4 km => 1hr (60 min) (60×60=120 sec)

1 km => 120 ÷ 55.4

1.21 km => (120 ÷ 55.4) × 1.21

= 2.6209 sec ( 5sf )

= 2.62 sec ( 3sf )

i hope you are able to understand my solution :))

Answer:

Explanation:

[tex]y(t) = 5t-5t^2[/tex]

When y(t) = 0, the ball is on the hand.

[tex]0=5t-5t^2\\0=5t(1-t)\\t=0,1[/tex]

It takes 1 second.

There are several different possibilities.

==>  If the 5.0 means 5 miles per hour, that's 2.24 meters per second, up.

The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (2.24/9.8) = 0.229 second.  Then, after is starts to fall, it takes the same amount of time to the person's hand.

Total time = 0.457 second.

==>  If the 5.0 means 5 meters per second, up . . .

The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (5.0/9.8) = 0.51 second.  Then, after is starts to fall, it takes the same amount of time to the person's hand.

Total time = 1.02 second.

==>  If the 5.0 means 5 km/minute, that's about 83.33 meters per second, up.

The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (83.33/9.8) = 8.503 seconds.  Then, after is starts to fall, it takes the same amount of time to the person's hand.

Total time = 17.01 seconds.

==>  If the 5.0 means 5 furlongs per fortnight, that's about 0.00083 meters per second, up.

The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (0.00083/9.8) = 0.000085 second.  Then, after is starts to fall, it takes the same amount of time to the person's hand.

Total time = 0.00017 second.

This is why all of your numbers always need their units.

Answer:

Beat frequency together = 2.95 Hz (Approx)

Explanation:

Given:

Frequency (F) = 294 H

Decrease in tension = 2%

Find:

Beat frequency together

Computation:

Tension = (100 - 2) / 100

Tension (T) = 0.98

Beat frequency together = Frequency (F) - (√T × F)

Beat frequency together = 294 - (√0.98 × 294)

Beat frequency together = 2.95 Hz (Approx)

The beat frequency heard when the two strings are played together is 2.95 Hz.

Given data:

The tuning frequency of the violin is, f = 294 Hz.

Decrement in the tension is, 2 %.

Since, tension is reduced at the rate of 2%. Then the new magnitude of tension on the string is,

T = (100 - 2 )/100

T = 0.98

Then the expression for the beat frequency heard when the two strings are played together is given as,

[tex]f_{b}=f -(\sqrt{T \times f})[/tex]

Solving as,

[tex]f_{b}=294-(\sqrt{0.98 \times 294})\\\\f_{b}=2.95\;\rm Hz[/tex]

Thus, we can conclude that the beat frequency heard when the two strings are played together is 2.95 Hz.

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did you ever get the answer?

Answer:

8.7

Explanation:

A=2.4m/s2                        1.5m/s + (2.4m/s2)(3s)

Vi=1.5m/s                               1.5m/s + 7.2

Vf=?                                           Vf= 8.7m/s

T=3s

Answer:

The answer is "26.3 m".

Explanation:

The positive value from the x-axis is to the direction east side

The negative value from the x-axis is to the direction west side

The positive value from the y-axis is to the direction upwards side

The negative value from the y-axis is to the direction down words side

The positive value from the z-axis is to the direction southside

The negative value from the z-axis is to the direction north side

If the value is i, j, and k are the unit of the given vectors, which can be defined as follows:

[tex]\hat i, \hat -i, \hat j, \hat -j, \hat k, \hat -k[/tex]

The displacements values:

[tex]\underset{d_1}{\rightarrow} = -14 \ \hat i\\\\\underset{d_2}{\rightarrow} = 22\ \hat j\\\\\underset{d_3}{\rightarrow} = -12 \ \hat k\\\\\underset{d_4}{\rightarrow} = 6 \ \hat i\\\\[/tex]

calculating the final displacement that is  [tex]\underset{d_5}{\rightarrow}[/tex]:

[tex]\Rightarrow \underset{d_1}{\rightarrow}+\underset{d_2}{\rightarrow}+\underset{d_3}{\rightarrow}+\underset{d_4}{\rightarrow}+\underset{d_5}{\rightarrow} =0\\\\\Rightarrow \underset{d_5}{\rightarrow} \ \ = -(\underset{d_1}{\rightarrow}+\underset{d_2}{\rightarrow}+\underset{d_3}{\rightarrow}+\underset{d_4}{\rightarrow})[/tex]

         [tex]=- (-14 \hat i+ 22 \hat j -12 \hat k+ 6 \hat i)\\\\=- (-8 \hat i+ 22 \hat j -12 \hat k)\\\\=8 \hat i- 22 \hat j +12 \hat k\\[/tex]

[tex]|\underset{d_5}{\rightarrow}|=\sqrt{8^2 +(- 22)^2 +(12)^2 }\\\\[/tex]

       [tex]= \sqrt{64 + 484 +144 }\\\\= \sqrt{208 + 484}\\\\= \sqrt{692}\\\\= 26.3 \ m[/tex]

Answer:

it is A  or 20x80 cm

Explanation: DID IT ON APEX

WOO

1.5 microseconds means . . .

-- 1.5 millionths of a second

-- 1.5 x 10⁻⁶ second

-- 0.0000015 second

Answer:

The  value is  [tex]u_1 =  236 \ m/s[/tex]

Explanation:

From the question we are told that

   The  mass of bullet  is  [tex]m_b  =  0.0125 \  kg[/tex]

   The  mass of the block is  [tex]M_B  =  0.240 \  kg[/tex]

    The  spring constant is  [tex]k  =  2.25*10^{3} \  N/m[/tex]

   The  amplitude is  [tex]A= 12.4 \ cm  =  0.124 \ m[/tex]

Generally according to the conservation of momentum is  

     [tex]m_b u_1 + M_B  u_2 =  (m_b + M_B) v[/tex]

given that the block was at rest we have that

        [tex]m_b u_1  =  (m_b + M_B) v[/tex]

Now the angular velocity of the both bodies is mathematically represented as

     [tex]w =  \sqrt{\frac{k}{M_B  + m_b} }[/tex]

          [tex]w = \sqrt{\frac{ 2.25*10^{3}}{ 0.0125 + 0.240 } }[/tex]

         [tex]w =  94.4 \  rad/s[/tex]

Given that the system after collision set into oscillation

The maximum  linear velocity of the system after impact  is mathematically represented as

       [tex]v  = A * w[/tex]

      [tex]v  =  94.4 *0.124[/tex]

      [tex]v  =  11.7 \  m/s[/tex]

From above equation

    [tex]u_1  =  \frac{(m_b +  M_B ) * v}{m_b}[/tex]

=>    [tex]u_1 =  \frac{(0.240 + 0.0125) *  11.7}{ 0.0125}[/tex]

=>   [tex]u_1 =  236 \ m/s[/tex]

Answer:

Car 2 is travelling in a much higher speed than Car 1.But they are travelling or meeting in the same acceleration