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Brrunno [24]
3 years ago
9

You launch a baseball through the air at speed 37 m/sec, at angle 45 degrees above horizontal. Neglect air resistance. What shap

e is the trajectory?
Physics
1 answer:
Marizza181 [45]3 years ago
6 0

Answer:

Parabolic motion

Explanation:

It is given that,

Speed with which a baseball is launched through the ar, v = 37 m/s

Angle of projection, \theta=45^{\circ}

We know that when an object is projected with some initial speed having some angle of projection, the path followed by the object is called projectile motion. The object will act as a projectile here. As a result the shape of the trajectory will be parabolic.

The equation of parabolic path is given by :

y=tan\theta\ x-\dfrac{gx^2}{2u^2\ cos^2\theta}

It is similar to,

y=ax+bx^2 which is the equation of parabola

Hence, this is the required solution.                                                                                                    

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According to Ohm's law, a circuit with a high resistance _____. will have a low electric current will have a high electric curre
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Remark
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Formula
I = E/R which comes from E = IR. To get to the derived formula, divide both sides by R
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Discussion
This is an inverse relationship. That means that as one goes up the other one will go down. 

So in this case you keep E constant and you manipulate R and look at your results for I

Case 1

Let us say that E = 10 volts
Let us also say the R = 10 ohms

I = E/R
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Case Two
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As the Resistance goes up, the current goes down. Answer: A
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Considering the factors that affect gravitational pull, in which location would the gravitational pull from the earth be SMALLES
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When something is hit harder how does the transverse wave change?
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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
3 years ago
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