A rectangular trough, 1.6 m long, 0.50 m wide, and 0.45 m deep, is completely full of water. One end of the trough has a small drain plug right at the bottom edge. Part A When you pull the plug, at what speed does water emerge from the hole

Answer: [tex]x(t) = 14e^{-2t} - 10e^{-3t}[/tex]

Explanation: In a mass-spring-damper system, the differential equation that rules the motion of the mass is: mx" + cx' + kx = 0

Using m = 4, k = 24 and c = 20, we have

4x" + 20x' + 24x = 0

Simplifying, we have

x" + 5x'+ 6x = 0

The characteristic equation of this differential is

[tex]r^{2} + 5r + 6 = 0[/tex]

The solutions for the quadratic equation are: [tex]r_{1} = -2[/tex] and [tex]r_{2} = -3[/tex]

Hence:

x(t) = [tex]C_{1}e^{-2t} + C_{2}e^{-3t}[/tex]

x'(t) = [tex]-2C_{1}E^{-2t} - 3C_{2}e^{-3t}[/tex]

To determine the constants, we have the initial conditions x(0) = 4 and

x'(0) = 2, then:

[tex]x(0) = C_{1} + C_{2} = 4\\ C_{1} = 4 - C_{2}[/tex]

[tex]x'(0) = -2C_{1} -3C_{2} = 2\\-2(4-C_{2}) -3C_{2} = 2\\C_{2} = -10\\C_{1} = 4 - C_{2}\\C_{1} = 14[/tex]

Substituing the constants:

[tex]x(t) = 14e^{-2t} - 10e^{-3t}[/tex]

The position function for this system is: [tex]x(t) = 14e^{-2t} - 10e^{-3t}[/tex]

Answer:

v = 4.08m/s₂

Explanation:

Answer:

a. 1000N/C

Explanation:

Data mentioned in the question

Electrical field magnitude = 1000 NC

Perpendicular distance = 0.1 m

Perpendicular distance = 0.2 m

Based on the above information, the electric field is

As we know that

[tex]E = \frac{\sigma}{2\times E_o}[/tex]

where,

[tex]\sigma[/tex] = surface charge density

E = distance from nearby point to sheet i.e be independent

The distance at 0.1 and 0.2, the electric field would remain the same

So,

Based on the above explanation, the first option is correct

Answer:

aₓ = 0 ,       ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

          vₓ = v₀ₓ = 1.10 m / s

          aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

         [tex]v_{y}[/tex]= v_{oy} -ay t

          ay = (v_{oy} -v_{y}) / t

          ay = (0 -10.9) / 1.6

          ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

Answer:

The maximum speed is  [tex]u = 22 \ m/s[/tex]

Explanation:

From the question we are told that

   The distance covered by the debris is [tex]R = 50 \ m[/tex]

The maximum range of the debris projectile is mathematically represented as

      [tex]R = \frac{ u^2 sin^2 \theta }{g}[/tex]

At maximum  [tex]\theta = 90 ^o[/tex]

Now making u which is the maximum speed at which debris was blown outward by the explosion.

  we  have

      [tex]u = \sqrt{\frac{R * g }{ sin ^2 \theta } }[/tex]

substituting values

          [tex]u = \sqrt{\frac{50 * 9.8 }{ [sin 90] ^2} }[/tex]

        [tex]u = 22 \ m/s[/tex]

Answer:

Work = 1167.54 J

Explanation:

The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:

Gain in K.E = Loss in P.E + Work

(0.5)(m)(Vf² - Vi²) - mgh = Work

where,

m = mass of boy = 60 kg

Vf = Final Speed = 8.5 m/s

Vi = Initial Speed = 1.6 m/s

g = 9.8 m/s²

h = height drop = 1.57 m

Therefore,

(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work

Work = 2090.7 J - 923.16 J

Work = 1167.54 J

Answer:

a. 94.54 N

b. 0.356 m/s^2

Explanation:

Given:-

- The mass of the inclined block, M = 100 kg

- The mass of the vertically hanging block, m = 10 kg

- The angle of inclination, θ = 20°

- The coefficient of friction of inclined surface, u = 0.3

Find:-

a) The magnitude of tension in the cable

b) The acceleration of the system

Solution:-

- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.

- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.

- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:

                        [tex]T - m*g = m*a\\\\[/tex]  ... Eq 1

Where,

              a: The acceleration of the system

- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.

- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.

