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ch4aika [34]
3 years ago
14

The motor of a four wheeler traveling along a muddy trail generates an average power of 7.05 104 W when moving at a constant spe

ed of 15 m/s. When pulling a log along the trail at the same speed, the engine must generate an average power of 8.15 104 W. What is the tension in the tow rope pulling the log?
Physics
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

F=733.33 N

Explanation:

Power is the work to make a motion in a determine time also the al the work is the force make in a distance so:

P=\frac{W}{t}

W=F*d

Replacing

P=F*\frac{d}{t}

Knowing that the velocity is:

v=\frac{d}{t}

So:

P=F*v

Solve to F to determinate the tension that in this case is the force in the rope

F=\frac{P}{v}

The power is the total power so:

P=8.15x10^4W-7.05x10^4W=11x10^3W

F=\frac{3kW}{15\frac{m}{s}}=733.33 N

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Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
On Earth, plasma exists in the ionosphere, in flames, and in chemical and nuclearexplosions. Matter in a controlled thermonuclea
nasty-shy [4]

The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.

<h3>How did the cost of developing thermonuclear power defended?</h3>

The  cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of  the occurrence of plasmain nature,

It should be noted that this  thermonuclear power with plasmabe  posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.

In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.

Therefore, option D is correct.

Read more about cost at:

brainly.com/question/25109150

#SPJ1

3 0
1 year ago
Jay pushes the box with a force of 25 Newtons. There are 5 Newtons of friction present. In what direction will the box accelerat
yKpoI14uk [10]

Answer: To the right

Explanation:

8 0
1 year ago
A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the
Mademuasel [1]

Answer:

a)

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b)

m = 48lb

c)

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)    (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s

Then, you obtain by replacing in (1):

6in = 0.499 ft

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b.

mass, m = 48lb

c.

b = 144.76 lb/s

8 0
3 years ago
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