There has long been an interest in using the vast quantities of thermal energy in the oceans to run heat engines. A heat engine needs a temperature difference, a hot side and a cold side. Conveniently, the ocean surface waters are warmer than the deep ocean waters. Suppose you build a floating power plant in the tropics where the surface water temperature is ~ 35.0 C. This would be the hot reservoir of the engine. For the cold reservoir, water would be pumped up from the ocean bottom where it is always ~ 5.00 C.
What is the maximum possible efficiency of such a power plant?

Given that,

Length of bar = 600 mm

Diameter of bar = 40 mm

Diameter of hole = 30 mm

Length of hole = 100 mm

Modulus of elasticity = 85 GN/m²

Load = 180 kN

We need to calculate the area of cross section without hole

Using formula of area

[tex]A=\dfrac{\pi\times d^2}{4}[/tex]

Put the value into the formula

[tex]A=\dfrac{\pi\times40^2}{4}[/tex]

[tex]A=1256.6\ mm^2[/tex]

We need to calculate the area of cross section with hole

Using formula of area

[tex]A=\pi\times\dfrac{(d_{b}^2-d_{h}^{2})}{4}[/tex]

Put the value into the formula

[tex]A=\pi\times\dfrac{(40^2-30^2)}{4}[/tex]

[tex]A=549.77\ mm^2[/tex]

We need to calculate the total contraction on the bar

Using formula of total contraction

Total contraction = contraction in bar without hole part + contraction in bar with hole part

[tex]Total\ contraction = \dfrac{F\times L_{1}}{A_{1}\times E}+\dfrac{F\times L_{2}}{A_{2}\times E}[/tex]

Where, F = load

L = length

A = area of cross section

E = modulus of elasticity

Put the value into the formula

[tex]Total\ contraction=\dfrac{180\times10^3}{85\times10^{3}}(\dfrac{500}{1256.6}+\dfrac{100}{549.77})[/tex]

[tex]Total\ contraction = 1.227\ mm^2[/tex]

Hence, The total contraction on the bar is 1.227 mm²

Answer:

A = Tan{-1} d/h

Explanation

Let the angle Theta be A

From identity the angle A has an opposite side d and an adjacent h.

From trigonometry ratio

Tan A = d/h

A = Tan{-1} d/h

Answer:

  22.5°

Explanation:

Short answer: The trajectories will hit the same target when the projectile is launched at complementary angles. The second angle is 90° -67.5° = 22.5°.

__

Longer answer: The horizontal speed of the snowball launched at angle α with speed s is ...

  vh = s·cos(α)

Then the horizontal distance at time t is ...

  x = vh·t

and the time taken to get to some distance x is ...

  t = x/vh = x/(s·cos(α))

The equation for the vertical motion of the projectile is ...

  y = -4.9t² +s·sin(α)·t

Substituting the above expression for t, we have y in terms of x:

  y = -4.9x²/(s·cos(α))² +(s·sin(α)·x)/(s·cos(α))

Factoring gives ...

  y = (x/(cos(α))(-4.9x/(s²·cos(α)) +sin(α))

The height y will be zero at x=0 and at ...

  0 = -4.9x/(s²·cos(α)) +sin(α)

  x = s²·sin(α)·cos(α)/4.9 = (s²/9.8)sin(2α)

So, for some alternate angle β, we want ...

  sin(2α) = sin(2β)

We know this will be the case for ...

  2β = 180° -2α

  β = 90° -α

The second snowball should be thrown at an angle of 90°-67.5° = 22.5° to make it hit the same point as the first.

Answer:

The magnification of the image is  [tex]M = -1.5[/tex]

Explanation:

From the question we are told that

      The object distance is  [tex]u = 14 cm[/tex]

      The image distance is  [tex]v = 21 \ cm[/tex]

The magnification of the image is mathematically represented as

        [tex]M = - \frac{v}{u}[/tex]

substituting values

       [tex]M = - \frac{21}{14}[/tex]

        [tex]M = -1.5[/tex]

The negative value means that the image is real , inverted and enlarged

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is the second option

Explanation:

 Generally the electric force exerted by the charge Q  on the  charge  (q) is mathematically represented as

           [tex]F_Q = \frac{kqQ}{r^2}[/tex]

