Humans living in highly populated areas are more inclined to
a. violence
b. socially isolate
C.appetite loss
d. all of the above

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:

No, the treatment won't work

Explanation:

Since the abscess is filled with gas, and is already putting pressure on the tooth nerve, applying moist heat will only aggravate the problem. The pressure of a given gas increases with temperature and so, the gas within the abscess will only put more pressure on the nerve due to the expansion of the gas within the abcess, causing her more discomfort.

Answer:

Use of telemetry and radar astronomy

Explanation:

An astronomical Unit (AU) is a unit of measuring distances in outer space, which is based on the approximate distance between the earth and the Sun.

After several years of trying to approximate the distance between the Sun and the Earth using several methods based on geometry and some other calculations, advancements in technology made available the presence of special motoring equipment, which can be placed in outer space to remotely monitor and measure the position of the sun.

The use of direct radar measurements to the sun (radar astronomy) have also made the determination of the AU more accurate.

A standard radar pulse of known speed is sent to the Sun, and the time with which it takes to return is measured,  once this is recorded, the distance between the Earth and the Sun can be calculated using

distance = speed X time.

However, most of these means have to be corrected for parallax errors

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:

Explanation:

From the formula

v² = u² + 2 a s , v is final velocity , u is initial velocity , a is acceleration and s is distance travelled .

u² / 4 = u² + 2 a x 19

- 3/4 u² = 2 a x 19

- (3 / 4) x 3.1²  = 2 x 19 x a

a = -  .18967 m /s²

deceleration due to friction = μg  where g is acceleration due to gravity and μ is coefficient of friction .

a =   μg

μ  =  a / g

=  .18967  / 9.8

= .019 .

Answer: oh i actually dont know

Explanation:

But i wish i can help SORRY!

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:

(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:

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:

D) Kinetic.

Explanation:

Kinetic energy is energy in motion.

Answer: D

Explanation: kinetic is also called energy of motion , it is defined at the work needed to accelerate a body of a given mass from rest to its stated velocity.

Answer:

C

Explanation:

because at the top there is the greatest energy stored(potential energy)

so its maximum, and minimum

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:

hi

Explanation:

hi

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:

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

Explanation:

Answer: Newton denoted using N

Examples: 25 N

Please rate 5 stars and vote as brainliest:)

(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)

Answer: transverse wave.

Explanation:

In a wave motion rope, the particles of the rope move in the perpendicular direction to the actual wave, so this is a transverse wave.  (we have a longitudinal wave when the movement of the particles is in the same direction than the wave propagation, an example of this is the waves in the surface of the water when you throw a rock in)

Now, the fact that he waved it only once does not mean that this is a pulse if the other end of the rope is connected to a fixed point when the wave reaches that point will be reflected (losing a bit of amplitude, but we still will have a wave)

so the correct option is a transverse wave.

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:

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:

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.

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:

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:

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:

In econ: When supply and Demand are equal

Explanation:

Equilibrium is when something evens out

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:

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

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: The density is greater at point C

Explanation: At that point

The convention current slowly cools off

Answer:

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

Explanation: