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Blababa [14]
3 years ago
7

Small rock substances that orbit the sun in groups is called

Physics
1 answer:
Sav [38]3 years ago
5 0
Asteroids revolve around the sun in groups. There is a belt of them between Mars and Jupiter.
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A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe
Levart [38]

Answer:

Coefficient of static friction will be equal to 0.642  

Explanation:

We have given acceleration a=6.3m/sec^2

Acceleration due to gravity g=9.8m/sec^2

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to a=\mu g, here \mu is coefficient of static friction and g is acceleration due to gravity

So \mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642

So coefficient of static friction will be equal to 0.642

3 0
3 years ago
Abcdefghijklmnopqrstuvwxyz
AlladinOne [14]

Answer:

lhgwljvqlivlajvliavpjavphvalhvqlhvqlhv

Explanation:

wlhfsougspuva???

6 0
2 years ago
Read 2 more answers
Which is the currrent SI unit to use in measuring the amount of material contained in a solid sphere
olya-2409 [2.1K]
Hey

SI unit for measure the amount of material contained in a solid sphere metre cube(m^3).

mean it have Volume .
3 0
3 years ago
Which of the following best describes why understanding a watershed and its boundaries is important in designing housing develop
vagabundo [1.1K]

Answer:

d

Explanation:

5 0
2 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

The compression of the net is 1.52 m

4 0
3 years ago
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