 Generally the electric force exerted by the charge 2Q  on the  charge  (q) is mathematically represented as

         [tex]F_{2Q} = \frac{kq2Q}{2r^2}[/tex]    

Now the net force exerted on q is

       [tex]F_{net} = \frac{kqQ}{r^2} - \frac{2k q Q}{4r^2}[/tex]

        [tex]F_{net} = \frac{4kqQ- 2kqQ}{4r^2}[/tex]

        [tex]F_{net} = \frac{kqQ}{2r^2}[/tex]

Looking at the resulting equation we see that [tex]F_{net} > 0[/tex]

This implies that the charge q would move to the right

Answer:

The visible light spectrum is the segment of the electromagnetic spectrum that the human eye can view. More simply, this range of wavelengths is called visible light. Typically, the human eye can detect wavelengths from 380 to 700 nanometers.

Explanation:

Answer:

Visible light, which travels at a dizzying 186,282 miles per second through space, is just one part of light's broad spectrum, which encompasses all electromagnetic radiation. We can detect visible light because of cone-shaped cells in our eyes that are sensitive to the wavelengths of some forms of light

Explanation:

Answer:

A. The magnitude of the net force exerted on the disk

B. The distance between the center of the disk and where the net force is applied to the disk

Explanation:

To determine the change in angular momentum of the disk after a stipulated time, one must measure the above options.

The radius of the disk is fixed and does not vary with the experiment, and the mass of the disk is also constant and known.

One must first measure the magnitude of the net force exerted on the disk, and determine the torque as a result of this torque from the distance between the center of the disk and the point where the net force is applied.  The above statement also points out the necessity of measuring the distance between the center of the disk and the point where the net force is applied on the disk, as both the torque, and the moment of inertia is calculated from this point.

torque T = Force time distance of point of action of force from mid point of the disk

T = F X r

T x t = Δ(Iω)

Where t is the time,

and Δ(Iω) is change in angular momentum.

Answer:

The correct answer will be:

(a) 2.2 rad/s

(b) 2.2 rad

Explanation:

The given values are:

Angular acceleration, [tex]\alpha = 0.35 \pi \ rad/s^2[/tex]

(a)...

At time t = 2.0 s,

The angular velocity will be:

⇒  [tex]\omega = \alpha t[/tex]

On putting the estimated values, we get

⇒     [tex]=0.35\pi \times 2.0[/tex]

∴ [[tex]\pi = 3.14[/tex]]

⇒     [tex]=0.35\times 3.14\times 2.0[/tex]

⇒     [tex]=2.2 \ rad/s[/tex]

(b)...

The angular displacement will be:

⇒  [tex]\theta = \frac{1}{2}\alpha t^2[/tex]

On putting the estimated values, we get

⇒     [tex]=\frac{1}{2}\times 0.35\pi\times (2.0)^2[/tex]

⇒     [tex]=\frac{1}{2}\times 0.35\times 3.14\times 4[/tex]

⇒     [tex]=1.099\times 2[/tex]

⇒     [tex]=2.2 \ rad[/tex]

Answer:

276 N and 225 N

Explanation:

Draw a free body diagram.  There are three forces on the hoist:

Weight force 350 N pulling down,

Tension force T₁ pulling up and left 50° from the horizontal,

Tension force T₂ pulling up and right 38° from the horizontal.

Sum of forces in the x direction:

∑F = ma

T₂ cos 38° − T₁ cos 50° = 0

T₂ cos 38° = T₁ cos 50°

T₂ = T₁ cos 50° / cos 38°

Sum of forces in the y direction:

∑F = ma

T₂ sin 38° + T₁ sin 50° − 350 = 0

T₂ sin 38° + T₁ sin 50° = 350

Substitute:

(T₁ cos 50° / cos 38°) sin 38° + T₁ sin 50° = 350

T₁ cos 50° tan 38° + T₁ sin 50° = 350

T₁ (cos 50° tan 38° + sin 50°) = 350

T₁ = 350 / (cos 50° tan 38° + sin 50°)

T₁ = 276 N

T₂ = T₁ cos 50° / cos 38°

T₂ = 225 N

Answer:

a = -0.05 m/s² (negative sign shows deceleration)

Explanation:

