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leva [86]
3 years ago
15

How is a projectile different from an object in free fall

Physics
1 answer:
aksik [14]3 years ago
7 0

Answer:

Explanation:

Projectile Motion. Projectile motion is different than free fall: it involves two dimensions instead of one. ... Balls traveling in two dimensions, only one of which experiences acceleration, require two sets of equations: one set for the x-direction and the other for the y-direction.

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Given theta = 7pi/6, find (sec theta, cos theta) ...?
uysha [10]
There is an identity of sin(2pi-x) = -sin(x) and cos(2pi-x) = cos(x). This is what we are going to use. 


<span>(7pi/6) = (2pi)-(pi/6)

 </span>
Therefore: 
<span>
1) sec(7pi/6) = 1/cos(2pi-(pi/6)) = 1/cos(pi/6) = 2sqrt(3)/3 
</span>
<span>2) cos(7pi/6) = cos(pi/6) = sqrt(3)/2


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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5 0
3 years ago
two soccer players are running for the same ball during a game. Without looking the two collide. One player has a mass of 45 kg
jeka57 [31]

Well yuh see I dont got a clue

6 0
4 years ago
In which situations is gravitational potential energy present?
Assoli18 [71]
Gravitational potential energy is associated with the shape or position of an object.
1.)When an object is placed at height h above ground, gravitational potential energy associated with it is given by,
P.E = mgh
2.)In projectile motion during upward motion, kinetic energy of object is converted into potential energy. 

6 0
3 years ago
What is the wavelength of a radiowave that has a frequency of 9.650 x10^7 Hz (cycles per second)?h = 6.6261 x10^-34 J/sc = 2.998
slava [35]

Hello!!

For this, we only need frequency and speed, so for calculate we applicate formula:

\boxed{\lambda = V/f}

\textbf{Being:}

\sqrt{} \lambda = Wavelength = \   ?

\sqrt{} V = Velocity = 2.998*10^{8} \ m/s

\sqrt{} f = Frequency = 9,650*10^{7} \ Hz

\text{Then let's \textbf{replace it according} we information:}

\lambda = 2,998*10^{8} \ m /s  / 9,650*10^{7} \ Hz

\lambda  =  3.106 \ m

\textbf{Result:}\\\text{The wavelength of the radiowave is \textbf{3.106 meters}}

4 0
3 years ago
Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation
gogolik [260]

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

3 0
3 years ago
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