Hanna is fascinated by the development of factories and the history of mass production. Which type of society BEST fits Hanna's interests?

A. hunting and gathering society

B. industrial society

C. agrarian society

D. horticultural society

Answer:

a) [tex]v=1*1.25=1.25\: m/s[/tex]

b) [tex]y(x,t)=2sin(2\pi( x-1.25 t)[/tex]  

c) [tex]y(3,10)=1.73 m[/tex]    

Explanation:

a) The speed of a wave is given by the following equation:

[tex]v=\lambda f[/tex]

Where:

λ is the wavelength

f is the frequency

[tex]v=1*1.25=1.25\: m/s[/tex]

b) The harmonic wave has the following equation:

[tex]y=Asin(kx-\omega t)[/tex]

A is the amplitude (2 m)

k is the wavenumber (2π/λ)  

ω is the angular frequency (2πf)

[tex]y(x,t)=2sin(2\pi x-2\pi*1.25 t)[/tex]  

[tex]y(x,t)=2sin(2\pi( x-1.25 t)[/tex]  

c) Here we need to find the heigth at x=3 m and t =10 s, so we need to find y(3,10).

[tex]y(3,10)=2sin(2\pi(3-1.25*10)[/tex]

[tex]y(3,10)=2sin(2\pi(3-1.25*10)[/tex]              

[tex]y(3,10)=1.73 m[/tex]              

I hope it helps you!

Answer:

    [tex]\frac{h_{liquid} }{ h_{body} }[/tex] = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          [tex]\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }[/tex]

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          [tex]\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }[/tex]

we substitute

           5/9 = [tex]\frac{h_{liquid} }{ h_{body} }[/tex]

Answer:

Are you talking about aurora borealis?

Explanation:

Answer:

W = 9800  N

Explanation:

Given that,

Mass of a car, m = 1000 kg

Acceleration of the car, a = 3 m/s²

We need to find the weight of the car. Weight of an object is given by the product of mass and acceleration due to gravity on the Earth.

W = mg

Put all the values,

W = 1000 kg × 9.8 m/s²

= 9800  N

So, the weight of the car is 9800  N.

Answer:

85 m/s

Explanation:

The formula for momentum is product of weight and velocity

p=mv  where m is mass and v is velocity

Given that;

Momentum = 153 kg*m/s

Weight = 1.8 kg

Velocity = ?

p=mv

153 = 1.8 * v

153/1.8 = v

85 m/s

Answer:

R₂ = V₂R₁/(E + V₂)

Explanation:

By voltage divider, V₂ = ER₂/(R₁ + R₂)

So, cross-multiplying, we have

V₂(R₁ + R₂) = ER₂

expanding the bracket, we have

V₂R₁ + V₂R₂ = ER₂

collecting like terms, we have

V₂R₁ = ER₂ + V₂R₂

factorizing, we have

V₂R₁ = (E + V₂)R₂

dividing through by (E + V₂), we have

R₂ = V₂R₁/(E + V₂)

Answer:

A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.

W = F•x cos ∅

If ∅ = 0°,

W = F•x ===> Maximum Work Done.

If ∅ = 45°,

W = F•x/√2

If ∅ = 90°,

W = 0

If ∅ = 180°,

W = –F•x ===> Minimum Work Done.

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5  x v + 11 x 15

where v is velocity of A after the collision .

5  x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

Answer:

22.48m

Explanation:

Given that the pool deck platform is 4 m above the ground.

The initial velocity of mortar in the upward direction, u=21 m/s

At the maximum height, the velocity of the mortar will become zero.

So, the final velocity, v=0

Acceleration due to gravity, g = 9.81 m/[tex]s^2[/tex] in the downward direction.

By using the equation of motion [tex]v^2=u^2+2as\\[/tex]

On putting all the values, we have

[tex]0^2=21^2+2(-9.81)s\\\\s=21^2/(2\times 9.81)[/tex]

s= 22.48 m

Hence, the mortar will reach a maximum height of 22.48m.

Answer:

Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons

Explanation:

From Plato. Hope this helps!

Answer:

Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons

Explanation:

Edmentum

Answer:

[tex]93\:\mathrm{kg}[/tex]

Explanation:

We can use Newton's Universal Law of Gravitation to solve this:

[tex]F=G\frac{m_1m_2}{r^2}[/tex], where [tex]F[/tex] is force, [tex]G[/tex] is gravitational constant [tex]6.67\cdot 10^{-11}[/tex], [tex]m_1[/tex]and [tex]m_2[/tex] represent the masses of both objects, and [tex]r[/tex] represents the distance between their center of masses.

Plugging in our given values, we have:

[tex]5.2\cdot 10^{-8}=6.67\cdot 10^{-11}\cdot \frac{97\cdot m_2}{3.4^2},\\m_2=92.9=\fbox{$93\:\mathrm{kg}$}[/tex](two significant figures).

