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vovikov84 [41]
3 years ago
12

How much tension must a rope withstand if it is used to accelerate 70.0 skier behind a boat at 1.50m/s^2?

Physics
1 answer:
SIZIF [17.4K]3 years ago
6 0
It will  be stand 46.67 all i did was divide both numbers but im not sure if im right but i hope i am hope i helped:)
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Kelvin, Celsius, and Fahrenheit are three types of<br> scales
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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
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Answer:

a) \Delta{t} = 5.39s

b) the motorcycle travels 155 m

Explanation:

Let t_2-t_1 = \Delta{t}, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}

where:

v_{m2} is the speed of the motorcycle at time 2

v_{c} is the velocity of the car (constant)

v_{0} is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

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The crest of one wave with an amplitude of 4 m meets up with the crest of a second
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Explanation:

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A rocket is fired at 100 m/s at an angle of 37, how many seconds did it take to get to the top?
Kobotan [32]

Answer:

6.14 s

Explanation:

The time the rocket takes to reach the top is only determined from its vertical motion.

The initial vertical velocity of the rocket is:

u_y = u sin \theta = (100)(sin 37^{\circ})=60.2 m/s

where

u = 100 m/s is the initial speed

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Now we can apply the suvat equation for an object in free-fall:

v_y = u_y +gt

where

v_y is the vertical velocity at time t

g=-9.8 m/s^2 is the acceleration of gravity

The rocket reaches the top when

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So by substituting into the equation, we find the time t at which this happens:

t=-\frac{u_y}{g}=-\frac{60.2}{-9.8}=6.14 s

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