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mojhsa [17]
3 years ago
10

Would u rather/ tiktoc famous or insta famous

Physics
2 answers:
IRISSAK [1]3 years ago
5 0
Answer: Both lollllll
Triss [41]3 years ago
3 0

Answer:

neither

Explanation:

You might be interested in
*PLEASE HELP*
Westkost [7]

Answer:

640.5

Explanation: i got it right on acellus

5 0
3 years ago
Read 2 more answers
The p wave on an electrocardiogram represents __________.
Harrizon [31]

It represents the depolarization of the atria.

<u>Explanation:</u>

The P wave speaks to the depolarization of the left and right chamber and furthermore relates to atrial compression. Carefully, the atria contract a brief moment after the P wave starts. Since it is so little, atrial repolarization is typically not unmistakable on ECG.

In an ordinary ECG, there's three particular waves. The primary wave is the P wave, which speaks to the depolarization of the atria. This happens directly before the atria agreement and drive blood into the ventricles. The following wave is known as the QRS wave.

3 0
3 years ago
A rope is pulling a sled is sloped at 25 degrees upward at the point of contact. The tension in the rope is 40.0N. What is the h
pshichka [43]
T = 40.0 N

angle = 25°

Trigonometric ratios:

sin(25°) = vertical component of the force / force

cos(25°) = horizontal component of the force / force

tan (25°) = vertical compoent of the force / horizontal component of the force.

From cos(25°) you can find the horizontal component of the force:

horizontal component of the force = force * cos(25°)

The force is the tension, 40.0 N.

horizontal component of the force = 40.0 N * cos (25°) = 36.25 N

Answer: 36.25 N
7 0
4 years ago
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo
Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              F = k\frac{|q_1|.|q_2|}{r^2}

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1 ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                         

                          q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

 

6 0
4 years ago
What is the speed of a wave with a frequency of 100 hz and a wave length of .5 m?
____ [38]

Answer:

Explanation:

There are 2 ways to help with this. Explain the details, which are fairly simple in this topic, or give the formula. My hope is that an explanation will last longer than memorizing the formula. I give you both.

If a wave has frequency, f, of 3 Hz, its period, T, is

1

3

s

. The wavelength,

λ

, is 5 meters. That means that in the time of one period, the wave travels 5 m.

In general,

S

p

e

e

d

=

distance

time

In applying this general definition of speed

↑

to a wave, we have

speed of the wave

=

wavelength

period

Note: we generally use v for speed of a wave. Using the variable names, then that last formula is written

v

=

λ

T

Since

T

=

1

f

, we can also say that

v

=

λ

⋅

f

So, using that last formula

v

=

5

m

⋅

3

H

z

=

15

m

s

Note: the unit Hz is equivalent to what it was called 100 years ago,

cycles

second

(

also cps

)

. Cycles is not a true unit, so the Hz contributed only the "per second" to the result

15

m

s

.

7 0
3 years ago
Read 2 more answers
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