All categories

2 years ago

3. A very light bamboo fishing rod 3.0 m long is secured to a boat at the bottom end. It is

Physics

1 answer:

netineya [11]2 years ago

7 0

The image in the attachment describes the situation of the fishing rod.

Answer: F = 10.8 N

Explanation: The image shows a fishing rod attached to an axis. To stay in equilibrium, Torque must be equal for the force of magnitude 18N and for the unknow force.

Torque (τ) is a measure of a force's tendency to cause rotation and, in physics, defined as:

τ = F.r.sin(θ)

F is the force acting on the object;

r is distance between where the torque is measured to where the force is applied;

θ is the angle between F and r;

For the fishing rod:

$\tau_{1} = \tau_{2}$

$F_{1}.r_{1}.sin(\theta) = F_{2}.r_{2}.sin(\theta)$

Assuming part (1) is related to unknown force:

$F = \frac{F_{2}.r_{2}.sin(\theta}{r_{1}.sin(\theta) }$

Replacing the corresponding values:

$F = \frac{18*1.8*sin(30)}{3*sin(30)}$

$F = \frac{18*1.8}{3}$

F = 10.8

The fishing line exert on the the rod a force of 10.8N.

You might be interested in

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

2 years ago

The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp

Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

3 0

1 year ago

To persuade my audience that the use of mobile communication devices by drivers—even when they are hands-free—is contributing to

pickupchik [31]

Answer:

specific purpose statement

Explanation:

It is a specific purpose statement made for a persuasive speech on the question of fact.

A specific purpose statement helps to build on the general purpose (that is to inform) and to make it more specific to the audience. So if the first speech is an informative speech, our general purpose is to inform our audience about a very specific realm of knowledge.

A specific purpose statement is given to audience to persuade on specific information.

8 0

2 years ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$