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20 days ago

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A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b

y the rope, and a horizontal drag forcc Fa is cxertcd by thc watcr on the ski. The water also exerts a vertical lift force L on the skier. Which of the following are correct relationships between the forces exerted on the skier-ski system? Select two answers. (A) T-Fd = ma (B) L-Fg=ma (C) L- =0 (D) T- Fd 0

1 answer:

Maru [2.3K]20 days ago

3 0

Answer:

The accurate equations are T-fg=ma and L-fg=0.

Options (A) and (C) are valid.

Explanation:

Given that:

Weight Fg = mg

Acceleration = a

Tension = T

Drag force = Fa

Vertical force = L

We are required to determine the correct relationships

Using the equilibrium equations

Horizontally,

The acceleration is a

$T-Fd=ma$ ...(I)

Vertically,

No acceleration occurs

$w=L$

$mg-L=0$

Substituting the value of mg

$L-fg=0$ ....(II)

Thus, the correct relationships are T-fg=ma and L-fg=0.

Options (A) and (C) are correct.

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A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Maru [2360]

Answer:

The final size is nearly the same as the initial size because the increase in size $1.055\times 10^{- 7}$ is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, $m_{p} = 1.67\times 10^{- 27} kg$

Initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This proton energy is $\simeq 250 GeV$

Thus, the speed of the proton, v $\simeq c$

The time to cover 1 km = 1000 m of distance is calculated as:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

According to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

3 0

10 days ago

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

serg [2593]

Answer:

$4.1\cdot 10^8 N$

Explanation:

To begin with, we must determine the pressure acting on the sphere, which is calculated using:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ denotes the atmospheric pressure

$\rho = 1000 kg/m^3$ represents the density of the water

$g=9.8 m/s^2$ signifies the acceleration due to gravity

$h=11,000 m$ indicates the depth

By substituting these values,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The sphere's radius is calculated as r = d/2 = 1.1 m/2 = 0.55 m

Thus, the sphere's total surface area can be expressed as

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

Consequently, the inward force acting on the sphere equals

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

1 day ago

Read 2 more answers

A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete

Sav [2230]

Answer:

407 steps

Explanation:

Based on the question,

P = mgh/t........... Equation 1

Where P stands for power, m is mass, g denotes gravity, h is height, and t represents time.

Rearranging the equation to solve for h, we have:

h = Pt/mg............. Equation 2

Providing values: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Given constant: g = 9.8 m/s²

By substituting into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

Calculating number of steps: 6525/16

The resulting number of steps = 407 steps

6 0

1 month ago

When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (

inna [2210]

The answer is (C) 4 beats per second. The number of beats is computed as the difference between the frequencies of the two tuning forks. Plugging in the frequency values yields a result. Thus, the number of observable beats per second will be 4.

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6 days ago

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

Sav [2230]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

(m)child(d)child)=(m)rock(d)rock)

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0

26 days ago