Which best describes the direction of heat?

Answer:

0.06563 N/m

The percentage difference between the calculated and tabulated value of water surface tension is 9.78%

Explanation:

Solution:-

- Surface tension is the ability of any fluid to resist any external force due to cohesive nature of the fluid molecules.

- Surface tension ( γ ) is defined as the force imparted per unit length on the fluid.

                          γ = F / L

Where,

             F: Force

             L: The length over which force is felt

- The mass ( M = 3.78g ) of ( n = 100 ) water droplets are carefully burette. We need to determine the mass of a single droplet as follows:

                        m = M / n

                        m = 3.78 / 100

                        m = 0.0378 g        

- Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity ( Weight of the droplet ):

                       F = m*g

                       F = 0.0378*9.81*10^-3

                       F = 0.000370818 N      

- The length over which the force is felt can be modeled into a circular patch with diameter ( d ) = equal to the diameter ( d ) of the single water drop. The length ( L ) is defined as the circumference of the patch:

                       L = π*d

                       L = π*( 0.0018 )

                       L = 0.00565 m

- The surface tension would be:

                       γ = F / L

                       γ = 0.000370818 / 0.00565

                       γ = 0.06563 N / m

- The given value of water's surface tension is given as follows:

                      γa = 0.07275 N/m

- To compare the two values we will determine the percentage difference between the value evaluated  and tabulated value as follows:

                   % [tex]diff = \frac{0.07275- 0.06563}{0.07275} * 100 \\\\diff = 9.78 \\\\[/tex]

- The percentage difference ( 9.78% ) between the two values is within  practical limits of 10%. Hence, the calculated can be accepted.

Hello!

The answer is Gravity. Lets take a look at the options:

A. Nope, this would be if the shuttle was going horizontal.

B. Again, this would be if the shuttle was going horizontal.

C. Yes! This is the only force that is being overcome by using the rocket thrusts to move up.

D. No, the shuttle is using its thrust rocks to push against gravity.

I hope this helps! :)

Answer: oh i actually dont know

Explanation:

But i wish i can help SORRY!

Answer:

hi

Explanation:

hi

Answer: The speed is 15.5 m/s

Explanation:

The kinetic energy can be written as:

K = (1/2)*m*v^2

where m is the mass and v is the speed.

Then we have that:

18 j = (1/2)*0.15kg*v^2

Now we solve this for v.

√(18*2/0.15) = v

15.5 m/s = v

Answer:

6.5 ly is equal to 6.15 x 10¹⁶ m

Explanation:

Light year is the unit of length. It can easily be calculated by multiplying the speed of light in vacuum withe no. of seconds in an year. Let us calculate the value of 1 light year in meters.

1 Light Year = (Speed of Light)(Seconds in 1 Year)

1 Light Year = (3 x 10⁸ m/s)(1 year)(365 days/year)(24 h/day)(3600 s/h)

1 Light Year = 9.46 x 10¹⁵ m

Hence, to get the value of 6.5 Light years, in meters, we will multiply both sides by 6.5.

6.5 Light Years = (6.5)(9.46 x 10¹⁵ m)

6.5 Light Years = 6.15 x 10¹⁶ m

OR

6.5 ly =  6.15 x 10¹⁶ m

Therefore, there are 6.15 x 10¹⁶ m in 6.5 ly

Answer:

(a) μ ≈ 0.57

(b) not enough information

Explanation:

Draw a free body diagram.

There are four forces acting on block B:

There are three forces acting on the node:

(a)

Sum of forces on block B in the y direction:

∑F = ma

N − Mb g = 0

N = Mb g

Sum of forces on block B in the x direction:

∑F = ma

T₁ − Nμ = 0

T₁ = Nμ

T₁ = Mb g μ

Sum of forces on the node in the y direction:

∑F = ma

T₂ sin 35° − Ma g = 0

T₂ sin 35° = Ma g

Sum of forces on the node in the x direction:

∑F = ma

T₂ cos 35° − T₁ = 0

T₂ cos 35° = T₁

T₂ cos 35° = Mb g μ

Divide the previous equation by this equation, eliminating T₂.

tan 35° = Ma g / (Mb g μ)

μ = Ma / (Mb tan 35°)

μ = 0.4 / tan 35°

μ ≈ 0.57

(b) T₂ sin 35° = Ma g = 0.4 Mb g

Without knowing the value of Mb, we cannot find the value of the tension force T₂.