- We will employ the static equilibrium of the inclined block in the normal direction and we have:

                        [tex]N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )[/tex]

- The frictional force ( F ) is proportional to the contact force ( N ) as follows:

                        [tex]F = u*N\\\\F = u*M*g *cos ( Q )[/tex]

- Now we will apply the Newton's second law of motion parallel to the plane as follows:

                       [tex]M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\[/tex] .. Eq2

- Add the two equation, Eq 1 and Eq 2:

                      [tex]M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}[/tex]

- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).

- Using equation 1, we determine the tension ( T ) in the cable as follows:

                     [tex]T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N[/tex]

Answer:

xf = 5.68 × 10³ m  

yf = 8.57 × 10³ m  

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0                            yi = 0

xf = ?                            yf = ?

vxi =  vicosθ               vyi = visinθ  

ax = 0                          ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t²     ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m  

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ²     ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s)  ²

yf = 8.57 × 10³ m  

Answer:

The mass of the hailstone is  [tex]m_2 = 0.05[/tex]

Explanation:

From the question we are told that

    The mass of the hockey puck is [tex]m_1 = 0.2 \ kg[/tex]

     The initial speed of the puck is [tex]u_1 = 10 \ m/s[/tex]

      The final speed of the puck is  [tex]v_f = 8 \ m/s[/tex]

According to the principles of linear momentum

      [tex]m_1u_1 + m_2u_2 = (m_1 + m_2)v_f[/tex]

At initial the velocity of the hailstone was zero so

       [tex]u_2 = 0 \ m/s[/tex]

So  

     [tex]m_1 u_1 = (m_1 + m_2 )v_f[/tex]

substituting values

        [tex]0.2 * 10 = (0.2 + m_2 )* 8[/tex]

=>     [tex]2 = (0.2 + m_2 )* 8[/tex]

=>     [tex]0.25 = 0.2 + m_2[/tex]

=>    [tex]m_2 = 0.05[/tex]

Answer:

λ = 8.88 x 10⁻⁷ m = 888 nm

Explanation:

The energy band gap is given as:

Energy Gap = E = 1.4 eV

Converting this to Joules (J)

E = (1.4 eV)(1.6 x 10⁻¹⁹ J/1 eV)

E = 2.24 x 10⁻¹⁹ J

The energy required for photovoltaic generation is given as:

E = hc/λ

where,

h = Plank's Constant = 6.63 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = ?

Therefore,

2.24 x 10⁻¹⁹ J = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.24 x 10⁻¹⁹ J)

λ = 8.88 x 10⁻⁷ m = 888 nm

Answer:

D Remains constant

Explanation:

D. (1m, 0.5m)

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

[tex]h=\sqrt{(36m)^2+(227m)^2}=229.83m[/tex]    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

[tex]y=y_o+v_ot+\frac{1}{2}gt^2[/tex]    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

[tex]3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s[/tex]

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

[tex]v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}[/tex]

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

[tex]\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°[/tex]

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

Answer: Smaller than ; larger than

Explanation:

When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of

N= mg +ma, (g is gravity and a is acceleration)

here ma is negative so the N= mg-ma

Hence, it feels smaller than its original weight.

When the elevator is moving downward , then the force acting will be positive in nature

N= mg+ma,

here ma will be positive so it feels larger the original weight of passenger.

Answer:

Explanation:

this is the answer to your question

Explanation:

1 b. homogenous.

2 b. chemical.

Hope this helps

Answer:

a) The distance from the cornea vertex to the retina is 2.37 cm

b) This distance is shorter than for the normal eye.

Explanation:

a) Let refractive index of air,

n(air) = x = 1

Let refractive index of lens,

n(lens) = y = 1.4

Object distance, s = 36 cm

Radius of curvature, R = 0.65 cm

The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.

Image distance, s' = ?

(x/s) + (y/s') = (y-x)/R

(1/36) + (1.4/s') = (1.4 - 1)/0.65

1.4/s' = 0.62 - 0.028

1.4/s' = 0.592

s' = 1.4/0.592

s' = 2.37 cm

Distance from the cornea vertex to the retina is 2.37 cm

(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.

Answer:

the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]

Explanation:

The change in length of a bar can be expressed with the relation;

[tex]\Delta L = L_f - L_i[/tex]   ---- (1)

Also ; the relative or fractional increase in length  is proportional to the change in temperature.