Answer:

the answer is 93kg

Explanation:

the answer is 93kg

Answer:

Explanation:

Just use the Force formula.

F = M . A

Acceleration Formula

A = V - Vo / Time

So...

F = 845 . (30 - 2 / 0.9)

F = 845 . 20

F = 16900 N

Answer:

I believe it is A. The human body would not function normally without essential body fat.

Explanation:

Answer:

Cccccccccccc

Explanation:

Answer:

Resistance of A is [tex]6\ \Omega[/tex] and B is [tex]3\ \Omega[/tex]

Explanation:

The voltage across both the resistances will be the same as they are connected in parallel.

V = Voltage = 6 V

[tex]I_B=2\ \text{A}[/tex]

Resistance is given by

[tex]R_B=\dfrac{V}{I_B}\\\Rightarrow R_B=\dfrac{6}{2}\\\Rightarrow R_B=3\ \Omega[/tex]

[tex]V_B=V_b-V_A\\\Rightarrow V_B=6-4\\\Rightarrow V_B=2\ \text{V}[/tex]

Series connection

[tex]V_A=4\ \text{V}[/tex]

The current is constant in series connection

[tex]I=\dfrac{V_B}{R_B}\\\Rightarrow I=\dfrac{2}{3}\ \text{A}[/tex]

[tex]R_A=\dfrac{V_A}{I}\\\Rightarrow R_A=\dfrac{4}{\dfrac{2}{3}}\\\Rightarrow R_A=6\ \Omega[/tex]

The resistance of A is [tex]6\ \Omega[/tex] and B is [tex]3\ \Omega[/tex].

Answer:

16 times as strong

Explanation:

From the question given above, the following assumptions were made:

Initial Force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = ¼r

Final force (F₂) =?

Next, we shall obtain a relationship between the force and the distance apart. This can be obtained as follow:

F = GM₁M₂ / r²

Cross multiply

Fr² = GM₁M₂

If G, M₁ and M₂ are kept constant, then,

F₁r₁² = F₂r₂²

Finally, we determine the new force as follow:

Initial Force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = ¼r

Final force (F₂) =?

Fr² = F₂ × (¼r)²

Fr² = F₂ × r²/16

Fr² = F₂r² / 16

Cross multiply

16Fr² = F₂r²

Divide both side by r²

F₂ = 16Fr² / r²

F₂ = 16F

From the calculations made above, we can see that the new force is 16 times the original force.

Thus, the new force is 16 times stronger.

Answer:

2.45 J

Explanation:

The following data were obtained from the question:

Mass (m) = 0.5 kg

Height (h) = 1 m

Kinetic energy (KE) =?

Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 1/2 = 0.5 m

Final velocity (v) =?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 0.5)

v² = 9.8

Take the square root of both side

v = √9.8

v = 3.13 m/s

Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:

Mass (m) = 0.5 kg

Velocity (v) = 3.13 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 3.13²

KE = 0.25 × 9.8

KE = 2.45 J

Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J

Answer:

W= 702 J

Explanation:

Given that,

A weight lifter lifts a 360-N set of weights from ground level to a position over his head, a vertical distance of 1.95 m.

We need to find the work the weight lifter do. Work done by an object is given by the formula as follows :

W = Fd

Putting all the values,

W = 360 N × 1.95 m

= 702 J

So, the required work done is 702 J.

Answer:

First, let's write the movement equations for this ball.

The only force acting on the ball will be the gravitational acceleration (because we ignore the air resistance) then the acceleration equation is:

a(t) = -9.8m/s^2

Where the minus sign is because the ball is falling down.

To get the velocity of the ball, we need to integrate over time to get:

v(t) = -(9.8m/s^2)*t + v0

Where v0 is the initial velocity of the ball. Because it falls from rest, we can conclude that the initial velocity is 0 m/s, then the velocity equation is:

v(t) = -(9.8m/s^2)*t

For the position equation we need to integrate again, here we get:

p(t) = -(1/2)*(9.8m/s^2)*t^2 + p0

Where p0 is the initial position. In this cse we know that the ball falls from a height of 3m, then po = 3m

The position equation is:

p(t) = -(1/2)*(9.8m/s^2)*t^2 + 3m

The ball will hit the ground when p(t) = 0m, then we need to solve the equation:

p(t) = -(1/2)*(9.8m/s^2)*t^2 + 3m = 0m

for t.

-(1/2)*(9.8m/s^2)*t^2 + 3m = 0m

3m = (1/2)*(9.8m/s^2)*t^2

3m*2 = (9.8m/s^2)*t^2

6m/(9.8m/s^2) = t^2

√(6m/(9.8m/s^2)) = t = 0.78s

The ball needs 0.78 seconds to hit the ground.

