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olga nikolaevna [1]

3 years ago

12

What name do you give to elements found in group eight on the periodic table

1 answer:

BartSMP [9]3 years ago

6 0

Answer:

Explanation:

Group 8A — The Noble or Inert Gases. Group 8A (or VIIIA) of the periodic table are the noble gases or inert gases: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). The name comes from the fact that these elements are virtually unreactive towards other elements or compounds.

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Does air resistence decrease with speed

dybincka [34]

Answer:

yes

Explanation:

because when you slow down, the resistance slows with the speed.

3 0

4 years ago

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How much is the velocity of a body when it travels 600m in 5 minutes?

maks197457 [2]

answer

so unit of velocity is m/s

displacement=600m

5minutes should be converted to seconds

5×60=300 seconds

so,

velocity= displacement÷time

= 600m ÷300s

=2m/s or 2ms^-1

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2 years ago

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Does the moon fall partly into earth's shadow when it is n "full"?

murzikaleks [220]

Yes, <span> the moon fall partly into earth's shadow when it is in its full size</span>

5 0

4 years ago

2. A stone is thrown horizontally at a speed of 6.0 m/s from the top of a cliff 78.4 m high.

Ostrovityanka [42]

Answer:

a) 8 seconds if you are using earth's gravity.

b) 48m if the velocity does not change

c) 9.8m/s

Explanation:

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

4 years ago