Answer:

Displacement, force, momentum, and velocity are all vectors

Explanation:

Vectors are quantities that show direction; which all of the above do. The rest of the terms are scalar quantities

Answer:

Target Marketing

Explanation:

The Marketing Concept is one of five concepts adopted by organizations when marketing their products to customers. The other four are the Production, Selling, Products, and Social Marketing concepts. The Marketing concept aims at satisfying the needs of the buyers through their products. This approach employs targeted marketing to gain competitive advantage over other firms.

Target Marketing would help Jodi sell colorful ribbons specifically to dancers. Jodi has a target market. These are the Dancers. So, after setting up her booth, Jodi would do well to concentrate her efforts on the dancers whom she hopes to sell the ribbons to. This would enable her establish competitive advantage over other sellers.

Answer:

angular acceleration = 1.67 rad/s²

Explanation:

given data

door wide = 1 m

initially ope angle = 30°

push force = 20 N

rotational inertia = 3.0 kg m²

solution

we apply force at middle so length will be here r1 = [tex]\frac{1}{2}[/tex]  = 0.5 m

and

now we get here torque that is express as

torque τ = Force × r1 × sin30   ......................1

put her value and we get

torque τ = 20 × 0.5 × sin30

torque τ = 5 Nm

and  we know

torque = rotational inertia × angular acceleration   .......................2

put her value and we get angular acceleration

angular acceleration = [tex]\frac{5}{3}[/tex]  

angular acceleration = 1.67 rad/s²

Answer:

Explanation:18kt alloy contains

i) 75% of gold

rhogold=19.3g/cm^3

=75/100×19.3

=14.475g/cm^3

ii) 16% of silver

rhosilver=10.5g/cm^3

=16/100×10.5

=1.68g/cm^3

iii) 9% of copper

rhocopper =8.90g/cm^3

=9/100×8.9

=0.801g/cm^3

Overall density of 18kt gold

=(0.801+1.68+14.475)g/cm^3

=16.956g/cm^3

=17g/cm^3 to 3s.f

Answer:

75mL

...............

The question is full of inconsistencies and conceptual errors. Students should not waste their time trying to answer it ... it will only mislead and confuse them.

The question is poor. It was written by someone unclear on the concepts. It should be ignored.

Answer:

Explanation:

Apply the law of conservation of energy

[tex]KE_i+PE_i=KE_f+PE_f[/tex]

[tex]Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)[/tex]

from the law of conservation of the linear momentum

[tex]m_1v_1=m_2v_2[/tex]

Therefore,

[tex]Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)[/tex]

[tex]=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ][/tex]

[tex]v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ][/tex]

Substitute the values in the above result

[tex]v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ][/tex]

[tex]=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s[/tex]

B)  the speed of the sphere with mass 107.0 kg is

[tex]v_2=\frac{m_1v_1}{m_2}[/tex]

[tex]=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s[/tex]

C)  the magnitude of the relative velocity with which one sphere is

[tex]v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s[/tex]

D) the distance of the centre is proportional to the acceleration

[tex]\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962[/tex]

Thus,

[tex]x_1=3.962x_2[/tex]

and

[tex]x_2=0.252x_1[/tex]

When the sphere make contact with eachother

Therefore,

[tex]x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r[/tex]

And

[tex]x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r[/tex]

The point of contact of the sphere is

[tex]32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m[/tex]

(A) The speed of the sphere of 27 kg is  [tex]1.2664 \times 10^{-5} \;\rm m/s[/tex].

(B)  The speed of sphere of mass 107 kg is  [tex]3.195 \times 10^{-6} \;\rm m/s[/tex].

(C)  The magnitude of the relative velocity with which one sphere is approaching to the other is  [tex]15.85 \times 10^{-6} \;\rm m/s[/tex].

(D)  The distance from the initial position of the center of the 27.0 sphere is 20.506 m.

Given data:

The mass of sphere 1 is,  [tex]m_{1} = 27.0 \;\rm kg[/tex].

The mass of sphere 2 is, [tex]m_{2} = 107.0 \;\rm kg[/tex].

The radius of each spheres are, r = 0.10 m.

The distance between the centers of each sphere is, d = 41.0 m.

(A)

In this part, we can apply the conservation of energy to find the speed at given distance of 26.0 m. So,

Total energy at initial = Total energy at final

[tex]\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{1}{2}(m_{1}v^{2}_{1}+m_{2}v^{2}_{2})[/tex]

Now, as per the conservation of momentum,

[tex]m_{1}v_{1}=m_{2}v_{2}\\\\v_{2}=\dfrac{m_{1}v_{1}}{m_{2}}[/tex]

Therefore,

[tex]\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{1}{2}(m_{1}v^{2}_{1}+m_{2} \times [m_{1}v_{1}/m_{2}]^{2})\\\\\\\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{m_{1}v^{2}_{1}}{2} \times (\dfrac{m_{1}+m_{2}}{m_{2}})[/tex]

Modifying as,

[tex]v^{2}_{1}=[\dfrac{2Gm^{2}_{2}}{m_{1}+m_{2}}] \times [\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}}][/tex]

Substitute the values in the above result

[tex]v^{2}_{1}=[\dfrac{2 \times 6.67 \times 10^{-11} \times 107^{2}}{27+107}] \times [\dfrac{1}{26}-\dfrac{1}{41}]\\\\v_{1}=\sqrt{1.6038 \times 10^{-5}}\\\\v_{1}=1.2664 \times 10^{-5} \;\rm m/s[/tex]

Thus, the speed of the sphere of 27 kg is  [tex]1.2664 \times 10^{-5} \;\rm m/s[/tex].

(B)

The sphere with mass 107 kg is calculated as,

[tex]v_{2}=\dfrac{m_{1}v_{1}}{m_{2}}\\\\v_{2}=\dfrac{27 \times 1.2664 \times 10^{-5} \;\rm m/s }{107}\\\\v_{2}=3.195 \times 10^{-6} \;\rm m/s[/tex]

Thus, the speed of sphere of mass 107 kg is  [tex]3.195 \times 10^{-6} \;\rm m/s[/tex].

(C)

The magnitude of the relative velocity with which one sphere is

[tex]v_{r} = v_{1} +v_{2}\\\\v_{r} =( 1.2664 \times 10^{-5})+ (3.195 \times 10^{-6})\\\\v_{r}=15.85 \times 10^{-6} \;\rm m/s[/tex]

Thus,  the magnitude of the relative velocity with which one sphere is approaching to the other is [tex]15.85 \times 10^{-6} \;\rm m/s[/tex].

(D)

The distance of the centre is proportional to the acceleration,

[tex]\dfrac{x_{1}}{x_{2}}=\dfrac{m_{2}}{m_{1}}\\\\\\\dfrac{x_{1}}{x_{2}}=\dfrac{107}{27}\\\\\\\dfrac{x_{1}}{x_{2}}=3.962\\\\\\x_{1}=3.962 \times x_{2}[/tex]

As per the given problem,

[tex]x_{1}+x_{2}+2R = d\\\\3.962x_{2}+x_{2}+(2R) = 41 \\\\x_{2} = 8.262-0.403R[/tex]

And,

[tex]x_{1}=0.252x_{2}\\\\x_{1}=0.252 \times (8.262-0.403R)\\\\x_{1}=32.747-1.597R[/tex]

Then for point of contact of the sphere:

[tex]x_{1} = x_{2}\\\\32.747-1.597R = 8.262-0.403R\\\\R =20.506 \;\rm m[/tex]

Thus, the distance from the initial position of the center of the 27.0 sphere is 20.506 m.

Learn more about the conservation of linear momentum here:

from the initial position of the center of the 27.0 sphere

Answer:

Explanation:

Energy stored in a capacitor = 1/2 C V² where , C is capacitance and V is potential of capacitor.

Putting the values

.5 x C x 12² = .0163

C = 226.4 x 10⁻⁶ F .

Frequency of L-C circuit = [tex]\frac{1}{2\pi}\sqrt{\frac{1}{LC} }[/tex]

where L is inductance of inductor and C is capacitance of capacitor.

putting the values

3585 = [tex]\frac{1}{2\pi}\sqrt{\frac{1}{L\times226.4\times10^{-6}} }[/tex]

506.87 x 10⁶ = [tex]\frac{1}{L\times226.4\times10^{-6}}[/tex]

[tex]L = \frac{1}{114755.37}[/tex]

= 8.7 x 10⁻⁶ H.

Answer:

Explanation:

If the dragster  attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .

So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .

v = u + a t

v₀ = 0 + a tmax

tmax = v₀ / a

The value of tmax is v₀ / a .

Answer:

More than 20% of the amazon has been destroyed, affects biodiversity.