In order, to find out the minimum average acceleration for a student starting at 5 m/s to slide to the end, we can use 3rd equation of motion. 3rd equation of motion is given as follows:

2as = Vf² - Vi²

where,

a = minimum acceleration required = ?

s = minimum distance covered = 250 m

Vf = Final Speed = 0 m/s (for minimum acceleration the student will barely cover 250 m and then stop)

Vi = Initial Velocity = 5 m/s

Therefore,

2a(250 m) = (0 m/s)² - (5 m/s)²

a = - (25 m²/s²)/(500 m)

a = -0.05 m/s² (negative sign shows deceleration)

Answer: The force at t = 1s is 14 N

Explanation:

I guess that the position equation actually is: (by looking at the units of C and D)

S = A - B*t + C*t^2 - D*t^3

As you may know by Newton's third law, if we want to find the force, we first need to find the acceleration, and before that, the velocity.

the velocity can be found by integrating over time, the velocity is:

V = 0 - B + 2*C*T - 3*D*t^2

For the acceleration we integrate again:

A = 0 + 2*C - 6D*T

and we know that F = m*A

then:

Force(t) = 1kg*(2*10m/s^2 - 6*1m/s^3*t)

We want the force at t = 1s, so we replace t by 1s.

F(1s) = 1kg*(20m/s^2 - 6m/s^2) = 14 N

Explanation:

We have,

Mass of a particle is 0.6 kg

Speed at A is 2 m/s

Kinetic energy at B is 7.5 J

(a) The kinetic energy at A is given by :

[tex]K_A=\dfrac{1}{2}mv^2\\\\K_A=\dfrac{1}{2}\times 0.6\times 2^2\\\\K_A=1.2\ J[/tex]

(b) Kinetic energy at B is given by

[tex]K_B=\dfrac{1}{2}mV^2\\\\V=\sqrt{\dfrac{2K_B}{m}} \\\\V=\sqrt{\dfrac{2\times 7.5}{0.6}} \\\\V=5\ m/s[/tex]

(c) The work done on the particle as it moves form A to B is given by work energy theorem as :

[tex]W=\dfrac{1}{2}m(V^2-v^2)\\\\W=\dfrac{1}{2}\times 0.6\times (5^2-2^2)\\\\W=6.3\ J[/tex]

Answer:

481 m

Explanation:

To fall 235 m, the time required is

t = √(2H/g)

t= √(2[tex]\times[/tex]235/9.8)

t=6.92 seconds.

The supplies will travel forward

6.92 [tex]\times[/tex] 69.4 ≈ 481 m

Therefore, the goods must be dropped 481  m in advance of the recipients.

Answer:

a. 6666.67 V/m

b. 2. the positively charged plat

Explanation:

a. The computation of the magnitude of the electric field between the plates is shown below:

As we know that

[tex]E = \frac{V}{d}[/tex]

[tex]= \frac{400 V}{0.06 m}[/tex]

= 6666.67 V/m

hence, the magnitude of the electric field is 6666.67 V/m

b. Based on this the higher potential is positively charged plate as the flow of the current goes from positive to negative and it is inverse in case of th electron

We simply applied the above formula  

Answer:

Momentum is mainly related to forces

Answer:

The nucleus of an atom contains the majority of the atom’s mass, and is composed of protons and neutrons, which are collectively referred to as nucleons. The much-lighter electrons orbit their atom’s nucleus. The Protons. Protons are positively charged particles found in an atom’s nucleus.

I hope u liked my answer. please mark me as branliest x

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

[tex] E = \frac{\Delta V}{d} [/tex]

Where:

E: is electric field

ΔV: is the potential difference = 3.95 mV  

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

[tex]E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m[/tex]

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

Answer:

Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.

Isotopes - ⦁ Atoms with the same number of protons, but different number of neutrons.

Nitrogen - ⦁ The name of the element with atomic number 7.

Atomic number - ⦁ The number of protons in the nucleus of an atom.

Answer:

a. Time = 16.11 s

b. Gauge Pressure = 1009400 Pa = 1 MPa  

c. Absolute Pressure = 1110725 Pa + 1.11 MPa

d. Force = 2.22 MN

Explanation:

a.

For the accelerated part of motion of submarine we can use equations of motion.

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

where,

t₁ = time taken during accelerated motion = ?

