PHYSICS COLLEGE

A plank 2.00 cm thick and 13.0 cm wide is firmly attached to the railing of a ship by clamps so that the rest of the board extends 2.00 m horizontally over the sea below. A man of mass 68.4 kg is forced to stand on the very end. If the end of the board drops by 5.20 cm because of the man's weight, find the shear modulus of the wood.

Answers

Answer 1

Answer:

Answer:

9.93 MPa

Explanation:

Given:

- mass of the man = 68.4 kg

- Deflection dx = 5.2 cm

- thickness of plank t = 2.0 cm

- width of plank w = 13.0 cm

- Length subtended L = 2.0 m

Find:

Shear Modulus of Elasticity S :

S = shear stress / shear strain

Shear stress = F / A

Shear stress = 68.4*9.81 / 0.02*0.13

Shear stress = 258078.4615 Pa

Shear strain = dx / L

Shear Strain = 0.052 / 2

Shear Strain = 0.026

Hence,

S = 258078.4615 / 0.026

S = 9.93 MPa

Related Questions

If George Washington had become "King of America," our government might have become?
what happens when an electric current passes through a coil of wire instead of a single straight peice of wire
A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initialspeed of 32 m/s, what is its new speed?A. 17 m/sB. 15 m/sC. 47 m/sD. 32 m/s
When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator accelerating upward at 2.0 m/s2 , how far does the spring stretch with the same box attached? (b) How fast and in which direction should the elevator accelerate for the spring stretch to be zero (that is, the spring returns to its relaxed length)?
A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels640 m in 40 s. For the 40 s interval, find (a) the average velocity, (b) the final velocity, and (c) the
acceleration.

Answers

Answer:

(a) The average velocity is 16 m/s

(b) The acceleration is 0.4 m/s^2

(c) The final velocity is 24 m/s

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+(a.t^2)/(2)\qquad\qquad [2]$

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

$\displaystyle \bar v=(x)/(t)=(640)/(40)=16$

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

$\displaystyle a=(v_f-v_o)/(t)$

$\displaystyle a= 2(x-v_ot)/(t^2)$

Since vo=8 m/s, x=640 m, t=40 s:

$\displaystyle a= 2(640-8\cdot 40)/(40^2)=0.4$

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

$v_f=8+0.4\cdot 40$

$v_f=8+16=24$

The final velocity is 24 m/s

Final answer:

The average velocity is 16 m/s, the final velocity is 8.0 m/s + (acceleration * 40 s), and the acceleration can be found by solving the equation 640 m = (8.0 m/s * 40 s) + (0.5 * acceleration * (40 s)^2.

Explanation:

To find the average velocity, we use the formula: average velocity = total displacement / total time. In this case, the total displacement is 640 m and the total time is 40 s, so the average velocity is 640 m / 40 s = 16 m/s.

To find the final velocity, we can use the formula: final velocity = initial velocity + (acceleration * time). In this case, the initial velocity is 8.0 m/s and the time is 40 s. Since the question states that it moves with constant acceleration, we can assume that the acceleration is the same throughout the 40 s interval. Therefore, the final velocity is 8.0 m/s + (acceleration * 40 s).

To find the acceleration, we can use the formula: total displacement = (initial velocity * time) + (0.5 * acceleration * time^2). In this case, the total displacement is 640 m, the initial velocity is 8.0 m/s, and the time is 40 s. Solving for acceleration, we have 640 m = (8.0 m/s * 40 s) + (0.5 * acceleration * (40 s)^2).

Learn more about Average velocity, Final velocity, Acceleration here:

brainly.com/question/21474168

#SPJ3

A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the lowest frequency associated with this portion of the spectrum?

Answers

Answer:

the lowest frequency is $7.5* 10^(14) Hz$

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν= $(v)/(L)$

frequency ν be lopwest when L will be heighest

ν(lowest)= $(3* 10^8)/(400* 10^-9)$

ν= $7.5* 10^(14) Hz$

Tell whether the statement below is a scalar or a vector

Answers

Answer:

1. Scalar

2.Vector

3. Scalar

4. Vector

5.Scalar

6.Scalar

7.Vector

8.Vector

9.Scalar

10.Scalar

11.Scalar

12. Vector

13.Scalar

Explanation:

Scalar refers to magnitude, and Vectors include magnitude with directions.

A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °

Answers

Answer: $\theta =45.73^(\circ)$

Explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

$T\cos\theta =mg$ -----1

$T\sin \theta =(mv^2)/(r)$ ------2

Divide 2 & 1

$tan\theta =(v^2)/(rg)$

$tan\theta =(4.65^2)/(2.15* 9.8)$

$tan\theta =1.026$

$\theta =45.73^(\circ)$

Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?

Answers

Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

v² = u² + 2as

v² = 44.50² + 2 x (-9.81) x 71.32

v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

v = u + at

24.10 = 44.50 - 9.81 t , t = 2.07 s

-24.10 = 44.50 - 9.81 t , t = 6.99 s

Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

s = ut + 0.5 at²

71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

1. A surfboarder rides a wave for 23.7 m at a constant rate of 4.1 m/s. How long did his triptake?

Answers

Answer:

His trip took 5.78 seconds

Explanation:

23.7m divided by 4.1m/s = 5.78048780488

Other Questions

A 133 kg horizontal platform is a uniform disk of radius 1.95 m and can rotate about the vertical axis through its center. A 62.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 28.5 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as J(r) = Br, where r is in meters, J is in amperes per square meter, and B = 2.35 ✕ 105 A/m3. This function applies out to the wire's radius of 2.00 mm. How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of 11.5 μm and is at a radial distance of 1.20 mm?

Two long ideal solenoids (with radii 20 mm and 30 mm, respectively) have the same number of turns of wire per unit length. The solenoid is mounted inside the larger, along a common axis. The magnetic field with in the inner solenoid is zero. The current in the inner solenoid must be: a. two-thirds the current in the outer solenoid b. one-third the current in the outer solenoid c. twice the current in the outer solenoid d. half of the current in the outer solenoid e. the same as the current in the outer solenoid