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>> No.4325882 [View]
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4325882

Hai /Sci/. Im doing some Kinematics work today to further expand my understanding of the topic and am stuck on this current question:

2. A ball was thrown from a balcony above a horizontal lawn. The velocity of projection was 10 ms-1 at an angle of elevation α, where tan α = ¾. The ball moved freely under gravity and took 3 s to reach the lawn from the instant when it was thrown. Calculate:

a) the vertical height above the lawn from which the ball was thrown

b) the horizontal distance between the point of projection and the point A at which the ball hit the lawn

c) the angle, to the nearest degree, between the direction of the velocity of the ball and the horizontal at the instant when the ball reached A.

I've easily worked out A and B but im having difficulties understanding C. Could anyone provide some assistance?

For the record (a) = 26.1 m and (b) = 24m.

Any help which points me in the right direction is appreaciated =]

>> No.4053006 [DELETED]  [View]
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4053006

Hai sci

I has a vector velocity p at time t

v=(3t\^{}2 - 12)i + 5tj

its asking me now :

with respect to a fixed origin o, the posiition vector of p when t=2 is 4j. Find the distance OP when t = 1. I integrated this vector to get

r=(t\^{}3-12t+c)i + (5/2 t\^{}2 + d)j

After i sub in t = 2 i then get

(-16+c) i + (10+d)j = 4j.

im tryna work out c and d to do this question what am id oing wrong? any advice? ty

>> No.4002965 [View]
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4002965

hi /sci/. Mechanics/Calculus faggot here. I fucking enjoy doing shit like this i have no idea why. Like everyone in my class hates it but when the teacher sets work i go and fucking do pages of it; not just the one question he sets. Im stumped on one question though. It states:

A particle moves in a straight line and its acceleration after t seconds is 6tms^-2. Its displacement at time t is x metres.Given that x=5 when t=1 and x=14 when t=2 find an expression for x as a fucntion of t and the value of x when t =3.

to solve this ive integrated 6t twice to get

t^3 +ct + d but now im fucking stumped. Ive tried subbing in the x#s and t and get a simultaneous equation but that doesnt seem to be helping me. any clues? thanks

>> No.3987686 [View]
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3987686

hai sci can you help me?

i has

7cosƟ + sinƟ = Rsin(Ɵ+alpha)?

so i expand it and get

7cosƟ + sinƟ = Rsin Ɵ cos(alpha) + RcosƟ sin(alpha)

so tan alpha = 1/7

but answers

7cosƟ + sinƟ = R[sin alpha cosƟ + cos alpha sin Ɵ]

so tan alpha = 7

wut am i doin wrong when i expand?

>> No.2921120 [View]
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2921120

>>2921080
Okay smartass, you are comparing a short mission versus a long term (I guess hundreds of thousands of years) deployment and terraforming Mars to support life.

two very different fucking things.

Protip: Life doesn't like to be constantly bathed in radiation for long periods of time. In fact it hates it so much it dies.

>> No.2881137 [View]
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2881137

Anyone wanna help me on this?


Q4) Two Forces P and Q act on a particle. The force P has magnitude 7N and acts due north. The resultant of P and Q is a force magnitude 10N acting in direction with beaing 120o. Find:

i) The magnitude of Q
ii) The direction of Q giving your answer as a bearing.

mag of q is 14.8 + = 8.66i - 12j

i) was piss easy but i dont get ii for shit. ive tried using tan -1 12 / 8.66 but i cant get it

been on this for over an hour

(1 hour 22 to be precise)

any help? :(

>> No.2760320 [View]
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2760320

>>2760080
does anyone know where I can get me some programs to spontaneously generate novels and books?

>> No.2743043 [View]
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2743043

Hai guise. Physic student fag here.


got exam on tuesday and just realised today so was doing questions online and after looking at a past one saw this:

A stone S is sliding on ice. The stone is moving along a straight line ABC, where AB = 24 m
and AC = 30 m. The stone is subject to a constant resistance to motion of magnitude 0.3 N.
At A the speed of S is 20 m s–1, and at B the speed of S is 16 m s–1. Calculate

(a) the deceleration of S,
(2)
(b) the speed of S at C.

I literally dont have any idea what to do.

Ive tried using suvat

but im not sure what displacement i use and stuff

any hepl? tnx

>> No.2710916 [View]
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2710916

>>2710884
>I'm as atheist as you are.
I sincerely doubt that.

>> No.2654802 [View]
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2654802

the curve y = 2+x - x^ 2 has a tangent at p (1,2)

the tangent meets the x and y axis at l and m

find the coordinates?

any help?

i tried finding the derivative subbing the coordinates in to get -3, form an equation and solve it but its wrong. any help pls?

>> No.2597969 [View]
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2597969

Hi /sci/

Was woondering if you guise could help me with this question

A cyclist and her bicycle have a combined mass of 78 kg. While riding on level
ground and using her greatest driving force, she is able to accelerate uniformly from
rest to 10 m s-1 in 15 seconds against constant resistive forces that total 60 N.

(a) Show that her maximum driving force is 112 N.

Ive tried working out impulse as 150 and final moomentum as 78V but apparnetly doing it wrong.


Can anyone help me please? thnx

>> No.2486475 [View]
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2486475

an arc subtends an angle of 1 radian at the centre of a circle, and a sector of area 72cm is bounded by this arc and the two radii. Find the radius of the circle.

any idea how to do this? Or what the diagram looks like? Thanks

>> No.2387909 [View]
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2387909

For -180<∅<180 solve tan²3∅=1 (use the general equations)?

Hi /sci/

Thakn for helping me on my previous work however im really stuck on this question and asking for help

From my own knowledge i did:

Tan 3∅ = +/- 1

3∅ = +/- 45

∅ = 15

And then by using the general equation

15+180n and -15 +180 n

the range is -180 < ∅ < 180

However It seems i have got it wrong when this is the only solution I can think of. Any help please?

The answers are +/- 45 15 75 105 135 165

>> No.2348302 [View]
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2348302

Need help with this trig shit

If s=sin ∅' and c=cos ∅' simplify:
1) (c^4-s^4)/(c^2-s^4)?
any help?

I keep getting

cos^2∅ - sin^2∅

and the answers 1 so it should be

cos^2∅ +sin^2∅

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