Mathematically;

ΔL/L_i ∝ kΔT

where;

k is replaced with ∝ (the proportionality constant )

[tex]\dfrac{ \Delta L}{L_i}=\alpha \Delta T[/tex]    ---- (2)

From (1) ;

[tex]L_f = \Delta L + L_i[/tex]  ---  (3)

from (2)

[tex]{ \Delta L}=\alpha \Delta T*{L_i}[/tex]  ---- (4)

replacing (4) into (3);we have;

[tex]L_f =(\alpha \Delta T*{L_i} ) + L_i[/tex]

On re-arrangement; we have

[tex]L_f = L_i + \alpha L_i (\Delta T )[/tex]

from the given question; we can say that :

[tex](L_f)_{brass}}} = (L_f)_{Al}[/tex]

So;

[tex]L_{brass} + \alpha _{brass} L_{brass}(\Delta T) = L_{Al} + \alpha _{Al} L_{Al}(\Delta T)[/tex]

Making the change in temperature the subject of the formula; we have:

[tex]\Delta T = \dfrac{L_{Al}-L_{brass}}{\alpha _ {brass} L_{brass}-\alpha _{Al}L_{Al}}[/tex]

where;

[tex]L_{Al}[/tex] = 10.02 cm

[tex]L_{brass}[/tex] = 10.00 cm

[tex]\alpha _{brass}[/tex] = 19 × 10⁻⁶ °C ⁻¹

[tex]\alpha_{Al}[/tex] = 24  × 10⁻⁶ °C ⁻¹

[tex]\Delta T = \dfrac{10.02-10.00}{19*10^{-6} \ \ {^0}C^{-1} *10.00 -24*10^{-6} \ \ {^0}C^{-1} *10.02}[/tex]

[tex]\Delta T[/tex] = −396.1965135 ° C

[tex]\Delta T[/tex] ≅ −396.20  °C

Given that the initial temperature [tex]T_i = 19^0 C[/tex]

Then ;

[tex]\Delta T = T_f - T_i[/tex]

[tex]T_f = \Delta T + T_I[/tex]

Thus;

[tex]T_f =(-396.20 + 19.0)^0 C[/tex]

[tex]\mathbf{T_f = -377.2^0 C}[/tex]

Thus; the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]

Answer:

0.15 m

Explanation:

First calculating the center of mass from the saw horse

[tex]\frac{3.15}{2} -1=0.575 m[/tex]

from the free body diagram we can write

Taking moment about the saw horse

55.9×9.81×y=14.5×0.575×9.81

y= 0.15 m

So, the painter walk from the saw horse on the right until the board begins to tip is 0.15 m far.

Answer:

When the spring in the ballistic pendulum is compressed, energy is stored up in the spring as potential energy. When the steel ball is launched by the spring, the stored up potential energy of the compressed spring is transformed and transferred into the kinetic energy of the steel ball as it flies off to hit its target. On hitting the soft target, some of the kinetic energy of the steel ball is transferred to the soft target (since they stick together), and they both start to swing together. During the process of swinging, the system's energy is transformed between kinetic and potential energy. At the maximum  displacement of the ball from its point of rest, all the energy is converted to potential energy of the system. At the lowest point of travel (at the rest point), all the energy of the system is transformed into kinetic energy. In between these two points, energy the energy of the system is a combination of both kinetic and potential energy.

In the end, all the energy will be transformed and lost as heat to the surrounding; due to the air resistance around; bringing the system to a halt.

Answer:

The two value of the wavelength for the out of tune guitar is  

[tex]\lambda _2 = (6.48,6.52) \ cm[/tex]

Explanation:

From the question we are told that

     The wavelength of the note is [tex]\lambda = 6.50 \ cm = 0.065 \ m[/tex]

     The difference in beat frequency is [tex]\Delta f = 17.0 \ Hz[/tex]

Generally the frequency of the note played by the guitar that is in tune is  

        [tex]f_1 = \frac{v_s}{\lambda}[/tex]

Where [tex]v_s[/tex] is the speed of sound with a constant value [tex]v_s = 343 \ m/s[/tex]

       [tex]f_1 = \frac{343}{0.0065}[/tex]

      [tex]f_1 = 5276.9 \ Hz[/tex]

The difference in beat is mathematically represented as

       [tex]\Delta f = |f_1 - f_2|[/tex]

Where [tex]f_2[/tex] is the frequency of the sound from the out of tune guitar

     [tex]f_2 =f_1 \pm \Delta f[/tex]

substituting values

      [tex]f_2 =f_1 + \Delta f[/tex]

      [tex]f_2 = 5276.9 + 17.0[/tex]  

     [tex]f_2 = 5293.9 \ Hz[/tex]

The wavelength for this frequency is

      [tex]\lambda_2 = \frac{343 }{5293.9}[/tex]

     [tex]\lambda_2 = 0.0648 \ m[/tex]

    [tex]\lambda_2 = 6.48 \ cm[/tex]

For the second value of the second frequency

     [tex]f_2 = f_1 - \Delta f[/tex]

     [tex]f_2 = 5276.9 -17[/tex]

      [tex]f_2 = 5259.9 Hz[/tex]

The wavelength for this frequency is

   [tex]\lambda _2 = \frac{343}{5259.9}[/tex]

   [tex]\lambda _2 = 0.0652 \ m[/tex]

   [tex]\lambda _2 = 6.52 \ cm[/tex]

This question involves the concepts of beat frequency and wavelength.

