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3 years ago

Which step should you take before you begin a scientific investigation?

Physics

2 answers:

Hoochie [10]3 years ago

6 0

Answer:

The correct answer is C.

Clarify any confusing information in the instructions.

Explanation:

The very first thing that we need before conducting a scientific investigation is a set of instructions which might include the following information:

-What we are experimenting on?

-How are we going to do it?

-What things are needed for it?

If there are any confusions in the instructions, the chances of making a mistake rises. So, it is very important to clarify the confusions found in the instructions before choosing things needed for the experiment and starting the experiment.

ivanzaharov [21]3 years ago

4 0

Answer:

Explanation:

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The motion of an electron is given by x(t)=pt3+qt2+r, with p = -2.3 m/s^3 ,q = +1.5 m/s^2 , and r = +9.0 m.A) Determine its velo

alexandr402 [8]

Answer:

$v(0)=0\\v(1)=-3.9\ m/s\\v(2)=-21.6\ m/s\\v(3)=-53.1\ m/s$

Explanation:

<u>Instant Velocity </u>

Given the position as a function of time x(t), the instant velocity is the derivative of the function:

$v(t)=x'(t)$

We are given the position as

$x(t)=-2.3t^3+1.5t^2+9$

The derivative of x is

$v(t)=x'(t)=-6.9t^2+3.0t$

A) Let's compute v(0)

$v(0)=-6.9(0)^2+3.0(0)=0$

$v(1)=-6.9(1)^2+3.0(1)$

$v(1)=-3.9\ m/s$

$v(2)=-6.9(2)^2+3.0(2)$

$v(2)=-21.6\ m/s$

$v(3)=-6.9(3)^2+3.0(3)$

$v(3)=-53.1\ m/s$

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3 years ago

Two men are trying to carry a wooden pole. If one of them is weaker than other, how can they carry the pole hence making small l

Romashka-Z-Leto [24]

Answer: Answer down below.

Explanation:

6 0

2 years ago

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Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy

Dominik [7]

a) 0.261 T

b) this field strength is obtainable with today's technology

Explanation:

The force experienced by a charged particle moving perpendicular to a magnetic field is given by

$F=qvB$

where

q is the charge

v is the velocity

B is the strength of the field

This force is perpendicular to the motion of the particle, which therefore moves in a circular path; and so, this force acts as centripetal force, so we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the circle

In this problem, we have:

$q=1.6\cdot 10^{-19}C$ (magnitude of the charge of antiprotons)

$v=5.00 \cdot 10^7 m/s$ (velocity)

$m=1.67\cdot 10^{-27}kg$ (mass of antiprotons)

r = 2.00 m (radius)

Therefore, we can re-arrange the equation and solve to find B, the magnetic field strength:

$B=\frac{mv}{qr}=\frac{(1.67\cdot 10^{-27})(5.00\cdot 10^7)}{(1.6\cdot 10^{-19})(2.00)}=0.261 T$

The strength of the magnetic field calculated in part A) is

$B=0.261 T$

This is indeed a very strong magnetic field. In fact, by comparison, the Earth's magnetic field has a strength of about

$B_{earth}=5\cdot 10^{-5} T$

However, there are current technologies available that are able to produce such strong fields. For instance, the superconducting magnets in the LHC (Large Hadron Collider) are able to produce magnetic fields of strength up to 8 Tesla (8 T).

Therefore, we can say that this field strength is obtainable with today's technology.

5 0

3 years ago

Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc

Vanyuwa [196]

Answer:

$1\cdot 10^{-8} N$ to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

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

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

$q_1=2nC=2\cdot 10^{-9}C$ is the magnitude of charge 1

$q_2=5nC =5\cdot 10^{-9}C$ is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

$F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N$

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

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Oksanka [162]

Mechanical or Electromagnetic

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3 years ago

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