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3 years ago

6

After the double-blind review policy was instituted, what percentage of published papers had a male first author?

1 answer:

Arte-miy333 [17]3 years ago

4 0

Answer:

hesadghtyou is you here boiii

Explanation:

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(quick help please??)A mechanic uses a mechanical lift to raise a car. The car weighs 10,200 N. The work required to do this was

GuDViN [60]

I believe It would be 2.2m

5 0

3 years ago

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.

Lelechka [254]

Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

Kinetic energy.
Potential energy.

Kinetic energy $=\frac{1}{2} mv^2$

Potential energy = $\frac{Kq_1q_2}{d}$

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

$=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}$

=(1280-337.5)J

=942.5 J

Total energy of a system remains constant.

Therefore,

E $=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}$

$\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750$

$\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750$

$\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}$

$\Rightarrow v= 1961.19$ m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

5 0

3 years ago

Who was th efirst person to come up with the idea of the atoms.

Gwar [14]

Answer:

Democritus

Explanation:

Democritus was a Greek philosopher who was the first person to use the term atom . However ,John Dalton was the first to adapt Democritus’ theory into the first modern atomic

model

7 0

2 years ago

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

From the question we are that

The mass of the particle $m = 1.8*10^{-3} kg$

The charge on the particle is $q = 1.22*10^{-8}C$

The velocity is $\= v = (3.0*10^4 m/s ) j$

The the magnetic field is $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced a force which is mathematically represented as

$F = q (\= v * \= B)$

Substituting value

$F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

$= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

$= (-5.966*10^4 N) \r k$

Note :

$i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\$

Now force is also mathematically represented as

$F = ma$

Making a the subject

$a = \frac{F}{m}$

Substituting values

$a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

$= (-0.33m/s^2)\r k$

$= 0.33m/s^2 * (- \r k)$

6 0

3 years ago

Which of the following is an example or an orginasm mantaining homeostasis

Kryger [21]

a cat grooming itself

8 0

3 years ago