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3 years ago

What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?

Physics

1 answer:

kvv77 [185]3 years ago

5 0

Answer:

$W = 78.48N\\W =0.0784kN\\W = 8kgf\\$

$W= 3.6lbf\\W= 115.2 lbm*ft/s2$

Explanation:

$m=8kg=3.6lb\\$

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

$W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\$

$W= 3.6lbf\\W= 3.6lbm *32.16 ft/s2 =115.2 lbm*ft/s2$

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An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a s

Dmitriy789 [7]

Answer:

0.17724 m/s²

Explanation:

D = Diameter of roll = Length of wing = 11 m

T = Time it takes to complete the circle = 35 s

Velocity

$v=\frac{2\pi R}{T}\\\Rightarrow v=\frac{\pi D}{T}\\\Rightarrow v=\frac{\pi\times 11}{35}\\\Rightarrow v=0.98735\ m/s$

Acceleration

$a=\frac{v^2}{R}\\\Rightarrow a=\frac{0.98735^2}{\frac{11}{2}}\\\Rightarrow a=0.17724\ m/s^2$

Acceleration of the tip of the plane is 0.17724 m/s²

3 0

3 years ago

A motorboat heads due east at 16 m/s across a river that flows due south at 9.0 m/s. a.) What is the resultant velocity of the b

alukav5142 [94]

Answer:

a)V=18.35 m/s (South -East)

b) t =7.41 m/s

c)D= 66.70 m

Explanation:

Given that

Velocity of boat in east direction = 16 m/s

Velocity of river = 9 m/s

a)The resultant velocity V

$V=\sqrt{16^2+9^2}\ m/s$

V=18.35 m/s (South -East)

We know that

Distance = Velocity x time

Lets t time takes to cross the river

136 = 18.35 x t

t =7.41 m/s

The distance covered downstream

We know that

Distance = Velocity x time

t= 7.41 s

D= 7.41 x 9 m

D= 66.70 m

3 0

3 years ago

Suppose that the acceleration of a model rocket is proportional to the difference between 100 ft/sec and the rocket's velocity.

sp2606 [1]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 80 ft/s

acceleration, a = 150 ft/s²

Let the time taken is t.

v = u + at

80 = 0 + 150 x t

t = 0.53 second

3 0

3 years ago

A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s

posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

$P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m$

The spring constant of the net is 20130.76 N

From Hooke's Law

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m$

The net would strech 0.03167 m

If h = 35 m

From energy conservation

$65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0$

Solving the above equation we get

$x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45$