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3 years ago

12

HELP 20 POINTSS!!

1 answer:

crimeas [40]3 years ago

3 0

Mass = force/acceleration
M=24/ 6
m=4 N

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What helps offset the ion diffusion through leak channels on neurons and helps return the membrane potential to its resting volt

Blizzard [7]

Explanation:

Resting potential is the static membrane potential of the dormant cells. It is opposite of the action potential or the graded membrane potential. Na+/K+ ATPase pumps helps offset the ion diffusion through leak channels on neurons and helps return the membrane potential to its resting voltage.

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3 years ago

Why did Thomson observe two glowing dots when he put neon gas into a<br> cathode-ray tube?

Artist 52 [7]

Answer:

Electrons accelerated to high velocities travel in straight lines through an empty cathode ray tube and strike the glass wall of the tube, causing excited atoms to fluoresce or glow.

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3 years ago

Which of the following is NOT a characteristic of a healthy diet?

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Answer: Drinking pop

Explanation:

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3 years ago

Find the coefficient of x3 y4 in the expansion of ( x+2y )^7.

yKpoI14uk [10]

Answer:

The coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.

Explanation:

The given expression is

$(x+2y)^7$

According to binomial expansion,

$(a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+...+^nC_{n-1}a^1b^{n-1}+^nC_0a^0b^n$

The r+1th term of the expansion is

$^nC_rx^{n-r}(2y)^r=^nC_r(2^r)x^{n-r}(y)^r$ ... (1)

In the term x³y⁴ the power of x is 3 and the power of y is 4. It means the value of r is 4 and the value n-r is 3.

$n-r=3$

$n-4=3\Rightarrow n=7$

Put n=7 and r=4 in equation (1)

$^7C_4(2^4)x^{7-4}(y)^4$

$\frac{7!}{4!(7-4)!}(16)x^3y^4$

$\frac{7\times 6\times 5\times 4!}{4!(3)!}(16)x^3y^4$

$560x^3y^4$

Therefore the coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.

5 0

4 years ago

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher

Scilla [17]

Answer:

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

Explanation:

The Coulomb's Law gives the force by the charges:

$\vec{F} = K\frac{q_1q_2}{r^2}\^r$

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

$F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)$

where Θ is the angle between $F_1$ and x-axis.

$F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

whereas

$F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

Finally, the x-component of the net force is

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

8 0

4 years ago