Recent content by JackandJones

Okay its just that the answer doesn't seem correct to me. Fnetx = 23.87 N Fnety = 0.26 N Using PT: Fnet = 23.87 N Then then using Trig Ratios, the Angle = 0.62 degrees... which doesn't seem correct..?

Okay perfect However could you please explain why I would not substract the F1y by the F2y? Because F2y is a downward motion, it would not be negative?

If I have Fnetx and Fnety Would I use pythegorean theorem to find Fnet, and then Trig ratio to get the angle?

Hey guys, I am having a little trouble determining the final answer to this question. Basically an object is being pulled by two forces and I am suppose to find the Net Force F1 = 12 N at an angle of 32 degrees (upwards) F2 = 15 N at an angle of 24 degrees (downwards) So what I did was...

No no no no. I understood what was being said: I just had to incorporate it into previous equations which I found challenging: Overall I figured it out and got this: which was correct on my exam. Ex: Δt = 42 m /34 m/s x cosθ 0 = v1 x sin x (42 m /34 m/s x cosθ) + ½ a(42 m /34 m/s x...

Okay awesome. Can you should me what the final answer is?

If Anyone could just tell me how to do part b with the numbers I have in the problem: it would be greatly appreciated! I really need to get this question done.

x = Vxt so x = velocity x time x = a x b I still don't get what you are trying to show me

x= Vxt Vx (32.33 horizontal velocity for 18 degree angle) x t (2.1 s) x = 67.89 m

Okay. But what i don't understand is, you are looking for a new angle u are only given the distance. Time and initial velocity change. So using your equation I can't see how to work it out to find the new angle given initial velocity for the older 18 degree angle and the new distance of 42 m.

Okay I see, can u give me an example?

Okay I have: Vvertical = 34.0 m/s x sin18 = 10.5 m/s Vhorizontal = 34 m/s x cos18 = 32.33 m/s Dvertical= v1xt + 1/2a x t^2 0 = (10.5 m/s x t - 4.9 m/s^2 x t ^2) t = 2.1 seconds Dhorizontal = Vh x t = 32.33 m/s x 2.1 s = 67.8 m Therefore distance traveled is 67.8 m, it will...

The Question: A water balloon is fired 34 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon's target (which has a radius of 1.0m) is painted on the asphalt 42m away from the water cannon. a) Will the balloon hit the target? Justify your...

I got 25 km N 65 degrees E...that can't be right, can u try and calculate the final answer?

Okay I tried doing that...but I am not sure it is correct. Could you show me some calculations of what you got? And/or show me a diagram?! I appreciate the help!