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3 years ago

Select the correct answer.

Physics

1 answer:

IgorC [24]3 years ago

7 0

Answer:

D. Newton's Third Law of Motion

Explanation:

Newton's law of gravity is definitely not applicable to your hands. So we can cross this bad boy out

Newton's First Law is F=MA (force equals mass times acceleration). This is basically the root of most physics but it isn't the reason for your hand being red after hitting a wall.

Newton's Second law deals with velocities and forces, so even though you are apply a force your are not changing the velocity of the wall much.

Newton's Third Law basically says that for whatever force you apply to an object, that object will apply an equal and opposite force back to you. This is why your hand gets red. When you slap the wall with all your strength, the wall hits your hand back with the same amount of force. The 2nd law can also be seen when you're trying to push a desk and it won't budge. You are pushing on it, but the desk is pushing back. (there are multiple other factors applicable like friction but we physicists like to ignore them :) )

I hope this helps!

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A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity

WARRIOR [948]

Answer:

The charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg

Explanation:

From the formulae

F = qvB and F = mv²/r

Where F is Force

q is charge

v is speed

B is magnetic field strength

m is mass

and r is radius

Then,

qvB = mv²/r

qB = mv/r

We can write that

q/m = v/rB ---- (1)

Also

From Electric force formula

F = Eq

Where E is the electric field

and magnetic force formula

F = Bqv

Since, electric force = magnetic force

Then, Eq = Bqv

E = Bv

∴ v = E/B

Substitute v = E/B into equation (1)

q/m = (E/B)/rB

∴ q/m = E/rB²

(NOTE: q/m is the charge to mass ratio)

From the question,

E = 3.10 ×10³ N/C

r = 4.20 cm = 0.0420 m

B = 0.360 T

Hence,

q/m = 3.10 ×10³ / 0.0420 × (0.360)²

q/m = 569517.9306 C/kg

q/m = 5.7 × 10⁵ C/kg

Hence, the charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg.

7 0

3 years ago

HELPPPPP MEEEEE PLEASE I NEED TO SUBMIT IN LESS THAN 10 MINSS

DerKrebs [107]

Answer:

X = 2146.05 m

Explanation:

We need to understand first what is the value we need to calculate here. In this case, we want to know how far from the starting point the package should be released. This is the distance.

We also know that the plane is flying a certain height with an specific speed. And the distance we need to calculate is the distance in X with the following expression:

X = Vt (1)

However we do not know the time that this distance is covered. This time can be determined because we know the height of the plain. This time is referred to the time of flight. And the time of flight can be calculated with the following expression:

t = √2h/g (2)

Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:

t = √(2*2500)/9.8

t = 22.59 s

Now replacing into (1) we have:

X = 95 * 22.59

This is the distance where the package should be released.

Hope this helps

6 0

3 years ago

Calculate the force on an object that has a mass of 12kg and an acceleration of 4m/s2.

iVinArrow [24]

<span>By Newton's second law of motion, we know that the resultant force acting on a body is directly proportional to the mass of the body and directly proportional to its acceleration. In system international (SI) units, the value of the constant of proportionality constant is 1. Therefore, the equation for Newton's second law of motion becomes: F = ma, where F is the resultant force, m is the mass and a is the acceleration of the object. Substituting the values of m and a into this formula, we get the result: F = 12 x 4 = 48. The SI unit for force is the Newton; therefore, <u>the answer is 48 Newtons.</u></span>

4 0

3 years ago

Read 2 more answers

If a sample of a radioactive isotope has a half-life of 1 year, how much of the original sample will be left at the end of the s

Tasya [4]

Answer:

1/4 of the original

Explanation:

That would be TWO half lives:

1/2 * 1/2 = 1/4 <======= 1/4 would be left

4 0

1 year ago

For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t

Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

$y=2+\frac{1}{x}$

$y=2+x^{-1}$

$y'=-x^{-2}$

$y'=-\frac{1}{x^{2}}$

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(1)^{2}}$

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-3=-1(x-1})$

$y-3=-1x+1$

$y=-x+1+3$

$y=-x+4$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-x+4=0$

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2)^{2}}$

$m=y'=-\frac{1}{4}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.5=-\frac{1}{4}(x-2})$

$y-2.5=-\frac{1}{4}x+\frac{1}{2}$

$y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}$

$y=-\frac{1}{4}x+3$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{4}x+3=0$

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2.5)^{2}}$

$m=y'=-\frac{4}{25}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.4=-\frac{4}{25}(x-2.5})$

$y-2.4=-\frac{4}{25}x+\frac{2}{5}$

$y=-\frac{4}{25}x+\frac{2}{5}+2.4$

$y=-\frac{4}{25}x+\frac{14}{5}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{4}{25}x+\frac{14}{5}=0$

and solve for x

$x=\frac{35}{20}$

so, when fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(4)^{2}}$

$m=y'=-\frac{1}{16}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.25=-\frac{1}{16}(x-4})$