Explanation:

Answer:

2.5 N (Centrifugal direction)

Explanation:

According to the Newton's Third Law, there is a reaction when an action exist. The reaction force has the same magnitude but opposite direction. Therefore,  the force that apple exerts on the Earth has a magnitude of 2.5 N and a centrifugal direction.

Answer:

1a) The direction to throw the boot is directly away from the closest shore.

2b) The magnitude of the force that the thrown boot exerts on the engineer = 391 N

3c) Time taken to reach shore = 8.414 s

Explanation:

1a) Newton's third law of motion explains that for every action, there is an equal and opposite reaction.

The force generated by throwing the boot in one direction is exerted back on the engineer as recoil in the opposite direction.

Hence, the best direction to throw the boot is opposite the direction that the engineer intends to move towards.

2b) Just as explained in (1a) above, the force exerted in one direction always has a reaction of the same magnitude in the opposite direction.

Hence, the force exerted by the boot on the engineer is equal to the force exerted by the engineer on the boot = 391 N.

3c) For this part, we analyze the total motion of the engineer.

The force exerted by the boot on the engineer initially accelerates the engineer until the engineer reaches a constant velocity dictated the impulse of the initial force (since impulse is equal to change in momentum), this constant velocity then takes the engineer all the way to shore, since the ice surface is frictionless.

The weight of the engineer = W = 588 N

W = mg

Mass of the engineer = (W/g) = (588/9.8) = 60 kg

Force exerted on the engineer by the thrown boot = F = 391 N

F = ma

Initial acceleration of the engineer = (F/m) = (391/60) = 6.52 m/s²

We can then calculate the distance covered during this acceleration

X₁ = ut + ½at₁²

u = initial velocity of the engineer = 0 m/s (the engineer was initially at rest)

t₁ = time during which the force acts = 0.576 s

a = acceleration during this period = 6.52 m/s²

X₁ = 0 + 0.5×6.52×0.576² = 1.08 m

For the second part of the engineer's motion, the velocity becomes constant.

So, we first calculate this constant velocity

Impulse = Change in momentum

F×t = mv - mu

F = Force causing motion = 391 N

t = time during which the force acts = 0.576 s

m = mass of the engineer = 60 kg

v = final constant velocity of the engineer = ?

u = initial velocity of the engineer = 0 m/s

391 × 0.576 = 60v

v = (391×0.576/60) = 3.7536 m/s.

The distance from the engineer's initial position to shore is given as 30.5 m

The engineer covers 1.08 m during the time the force causing motion was acting.

The remaining distance = X₂ = 30.5 - 1.08 = 29.42 m

We can then calculate the time taken to cover the remaining distance, 29.42 m at constant velocity of 3.7536 m/s

X₂ = vt₂

t₂ = (X₂/v) = (29.42/3.7536) = 7.838 s

Time taken to reach shore = t₁ + t₂ = 0.576 + 7.838 = 8.414 s

Hope this Helps!!!

Answer:

B) Aerosol

The particles from the pigeon droppings are likely to be transmitted from the air vents directly into Martyn's respiratory system

Answer:

aerosol

Explanation:

Answer:

a) 103.176 m / min

b) 1751.28 meters

Explanation:

Given:-

- Emma's and Lily's velocities ( E(t) and L(t) ) are given as functions respectively:

                         [tex]E(t) = \frac{7510}{t^2-7t + 80.22} \\\\L ( t ) = 12t^3*e^-^0^.^5^t[/tex]

- Where, E ( t ) and L ( t ) are given in m / min

- Both run for a total time of 15 minutes in the same direction along the straight track defined by the absolute interval:

                         ( 0 ≤ t ≤ 15 ) mins                  

- It is known that Emma is 10 meters ahead of Lily at time t = 0.

Find:-

a) Find the value of [tex]\frac{1}{6}*\int\limits^8_2 {E(t)} \, dt[/tex] using correct units, interpret the meaning of

b) What is the maximum distance between Emma and Lily over the time interval 0 ≤ t ≤ 15?  