The possible wavelengths of the out-of-tune guitar are "6.48 cm" and "6.52 cm".

The beat frequency is given by the following formula:

[tex]f_b=|f_1-f_2|\\\\[/tex]

f₂ = [tex]f_b[/tex] ± f₁

where,

f₂ = frequency of the out-of-tune guitar = ?

[tex]f_b[/tex] = beat frequency = 17 Hz

f₁ = frequency of in-tune guitar = [tex]\frac{speed\ of\ sound\ in\ air}{\lambda_1}=\frac{343\ m/s}{0.065\ m}=5276.9\ Hz[/tex]

Therefore,

f₂ = 5276.9 Hz ± 17 HZ

f₂ = 5293.9 Hz (OR) 5259.9 Hz

Now, calculating the possible wavelengths:

[tex]\lambda_2=\frac{speed\ of\ sound}{f_2}\\\\\lambda_2 = \frac{343\ m/s}{5293.9\ Hz}\ (OR)\ \frac{343\ m/s}{5259.9\ Hz}\\\\[/tex]

λ₂ = 6.48 cm (OR) 6.52 cm

Learn more about beat frequency here:

https://brainly.com/question/10703578?referrer=searchResults

Answer:

the compression of the spring is 0.5613 m

Explanation:

Given;

mass of bullet, m₁ = 0.01 kg

mass of block, m₂ = 1.65 kg

initial velocity of the block, u₂ = 0

initial velocity of the bullet before hitting the block, u₁ = 300 m/s

Final speed of the bullet-block system after collision, v = ?

spring constant, K = 17.2 N/m

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v( m₁ + m₂)

0.01 x 300 + 1.65 x 0 = v(0.01 + 1.65)

3 = 1.66v

v = 3 / 1.66

v = 1.807 m/s

Apply the principle of conservation of mechanical energy to determine the compression of the spring;

KE₁ + PE₁ = KE₂ + PE₂

¹/₂mu² + ¹/₂Kx₁² = ¹/₂mv² + ¹/₂Kx₂²

where;

m is mass of bullet and block embedded together

u is the initial velocity of the  bullet-block system = 1.807 m/s

v is the final velocity of the  bullet-block system = 0

x₁ is the initial compression of the spring = 0

x₂ is the final compression of the spring = ?

¹/₂(1.65 + 0.01) (1.807)² + ¹/₂(17.2)(0)² = ¹/₂(1.65 + ).01)(0)² + ¹/₂(17.2)(x₂)²

2.71 + 0 = 0 + 8.6(x₂)²

(x₂)² = 2.71 / 8.6

(x₂)² = 0.3151

x₂ = √0.3151

x₂ = 0.5613 m

Therefore, the compression of the spring is 0.5613 m

Answer:

12.25m/s

Explanation:

[tex]d=v_ot+\dfrac{1}{2}at^2[/tex]

Since the initial velocity of the dropped rock is 0, you can write this as:

[tex]d=\dfrac{1}{2}(9.8)(3)^2=44.1m[/tex]

Now, you can set up the equation for the thrown rock:

[tex]44.1=v_o(2)+\dfrac{1}{2}(9.8)(2)^2 \\\\\\44.1=2v_o+19.6 \\\\\\2v_o=24.5 \\\\\\v_o=12.25m/s[/tex]

Hope this helps!

Answer:

1.424 μC

Explanation:

I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),

tension T (acting along the string - to the pivot point), and

F (electric force – acting along the line connecting the charges).

We then have something like this

x: T•sin α = F,

y: T•cosα = mg.

Dividing the first one by the second one we have

T•sin α/ T•cosα = F/mg, ultimately,

tan α = F/mg.

