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3 years ago

15

Plz plz please answer it fast urgent

2 answers:

marusya05 [52]3 years ago

8 0

Answer:

A.-2

B.its 3/20=0.15

C.3/4=0.75

Reil [10]3 years ago

4 0

Answer - 1) -6/3

2)3/20

3)3/4

Hope this may helps you

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Solve for t.<br><br>3/4t = 1/4

Vlad1618 [11]

I hope this helps you

3 0

4 years ago

1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

As

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

so

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$

$\mathrm{Cancel\:the\:common\:factor:}\:10^{-12}$

$=\frac{12.6}{35.4\pi \left(10^{-2}\times \:2.5\right)^2}$

$=\frac{12.6}{0.025^2\times \:35.4\pi }$ ∵ $\left(10^{-2}\times \:2.5\right)^2=0.025^2$

$=\frac{12.6}{0.022125\pi }$ ∵ $35.4\pi\times 0.025^2=0.022125\pi$

$=\frac{12.6}{0.06950 }$

F $=181.274$ N

Therefore, the magnitude of the electric force between these two objects will be: 181.274 N.

i.e. F $=181.274$ N

7 0

3 years ago

29. New grass seeds grow rapidly. A grass seed has grown to 0.3 millimeters tall. Tomorrow it will be 0.9 millimeters

Cerrena [4.2K]

Answer:

The answer is a(n)=12+(n-1)11; 166 millimeter

3 0

3 years ago

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second

Molodets [167]

Given:

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second.

To find:

The equation for the given situation if the sum of the numbers is 77.

Solution:

a. Let the first number in the set is x.

The second is 3 more than the first. So, the second number is $(x+3)$ .

The 3rd number is a square of the second. So, the third number is $(x+3)^2$ .

Therefore, the first, second and third numbers are $x,(x+3),(x+3)^2$ respectively.

b. The sum of the numbers is 77.

First number + Second number + Third number = 77

c. So, the equation in terms of x is:

$x+(x+3)+(x+3)^2=77$

Therefore, the required equation is $x+(x+3)+(x+3)^2=77$ .

d. On simplification, we get

$x+x+3+x^2+6x+9=77$ $[\because (a+b)^2=a^2+2ab+b^2]$

$x^2+(6x+x+x)+(3+9)=77$

$x^2+8x+12=77$

Subtract 77 from both sides.

$x^2+8x+12-77=77-77$

$x^2+8x-65=0$

Therefore, the simplified form of the required equation is $x^2+8x-65=0$ .

6 0

3 years ago

Square root of 75000

Oduvanchick [21]

The square root of 75000 is 273.86

7 0

4 years ago

Read 2 more answers