Vf = final velocity = 4 m/s

Vi = Initial Velocity = 0 m/s   (Since, it starts from rest)

a = acceleration = 0.3 m/s²

Therefore,

t₁ = (4 m/s - 0 m/s)/(0.3 m/s²)

t₁ = 13.33 s

Now, using 2nd equation of motion:

d₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

where,

d₁ = the depth covered during accelerated motion

Therefore,

d₁ = (0 m/s)(13.33 s) + (0.5)(0.3 m/s²)(13.33 s)²

d₁ = 88.89 m

Hence,

d₂ = d - d₁

where,

d₂ = depth covered during constant speed  motion

d = total depth = 100 m

Therefoe,

d₂ = 100 m - 88.89 m

d₂ = 11.11 m

So, for uniform motion:

s₂ = vt₂

where,

v = constant speed = 4 m/s

t₂ = time taken during constant speed  motion

11.11 m = (4 m/s)t₂

t₂ = 2.78 s

Therefore, total time taken by submarine to move down 100 m is:

t = t₁ + t₂

t = 13.33 s + 2.78 s

t = 16.11 s

b.

The gauge pressure on submarine can be calculated by the formula:

Pg = ρgh

where,

Pg = Gauge Pressure = ?

ρ = density of salt water = 1030 kg/m³

g = 9.8 m/s²

h = depth = 100 m

Therefore,

Pg = (1030 kg/m³)(9.8 m/s²)(100 m)

Pg = 1009400 Pa = 1 MPa

c.

The absolute pressure is given as:

P = Pg + Atmospheric Pressure

where,

P = Absolute Pressure = ?

Atmospheric Pressure = 101325 Pa

Therefore,

P = 1009400 Pa + 101325 Pa

P = 1110725 Pa + 1.11 MPa

d.

Since, the force to open the door must be equal to the force applied to the door by pressure externally.

Therefore, the  force required to open the door can be found out by the formula of pressure:

P = F/A

F = PA

where,

P = Absolute Pressure on Door = 1110725 Pa

A = Area of door = 2 m²

F = Force Required to Open the Door = ?

Therefore,

F = (1.11 MPa)(2 m²)

F = 2.22 MN

Answer:

[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]

[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]

Explanation:

The given differential equation is

[tex]T\frac{d^2y}{dx^2} + \rho w ^2y=0[/tex] and y(0) = 0, y(L) =0

where T and ρ  are constants

The given rewrite as

[tex]\frac{d^2y}{dx^2} + \frac{\rho w^2}{T} y=0[/tex]

auxiliary equation is

[tex]m^2+ \frac{\rho w^2}{T} =0\\\\m= \pm\sqrt{\frac{\rho}{T} } wi[/tex]

Solution of this de is

[tex]y(x)=C_1 \cos \sqrt{\frac{\rho}{t} } wx + C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]

y(0)=0 ⇒ C₁ = 0

[tex]y(x) = C_2 \sin \sqrt{\frac{\rho}{T} } wx[/tex]

y(L) = 0 ⇒

[tex]C_2 \sin \sqrt{\frac{\rho}{T} } wL=0[/tex]

we need non zero solution

⇒ C₂ ≠ 0 and

[tex]\sin \sqrt{\frac{\rho}{T} } wL=0[/tex]

[tex]\sin \sqrt{\frac{\rho}{T} } wL=0 \rightarrow \sqrt{\frac{\rho}{T} } wL=n \pi[/tex]

[tex]w_n = \sqrt{\frac{T}{\rho} } \frac{n \pi}{L}[/tex]

solution corresponding these [tex]w_n[/tex] values

[tex]y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x[/tex]

[tex]y_n(x) = C_n \sin \frac{n \pi x}{L}[/tex]

Answer:

This statement is false , linear momentum is the product of mass and velocity

Answer:

Linear momentum is defined as the product of a system's mass multiplied by its velocity. In symbols, linear momentum is expressed as p = mv. ... Thus the greater an object's mass or the greater its velocity, the greater its momentum. Momentum p is a vector having the same direction as the velocity v.

Explanation:

hope this helps u stay safe

Answer:

1:4

Explanation:

We have, two identical objects A and B fall from rest from different heights to the ground.

