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marin [14]
3 years ago
11

Where does every piece of matter begin?

Physics
1 answer:
Margarita [4]3 years ago
6 0
Every piece of matter begins “Out of this world”
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1. Which of these is not a natural fibre?<br>a. leather<br>b. jute<br>C.Wool<br>d. cotton​
Oksanka [162]

Answer:

leather

Explanation:

plz mark as brainliest......hope it helps

6 0
3 years ago
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Please help.. I'm desperate.
rosijanka [135]

Answer:

Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry.

Explanation:

7 0
3 years ago
It takes you 8.3 min to walk with an average velocity of 1.6 m/s to the north from the bus stop to the museum entrance. How far
Anarel [89]

solution:

1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)  

96 m/min x 8.3 min = 796.8 m

s=ut +\frac{1}{2}at^2\\there is no accleration mentioned so,\\s= uv\\8.3\times60=498(s)\\510\times1.6=816(m)


3 0
3 years ago
A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

4 0
3 years ago
Which shows the formula for converting from kelvins to degrees Celsius? °C = (9/5 × K) +32 °C = 5/9 × (K – 32) °C = K – 273 °C =
madam [21]

The freezing point of the water is 0 C , and it equals to 273 K

Then, To convert from Kelvins degrees to Celsius degrees we use the relation

K = C + 273

Also,

C = K - 273

5 0
3 years ago
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