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2 years ago

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A teacher uses the model that little invisible gremlins speed up or slow down objects and the direction they push gives the dire

ction of acceleration. This is most similar to A) Newton's first law B) Newton's third law C) Newton's second law D)Conservation of momentum

2 answers:

Lady_Fox [76]2 years ago

6 0

The answer is C) Newton's second law

Vlada [557]2 years ago

4 0

Newtons second law.. <span>The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.</span>

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1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False

Studentka2010 [4]

1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is?
B.6

3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.

4 0

2 years ago

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A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0

2 years ago

Point m is located a distance 2d from the midpoint between the two wires. find the magnitude of the magnetic field b1m created a

Tema [17]

Note: The diagram referred to in the question is attached here as a file.

Answer:

The magnitude of the magnetic field is $B = \frac{0.071 \mu I}{d}$

Explanation:

The magnetic field can be determined by the relationship:

$B = \frac{\mu I}{2\pi R}$ ...............(1)

Were I is the current flowing through the wires

The distance R from point 1 to m is calculated using the pythagora's theorem

$R = \sqrt{d^{2} + (2d)^{2} }$

$R = \sqrt{5d^{2} } \\R = d\sqrt{5}$

Substituting R into equation (1)

$B = \frac{\mu I}{2\pi d\sqrt{5} }$

$B = \frac{0.071 \mu I}{d}$

3 0

2 years ago

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

<u>Answer:</u>

Cannonball will be in flight before it hits the ground for 2.02 seconds

<u>Explanation:</u>

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

2 years ago

If period of the pendulum in preceding sample problem were 24s how tall would the tower be ?

frutty [35]

Answer:

So length of pendulum is 143.129 m

Explanation:

We have given period of simple pendulum is 2 sec

We have to find the length of simple pendulum

Let the length of pendulum is l

Acceleration due to gravity $g=9.8m/sec^2$ is

Time period is given by $T=2\pi \sqrt{\frac{l}{g}}$

So $24=2\times 3.14\times \sqrt{\frac{l}{9.8}}$

$\sqrt{\frac{l}{9.8}}=3.821$

Squaring both side

${\frac{l}{9.8}}=14.60$

l =143.129 m

So length of pendulum is 143.129 m

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2 years ago

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