Solution:-

- The average value of a function f ( x ) over an interval [ a , b ] is determined using calculus via the following relation:

                          [tex]f_a_v_g = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx[/tex]

- The first part of the question is asking us to determine the average velocity of Emma over the time interval of ( 2 , 8 ). Therefore, ( E_avg ) can be determined using the above relation:

                         [tex]E_a_v_g = \frac{1}{8 - 2}*\int\limits^8_2 {E(t)} \, dt\\\\E_a_v_g = \frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\\[/tex]

- We will evaluate the integral formulation above to determine Emma's average velocity over the 2 ≤ t ≤ 8 minute range:

                         [tex]E_a_v_g = \frac{1}{6}*\int\limits^8_2 {\frac{7510}{t^2 - 7t + 80.22} } \, dt\\\\E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{50t^2 - 350t + 4011} } \, dt\\\\[/tex]

- Complete the square in the denominator:

                          [tex]E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{(5\sqrt{2}*t - \frac{35}{\sqrt{2} })^2 + \frac{6797}{2} } } \, dt\\\\[/tex]

- Use the following substitution:

                          [tex]u = \frac{5*(2t - 7 )}{\sqrt{6797} } \\\\\frac{du}{dt} = \frac{10}{\sqrt{6797} } \\\\dt = \frac{\sqrt{6797}}{10}.du[/tex]

- Substitute the relations for (u) and (dt) in the above E_avg expression.

                          [tex]E_a_v_g = \frac{1}{6}*37550\int {\frac{\sqrt{6797} }{5*(6797u^2 + 67997) } } \, du\\\\E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}} \int {\frac{1 }{(u^2 + 1) } } \, du[/tex]

- Use the following standard integral:

                          [tex]arctan(u) = \int {\frac{1}{u^2 + 1} } \, du[/tex]

- Evaluate:

                         [tex]E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}}* arctan ( u ) |[/tex]

- Apply back substitution for ( u ):

                        [tex]E_a_v_g = \frac{1}{6}*[\frac{75100* arctan ( \frac{5*(16 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } - \frac{75100* arctan ( \frac{5*(4 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } ]\\\\[/tex]

- Plug in the limits and find Emma's average velocity:

                        [tex]E_a_v_g = 151.82037*[arctan (0.54582 ) - arctan ( -0.18194 ) ]\\\\E_a_v_g = 103.176 \frac{m}{min}[/tex]

Answer: Emma's average speed over the interval ( 2 ≤ t ≤ 8 ) is 103.179 meters per minute.

- The displacement S ( E ) of Emma from time t = 0 till time ( t ) over the absolute interval of  0≤t≤15 is given by the relation:

                    [tex]S (E) = S_o + \int\limits^t_0 {E(t)} \, dt\\\\S ( E ) = 10 + \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } |_0^t\\\\S ( E ) = 10 + [ \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - \frac{75100*arctan( \frac{5*(0 - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } ]\\\\S ( E ) = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } + 375.71098\\[/tex]

- The displacement S ( L ) of Lily from time t = 0 till time ( t ) over the absolute interval of  0 ≤ t ≤ 15 is given by the relation:

                    [tex]S (L) = \int\limits^t_0 {L(t)} \, dt\\\\S (L) = \int\limits^t_0 ({12t^3 *e^-^0^.^5^t } )\, .dt\\[/tex]

 Apply integration by parts:

  [tex]S ( L ) = 24t^3*e^-^0^.^5^t - 64*\int\limits^t_0 ({e^-^0^.^5^t*t^2} \,) dt\\[/tex]

 Re-apply integration by parts 2 more times:

 [tex]S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2\int\limits^t_0 ({e^-^0^.^5^t*t} \,) dt ]\\[/tex]             [tex]S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2*( -2t*e^-^0^.^5^t - (4e^-^0^.^5^t - 4 ) ]\\\\[/tex]    

[tex]S ( L ) = e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) - 512 \\[/tex]

- The distance between Emma and Lily over the time interval 0 < t < 15 mins can be determined by subtracting S ( L )  from S ( E ):

                    [tex]S = S ( E ) - S ( L )\\\\S = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) + 887.71098\\[/tex]

- The maximum distance ( S ) between Emma and Lily is governed by the critical value of S ( t ) for which function takes either a minima or maxima.

- To determine the critical values of the function S ( t ) we will take the first derivative of the function S with respect to t and set it to zero:

              [tex]\frac{dS}{dt} = \frac{d [ S(E) - S(L)]}{dt} \\\\\frac{dS}{dt} = E(t) - L(t) \\\\\frac{dS}{dt} = \frac{7510}{t^2 - 7t+80.22} - 12t^3*e^-^0^.^5^t = 0\\\\( 12t^5 - 84t^4 + 962.64t^3) *e^-^0^.^5^t - 7510 = 0\\\\t = 4.233 , 11.671[/tex]

- We will plug in each value of t and evaluate the displacement function S(t) for each critical value:

Answer:

a point representing the mean position of the matter in a body or system.

