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Vinil7 [7]
3 years ago
13

PLEASE AWNSER CORRECTLY AND ILL GIVE BRAINLIEST

Physics
1 answer:
Lera25 [3.4K]3 years ago
4 0
Here you go hope this helps

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What is a molecule called when it has more than one element?
Alinara [238K]
The correct answer is compound
4 0
3 years ago
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A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.
Inessa [10]

Hi there!

We can use Newton's Second Law:
\Sigma F = ma

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}

6 0
2 years ago
What is the amount of charge when 13.5a is flowing for 2 1/2 hours
abruzzese [7]

Answer:

Quick Answer:

a.  8.9Ω

b.  1.2×104 C

Explanation:

4 0
3 years ago
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In Lesson 20, a magnesium strip was used to ignite the thermite reaction. When magnesium is placed in a flame from a small blow
nadezda [96]

Answer:

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

Explanation:

The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:

2Mg(s)+O_2(g)\rightarrow 2MgO(s)

Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.

[Mg]=1s^22s^22p^63s^2

In order to attain noble gas configuration it will loose two electrons.

[Mg]^{2+}=1s^22s^22p^6

Mg\rightarrow Mg^{2+}+2e^-...[1]

Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.

[O]=1s^22s^22p^4

In order to attain noble gas configuration it will gain two electrons.

[O]^{2-}=1s^22s^22p^6

O+2e^-\rightarrow O^{2-}..[2]

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

6 0
4 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
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