Since we already know that

q1=q2=q, and

r=2•L•sinα,

k=9•10^9 N•m²/C²

Remember,

F =k•q1•q2/r², if we substitute for r, we have

F = k•q²/(2•L•sinα)².

tan α = F/mg =

= k•q²/(2•L•sinα)² •mg.

q = (2•L•sinα) • √(m•g•tanα/k)=

=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =

q = 0.486 • √(8.61•10^-12)

q = 0.486 • 2.93•10^-6

q = 1.424•10^-6 C

q = 1.424 μC.

Answer:

46.67 N  Upwards (with a clockwise moment)

Explanation:

length of board = 2.5 m

weight of board = 120 N

the board has two supports,  say support A and support B

support A is at one end,

support B is at 100 m from the other end.

weight of diver = 100 N

diver stands on the other end of the board.

Magnitude of support A at the end of the board

To get the magnitude and force exerted by the support at the end of the board (support A, we take moment of the forces about support B

Moment of a force is the product of force and perpendicular distance of the force about a center.

The weight of the board acts at the center of the board (1.25 m from each end of the board). That is 2.5 m from the support B.

moment of board's weight about support B is  120 x 0.25 = 30 N-m

The moment due to the weight of the board acts anticlockwise.

Weight of the diver acts at the opposite side of the board, and it acts 1 m from support B.

Moment of diver about support B is 100 x 1 = 100 N-m

Th moment due to the diver acts clockwise.

The moment due to the reaction at support A acts at a distance 1.5 m from support B

If the reaction force on support A is Fa, then the reaction about support B is Fa x 1.5 = 1.5Fa.

The moment due to support A acts clockwise.

According to moment laws, the total clockwise movement must be equal to the total anticlockwise movement.

Total clockwise movements = 100 N-m + 1.5Fa

Total anticlockwise moment = 30 N-m

according to moment laws,

100 + 1.5Fa = 30

1.5 Fa = 30 - 100 = -70

Fa = -70/1.5 = -46.67 N

The magnitude of the force exerted at support A is equal but opposite to the reaction at support A and is equal to 46.67 N

Answer:

When the Net Forces are equal to 0

Explanation:

Momentum of a body can be defined as product of mass and velocity. It is in the same direction as in velocity. When the momentum of a body doesn't change, it is said to be conserved. If the momentum of a body is constant, the the net forces acting on a body becomes zero. When net forces acting on a body is zero, it means that no kinetic energy is being lost or gained, hence the kinetic energy is also conserved. If no energy is being gained or lost, it means that the body will experience no energy.

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

Vi = 0.105 m/s

Answer:

The tension on the string is  [tex]T = 43.302 \ N[/tex]

Explanation:

From the question we are told that

    The mass of the rock is [tex]m_r = 5.00 \ kg = 5000 \ g[/tex]

       The density of the rock is [tex]\rho = 4300 \ kg/m^3 = 4.3 g/dm^3[/tex]

Generally the volume of the rock is mathematically evaluated as

          [tex]V = \frac{m_r}{\rho}[/tex]

substituting values

        [tex]V = \frac{5000}{4.3}[/tex]

       [tex]V = 1162.7 \ dm^3[/tex]

The volume of the rock immersed in water is

      [tex]V_w = \frac{V}{2}[/tex]  

substituting values

     [tex]V_w = \frac{1162.7 }{2}[/tex]

     [tex]V_w = 581.4 \ dm^3[/tex]

mass of water been displaced by the this volume is

     [tex]m_w = V_w[/tex]     According to Archimedes principle

=>   [tex]m_w = 581.4 \ g[/tex]

     [tex]m_w = 0.5814 \ kg[/tex]

The weight of the water displace is  

      [tex]W _w = m_w * g[/tex]

      [tex]W _w = 0.5814 * 9.8[/tex]

      [tex]W _w = 5.698 \ N[/tex]

The actual weight of the rock is  

      [tex]W_r = m_r * g[/tex]

     [tex]W_r = 5.0 * 9.8[/tex]

     [tex]W_r = 49.0 \ N[/tex]

The tension on the string is

       [tex]T = W_r - W_w[/tex]

substituting values

       [tex]T = 49.0 - 5.698[/tex]

       [tex]T = 43.302 \ N[/tex]

Answer:

563.64 m

Explanation:

Given that as per the question

x = 5 cm = 0.05 m

D = 4.2 × 107 m

d = smallest aperture size

As per the situation the solution of the smallest aperture telescope that she can get away with is below :-

We will use Rayleigh's diffraction limit which is

[tex]d\frac{x}{D} = 1.22\lambda[/tex]

The equation will be

[tex]d\frac{0.05}{4.2\times 10^7} = 1.22[550\times 10^{-9}][/tex]

d = 563.64 m

So, the answer is d = 563.64 m