Explanation:

Answer:

It can easily be transformed from high voltages to low voltages

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

[tex]Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C[/tex]

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

[tex]\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J[/tex]

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

[tex]\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}[/tex]

hence, the speed of the spheres is 37.45 m/s

Answer:

3.16 m·s⁻¹ at an angle of 71.6°  

Explanation:

Assume that the diagram is like Fig. 1 below.

The boat is heading straight across the river and the current is directed straight downstream.

We have two vectors at right angles to each other.

1. Calculate the magnitude of the resultant

We can use the Pythagorean theorem (Fig. 2).

R² = (3 m·s⁻¹)² + (1 m·s⁻¹)² = 9 m²·s⁻² + 1 m²·s⁻² = 10 m²·s⁻²

R = √(10 m²·s⁻²) ≈ 3.16 m·s⁻¹

2. Calculate the direction of the resultant

The direction of the resultant is the counterclockwise angle (θ) that it makes with due East .

tanθ = opposite/adjacent = 3/1 = 3

θ = arctan 3 = 71.6°

To an observer at point O, the velocity of the boat is 3.16 m·s⁻¹ at an angle of 71.6°.

Answer:

Explanation:

Elasticity is the tendency of a material to regain its original shape and size after being deformed by an external force. Hooke's law of elasticity state that; provided the elastic limit of a material is not exceeded, the extension is proportional to the force applied.

The given graph shows the elastic property of the rabbit's tendon when a force (load) is applied.

At point 1 on the graph, a given value of load was applied to the tendon. At this point, Hooke's law is obeyed i.e the tendon supports the load.

At point 2, the value of the load was increased and tendon obeys Hooke's law. This implies that as the load is increased, the the tendon was able to support it.

At point 3, a further increase in the value of load causes the elastic limit of the tendon to be exceeded. This means that the tendon would not return to its original shape and size if the load is removed, Hooke's law is no more obeyed.

At point 4, the tendon breaks because it can no longer sustain any value of load.

Answer:

The precision will change by a factor of 1/4

Explanation:

Check attachment

Answer:

E=1/2CV^2

where E is potential energy, C is capacitance and V is the voltage.

E=1/2×125×9^2

E=1/2×125×81

E=5062.5

E=5100J to 2 significant figures.

Answer:

5.1

Explanation:

ed2020

Answer:

γ = 0.06563 N / m

9.78% difference

Explanation:

Solution:-

- Surface tension is the ability of any fluid to resist any external force which causes a decreases in surface area of the impact area due to inward compressive forces. These compressive forces occur due to cohesive nature of the fluid molecules.

- Mathematically, surface tension ( γ ) is defined as the force felt per unit length by the fluid.

                           γ = F / L

Where,

              F: Force imparted

              L: The length over which force is felt

- We are given the mass ( M ) of ( n = 100 ) water droplets to e 3.78 g. The mass of a single droplet ( m ) can be evaluated as follows:

                         m = M / n

                         m = 3.78 / 100

                        m = 0.0378 g        

- The force ( F ) imparted by a single drop of water from the burette can be determined from the force balance on a single droplet. Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity:

                        F = m*g

                        F = 0.0378*9.81*10^-3

                        F = 0.000370818 N      

- The length over which the force is felt can be magnified into a circular area with diameter equal to that of a single droplet ( d ). The circumferential length ( L ) of the droplet would be as follows:

                        L = π*d

                        L = π*( 0.0018 )

                        L = 0.00565 m

- Then the surface tension would be:

                        γ = F / L

                        γ = 0.000370818 / 0.00565

                        γ = 0.06563 N / m

- The tabulated value of water's surface tension is given as follows:

                       γa = 0.07275 N/m

- We will determine the percentage difference between the value evaluated  and tabulated value as follows:

                     [tex]p.diff = \frac{gamma_a - gamma}{gamma_a} * 100\\\\p.diff = \frac{0.07275- 0.06563}{0.07275} * 100 \\\\p.diff = 9.78 %[/tex]

- The %difference between is within the allowable practical limits of 10%. Hence, the evaluated value ( γ = 0.06563 N / m ) can be accepted with 9.78% error.

Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         Δ[tex]P_{y}[/tex] = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

         ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

          P = Pₓ / [tex]P_{y}[/tex]

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s