The correct answer is (d) the weight of steel ball is much larger than air resistance.

since the density of steel ball is quite higher than that of air. The weight of even a small steel ball will be much larger than the air resistance acting opposite to the motion of the steel ball. Hence in the case of a freely falling steel ball the air resistance can be neglected.

Learn more about air resistance:

https://brainly.com/question/2575108

Answer:

3.72 N / kg .

Explanation:

The weight of the object can be calculated using the expression below

Weight = mg

Weight= 785 Newtons

M = mass

g= acceleration due to gravity on Earth is 9.8 m/s² .

So, substitute the values we have,

So on Earth . . .

785 N = m x 9.8

mass= 785/9.8

=80.1kg

We can calculate when on On Mars, as

Weight= mg

298N= 80.1 kg x g( on Mars)

Acceleration of gravity of that Mars =

298 N/ 80.1

= 3.72 N / kg .

Hence, the magnitude of the gravitational acceleration on the surface of Pluto is 3.72 N / kg .

Answer:

the answer is 72.0 units:) thank me later!!!!!!!

Answer:

Wmoon = 131 [N]

Explanation:

We know that the weight of a body is equal to the product of mass by gravitational acceleration.

Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.

[tex]w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N][/tex]

The object would have a centripetal acceleration a of

a = (12 m/s)² / (0.7 m) ≈ 205.714 m/s²

so that the required tension in the string would be

T = (0.5 kg) a ≈ 102.857 N ≈ 100 N

(rounding to 1 significant digit)

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

The total energy of the composite system at any time after the collision is 4.7334 J and initial velocity of the bullet is 225.4 m/s.1`

Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.

The mass of the bullet is 30g and the mass of the wood block is 4.8 kg. Thus the total mass of these two is,

[tex]m=0.03+4.8\\m=4.83 \rm kg[/tex]

As the compound system of the block plus the bullet rises to a height of 10 cm. Thus, the speed of it is,

[tex]v=\sqrt{2gh}\\v=\sqrt{2\times9.81\times0.1}\\v=1.4\rm m/s[/tex]

Now, the total energy of the composite system at any time after the collision is equal to the kinetic energy. Therefore,

[tex]E=\dfrac{1}{2}mv^2\\E=\dfrac{1}{2}4.83(1.4)^2\\E=4.7334\rm J[/tex]

Total energy of the composite system at any time after the collision is 4.7334 J.

The block was at rest initial, thus it has initial velocity of it is zero. Thus, the initial momentum of the bullet is equal to the final momentum of bullet and block. Therefore,

[tex]0.03\times u_o=4.83\times1.4\\u_o=225.4\rm m/s[/tex]

Thus, the total energy of the composite system at any time after the collision is 4.7334 J and initial velocity of the bullet is 225.4 m/s.1`

Learn more about the conservation of momentum here;

https://brainly.com/question/7538238

Answer:

ozone layer

Explanation:

The ozone layer forms a thin shield in the upper atmosphere protecting life on earth from UV rays. Ozone is a naturally occurring gas that is found in two layers of the atmosphere

Answer:

a) Average speed = 200 Km/hr

b) Average velocity = 0 m/s

c) Total distance = 800 Km

d) Total displacement = 0 Km

Explanation:

a) Let the speed of the plane from Addis Ababa to Gamble be represented by [tex]S_{1}[/tex], and from Gamble to Addis Ababa by [tex]S_{2}[/tex].

Average speed = [tex]\frac{S_{1} + S_{2} }{2}[/tex]

[tex]S_{1}[/tex] = [tex]\frac{distance}{time}[/tex] = [tex]\frac{400}{2}[/tex]

        = 200 Km/hr

[tex]S_{2}[/tex] = [tex]\frac{distance}{time}[/tex] = [tex]\frac{200}{1}[/tex]

       = 200 Km/hr

Average speed = [tex]\frac{200 + 200}{2}[/tex]

        = 200 Km/hr

b) Velocity = [tex]\frac{displacement}{time}[/tex]

Average velocity = [tex]\frac{total displacement}{total time taken}[/tex]

Since the plane flies from Addis Ababa to Gamble, then back to Addis Ababa, its total displacement is zero.

So that,

Average velocity = 0 m/s

c) Total distance = 400 Km + 400 Km

                           = 800 Km

d) Total displacement = 400 Km + (-400 Km)

                           = 0 Km

Answer: the answer is a

Explanation:

Answer:

v = 0.41 m/s

Explanation:

       [tex]\Delta K + \Delta U = W_{ffr} (1)[/tex]

       [tex]\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)[/tex]

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

       [tex]W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)[/tex]

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

        [tex]\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)[/tex]

        [tex]\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)[/tex]

    [tex]\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)[/tex]