Object B takes twice as long as object A to reach the ground. It is required to find the ratio of the heights from which A and B fell. Let [tex]h_A\ \text{and}\ h_B[/tex] are the height for A and B respectively. So,

[tex]\dfrac{h_A}{h_B}=\dfrac{(1/2)gt_A^2}{(1/2)gt_B^2}\\\\\dfrac{h_A}{h_B}=\dfrac{t_A^2}{t_B^2}[/tex]

We have,

[tex]t_B=2t_A[/tex]

So,

[tex]\dfrac{h_A}{h_B}=\dfrac{t_A^2}{(2t_B)^2}\\\\\dfrac{h_A}{h_B}=\dfrac{1}{4}[/tex]

So, the ratio of the heights from which A and B fell is 1:4.

Answer:

v= 0.94 m/s

Explanation:

In order to calculate the speed of the waves, you use the following formula for the speed of the waves:

[tex]v=\lambda f[/tex]          (1)

v: speed of the wave

λ: wavelength of the wave

f: frequency of the wave

The frequency is calculated by using the information about the time that boat takes to travel from its highest point to its lowest point. This time is a half of a period:

[tex]\frac{T}{2}=3.5s\\\\T=7.0s[/tex]

Then, the frequency is:

[tex]f=\frac{1}{T}=\frac{1}{7s}=0.142Hz[/tex]  

The wavelength of the wave is the distance between crests of the wave

[tex]\lambda=6.6m[/tex]

With the values of the frequency and the wavelength, you can find the speed of the wave by using the equation (1):

[tex]v=(6.6m)(0.142Hz)=0.94\frac{m}{s}[/tex]

The speed of the wave is 0.94m/s

Answer:

v_r = 1.268 × 10⁸ mi/hr

Explanation:

We are given;

wavelength of the red light; λr = 693 nm = 693 × 10^(-9) m

wavelength of the yellow light; λy = 582 nm = 582 × 10^(-9) m

Frequency is given by the formula;

f = v/λ

Where v is speed of light = 3 x 10^(8) m

Frequency of red light; f_o = [3 x 10^(8)]/(693 × 10^(-9)) = 4.33 x 10¹⁴ Hz

Similarly, Frequency of yellow light;

f = [3 x 10⁸]/(582 × 10^(-9)) = 5.15 x 10¹⁴ Hz

To find the speed of the car, we will use the formula;

f = f_o[(c + v_r)/c)]

Where c is speed of light and v_r is speed of car.

Making v_r the subject;

cf/f_o = c + v_r

v_r = c(f/f_o - 1)

So, plugging in the relevant values, we have;

v_r = 3 × 10⁸[((5.15 x 10¹⁴)/(4.33 x 10¹⁴)) - 1]

v_r = 3 × 10⁸(0.189)

v_r = 5.67 x 10⁷ m/s

Converting to mi/hr, 1 m/s = 2.23694 mile/hr

So, v_r = 5.67 × 10⁷ × 2.23694

v_r = 1.268 × 10⁸ mi/hr

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  point charge is  [tex]Q_z = -0.0912 \ \mu C[/tex]

The inner shell is  [tex]Q_t = 0.4168 \ \mu C[/tex]

The outer shell is  [tex]Q_w = -0.6514 \ \mu C[/tex]

Explanation:

From the question we are told that

    The inner radius of thin first spherical conducting shell is  [tex]r_1[/tex]

    The outer radius of thin first spherical conducting shell is  [tex]r_2[/tex]

    The inner radius of second thin spherical conducting shell is [tex]R_1[/tex]

    The outer radius of second thin spherical conducting shell is [tex]R_2[/tex]

     The magnetic flux for different region is  [tex]\phi = -10.3 *10^3 N\cdot m^2 /C \ for \ r < r_1[/tex]

    The magnetic flux for first shell is [tex]\phi = 36 * 10^3 N \cdot m^2 /C \ for \ r_2 < r <R_1[/tex]

     The magnetic flux for second shell is [tex]\phi = -36 * 10^3 N \cdot m^2 /C \ for \ r <R_1[/tex]

The magnitude of the point charge is mathematically represented as

                [tex]Q_z = \ \phi_z * \epsilon _o[/tex]

               [tex]Q_z = -10.3*10^{3} * 8.85 *10^{-12}[/tex]

