Ask question

Ask question

All categories

Hunter-Best [27]

2 years ago

5

A fireman clings to a vertical ladder and directs the nozzle of a hose horizontally toward a burning building. The rate of water

flow is 6.31kg/s, and the nozzle speed is 12.5 m/s. The hose passes vertically between the fireman’s feet, which are 1.30 m below the nozzle. Choose the origin to be inside the hose between the fireman’s feet. What torque must the fireman exert on the hose? (This could also be stated, what is the rate of change of the angular momentum of the water?)

1 answer:

Aleks [24]2 years ago

4 0

Answer:

A torque of 102.5375 Nm must be exerted by the fireman

Explanation:

Given:

The rate of water flow = 6.31 kg/s

The speed of nozzle = 12.5 m/s

Now, from the Newton's second law we have

The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction

thus we have,

F = 6.31 kg/s x 12.5m/s

or

F = 78.875 N

Now,

The torque (T) exerted by water force about the fireman's will be

T = (F x d)

or

T = 78.875 N x 1.30 m

T = 102.5375 Nm

hence,

<u>A torque of 102.5375 Nm must be exerted by the fireman</u>

You might be interested in

A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ϕ1 with

strojnjashka [21]

Answer:

$x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

Explanation:

Let 'F₁' and 'F₂' be the forces applied by left and right wires on the bar as shown in the diagram below.

Now, the horizontal and vertical components of these forces are:

$F_{1x} = -F_1cos(\phi_1)\\F_{1y}=F_1sin(\phi_1)\\\\F_{2x}=F_2cos(\phi_2)\\F_{2y}=F_2sin(\phi_2)$

As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,

$\sum F_x=0\\F_{1x}=F_{2x}\\F_1cos(\phi_1)=F_2cos(\phi_2)\\\frac{F_1}{F_2}=\frac{cos(\phi_2)}{cos(\phi_1)}-------1$

Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.

At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.

Therefore, the net torque by the forces $F_{1y}\ and\ F_{2y}$ will be zero. This gives,

$-F_{1y}\times x + F_{2y}(L-x) = 0\\F_{1y}\times x=F_{2y}(L-x)\\x=\frac{F_{2y}(L-x)}{F_{1y}}$

But, $F_{1y}=F_1sin(\phi_1)\ and\ F_{2y}=F_2sin(\phi_2)$

Therefore,

$x=\frac{F_2sin(\phi_2)(L-x)}{F_1sin(\phi_1)}\\\textrm{From equation (1),}\frac{F_2}{F_1}=\frac{cos(\phi_1)}{cos(\phi_2)}\\\therefore x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times (L-x)\\x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times L-\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times x\\\\$

$x(1+\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L\\x(1+\frac{cos(\phi_1)}{sin(\phi_1)}\times \frac{sin(\phi_2}{cos(\phi_2)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L$

We know,

$tan(\phi)=\frac{sin(\phi)}{cos(\phi)}\\\\cot(\phi)=\frac{cos(\phi)}{sin(\phi)}$

∴ $x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

6 0

2 years ago

Which activity is a method of habitat restoration?

irina1246 [14]

A. cleaning up after environmental disasters

7 0

2 years ago

Read 2 more answers

Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end

iragen [17]

Answer:

The workdone is $W_d =-4400J$

Explanation:

The free body diagram is shown on the first uploaded image

From the question we are given that

The force is on the force gauge $F = 2750 N$

The distance that Magnus pulled the bus $d = 1.60m$

Generally the workdone by the tension force on Magnus is

$Workdone = Force * displacement \ in \ the \ direction \ of \ force$

$W_d = F * (-d)$

This negative sign show that is tension force is in the opposite direction to Magnus movement (i.e the movement of the bus )

Substituting value we have

$Workdone = - 2750 * 1.60$

$=-4400 J$

7 0

2 years ago

A tennis player's racket applies an average force of 200. newtons to a tennis ball for 0.025 second. The average force exerted o

Sever21 [200]

Answer:

200N

Explanation:

6 0

2 years ago

An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin

fgiga [73]

Given that,

The electric field is given by,

$\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}$

Suppose, B is the amplitude of magnetic field vector.

We need to find the complete expression for the magnetic field vector of the wave

Using formula of magnetic field

Direction of $(\vec{E}\times\vec{B})$ vector is the direction of propagation of the wave .

Direction of magnetic field = $\hat{j}$

$B=B_{0}\sin(kx-\omega t)\hat{k}$

We need to calculate the poynting vector

Using formula of poynting

$\vec{S}=\dfrac{E\times B}{\mu_{0}}$

Put the value into the formula

$\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}$

$\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

Hence, The poynting vector is $\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

7 0

2 years ago