               [tex]Q_z = -9.115*10^{-8} \ C[/tex]

               [tex]Q_z = -0.0912 \ \mu C[/tex]

Considering the inner shell

        [tex]Q_a = \phi_a * \epsilon _o[/tex]

=>    [tex]Q_a = 36 .8 * 10^3 * 8.85*10^{-12}[/tex]

      [tex]Q_a = 32.56*10^{-8} \ C[/tex]

       [tex]Q_a =0.326} \ \mu C[/tex]

Charge on the inner shell is

       [tex]Q_t = Q_a - Q_z[/tex]

                    [tex]Q_t = 0.326} \ \mu - ( -0.0912 \ \mu)[/tex]

                      [tex]Q_t = 0.4168 \ \mu C[/tex]

Considering the outer  shell

     [tex]Q_y = \phi_y * \epsilon_o[/tex]

=>    [tex]Q_y = -36.8 *10^{3} * 8.85*10^{-12}[/tex]

        [tex]Q_y = -32.56*10^{-8} \ C[/tex]

        [tex]Q_y = - 0.326} \ \mu C[/tex]

Charge on the outer shell is

      [tex]Q_w = Q_y - Q_z[/tex]

      [tex]Q_w =- 0.326} \ \mu - ( -0.0912 \ \mu)[/tex]

       [tex]Q_w = -0.6514 \ \mu C[/tex]

Answer:

The heat rate is  

Explanation:

From the question we are told that

  The surface emissivity is  [tex]e=0.1[/tex]

   The length is  [tex]L = 0.1 \ m[/tex]

    The width is  [tex]W = 0.1 \ m[/tex]

     The surface temperature of one side is  [tex]T_1 = 500 \ K[/tex]

     The temperature of the quiescent air [tex]T_c = 300 \ K[/tex]

      The temperature of the surrounding is  [tex]T_s = 300 \ K[/tex]

The heat rate from the flat plate is mathematically represented as

         [tex]Q = \sigma A e (T_1^4 - T_a^4)[/tex]

Where [tex]\sigma[/tex] is the quiescent air Stefan-Boltzmann constant  and it value is

       [tex]\sigma = 5.67*10^{-8} m^{-2} \cdot K^{-4}[/tex]

     A is the area which is mathematically evaluated  as

           [tex]A = W * L[/tex]

substituting values

           [tex]A = 0.1 * 0.1[/tex]

           [tex]A = 0.01 \ m^2[/tex]

substituting values

          [tex]Q = 5.67 *10^{-8} * (0.01) *(500^4 -300^4)[/tex]

          [tex]Q =3.045 \ Watt[/tex]

Answer:

A) N = (M + mf + md) g

B) d = md L / (2 mf)

Explanation:

Draw a free body diagram.  There are four forces acting on the board.

Weight force M g pulling down at the center of the board.

Normal force N pushing up at the center of the board.

Weight force mf g pulling down a distance d from the center.

Weight force md g pulling down at the end of the board.

Sum of forces in the y direction:

∑F = ma

N − M g − mf g − md g = 0

N = (M + mf + md) g

Sum of moments about the center of the board:

∑τ = Iα

md (L/2) − mf d = 0

d = md L / (2 mf)

The magnitude of the upward force n exerted by the support on the board will be the same with the magnitude of the weight of the seesaw. Which is

W = 9.8M Newton

The father should sit at L/4 at the other end in order to balance the system at rest

Since the seesaw is of a uniform board, the center of gravity will act at the center. If it is of length L, the support called the fulcrum will be positioned at L/2.

If the daughter is a distance L/2 from the center, Then, she must be positioned at L/4 from the one edge of the board.

A. The magnitude of the upward force n exerted by the support on the board will be the same with the magnitude of the weight of the seesaw. Which is

W = mg

W = 9.8M Newton

The weight of both the father and the daughter will act downward. so, they will not be included.

The father should sit at L/4 at the other end in order to balance the system at rest

Learn more here: https://brainly.com/question/15651723

Answer:

Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve. Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner.

The acceleration is not aligned with the direction of travel because the change in velocity is at a tangent (directed away) to the direction of travel of the bus.

Answer:

1800 J

Explanation:

Energy is conserved, so the maximum kinetic energy equals the change in gravitational energy.