Physical Application of Cal (1 Viewer)

[lyounamu]

A metal bar has a temperature of 1340°C and cools to 1010°C in 12 minutes, when the
surrounding temperature is 25°C. Find how much longer it will take the bar to cool to 60°C,
​

giving your answer correct to the nearest minute.��*​

What's your answer to this? I got 150.5965769 minutes but the answer is different. Answer is 139 minutes. I don't really need the working out, please just tell me what you got. I want to see where I may have made a mistake.

EDIT:

A ball is thrown horizontally from a point O on the edge of a cliff which is 20 metres above a beach and hits the beach at the (-20, 60). Taking g = 10ms-2 find the speed of projection of the second ball. (Standard results about projectile motion can be quoted without proof.)

I got V = 30 but the answer says 60. I am sure mine is right but...

Thanks in advance.
​

 

[undalay]

For the first question, it asks how much longer, so u need to take away 12 from ur 150.6

 

[lyounamu]

For the first question, it asks how much longer, so u need to take away 12 from ur 150.6

Ahhhhhhhhhhhhhhhh~~~

I will seriously have to read the damn question.

 

[Aerath]

I got v = 30, too.

x = vt (cos0 = 1)
y = -5t²+ 20

Since range, y = 0
-5t²+ 20 = 0
t = 2 (t>0)

sub t = 2 and x = 60 (range) into
x = vt
60 = 2v
v = 30m/s

 

[lyounamu]

I got v = 30, too.

x = vt (cos0 = 1)
y = -5t²+ 20

Since range, y = 0
-5t²+ 20 = 0
t = 2 (t>0)

sub t = 2 and x = 60 (range) into
x = vt
60 = 2v
v = 30m/s

Well, now I am assured that my answer may be valid. Thanks.

 

[Aerath]

But there's a seed of doubt in my mind.....how do we know that @ = 0? I just assumed @ = 0, because I didn't know how to figure it out.
[Which isn't the most mathematically correct method]

 

[lyounamu]

But there's a seed of doubt in my mind.....how do we know that @ = 0? I just assumed @ = 0, because I didn't know how to figure it out.
[Which isn't the most mathematically correct method]

Sorry, I forgot to add that the ball was thrown horizontally. I just copied and pasted it but... I guess I accidentally erased it.

 

[undalay]

"A ball is thrown horizontally from a point O on the edge of a cliff which is 20 metres above a beach and hits the beach at the (-20, 60). Taking g = 10ms-2 find the speed of projection of the second ball. (Standard results about projectile motion can be quoted without proof.)"

What second ball?

 

[lyounamu]

"A ball is thrown horizontally from a point O on the edge of a cliff which is 20 metres above a beach and hits the beach at the (-20, 60). Taking g = 10ms-2 find the speed of projection of the second ball. (Standard results about projectile motion can be quoted without proof.)"

What second ball?

There was part one to the question but that question is not related to this one. So I omitted it. (-20, 60) was actually the point that I worked out from the part 1 (which was the right answer).

This question is totally invalid for this one. But if you want, here we go:

A ball is thrown from a point O on the edge of a cliff which is 20 metres above a beach. The
​

ball is thrown with speed 15 2 ms​

-1​

at an angle of 45° above the horizontal. Taking​

g = 10ms-2 show that the ball hits the beach at a point 60 metres along the beach.

 

[undalay]

Mkay, ur right the first part is useless. imo your v=30 is correct.

 

[lyounamu]

A particle is projected from a point, O, on ground level with the velocity of 20 metres per second at an
angle of 60° to the horizontal. After a time T seconds, it reaches a point P, on its upward path, where
the direction of the flight is at 30° to the horizontal. Taking the acceleration due to gravity, g, to be
10m/s,
i. find T

Full solutions will be greatly appreciated. Thanks!!!​

​

 

[undalay]

Initially:
velocity x = 20cos60 = 10
velocity y = 20sin 60 = 10rt3


let x and y, be x and y velocities at time T

tan 30 = y/x
x= initial x velocity = 10

1/rt3 = y/x
y = 10/rt3 at time T

a = -10
v= vsin@ - 10t AT t = 0, y = 10/rt3
10/rt3 = 10rt3 - 10t
-t = 1/rt3 - rt3
t ~= 1.1547

Is that right? hope my wotrking makes sense.

 

[lyounamu]

Initially:
velocity x = 20cos60 = 10
velocity y = 20sin 60 = 10rt3


let x and y, be x and y velocities at time T

tan 30 = y/x
x= initial x velocity = 10

1/rt3 = y/x
y = 10/rt3 at time T

a = -10
v= vsin@ - 10t AT t = 0, y = 10/rt3
10/rt3 = 10rt3 - 10t
-t = 1/rt3 - rt3
t ~= 1.1547

Is that right? hope my wotrking makes sense.

You are awesome! Thanks.

 

[breaking]

read the title and thought you were talking about the person cal.
hehe.

haha i know right

 

[lyounamu]

Initially:
velocity x = 20cos60 = 10
velocity y = 20sin 60 = 10rt3


let x and y, be x and y velocities at time T

tan 30 = y/x
x= initial x velocity = 10

1/rt3 = y/x
y = 10/rt3 at time T

a = -10
v= vsin@ - 10t AT t = 0, y = 10/rt3
10/rt3 = 10rt3 - 10t
-t = 1/rt3 - rt3
t ~= 1.1547

Is that right? hope my wotrking makes sense.

What is this? Shouldn't v be the velocity of the projection?/

EDIT: Don't worry. I get it now. I am just a bit tired. My head cannot absorb information very well.

To shortykatt: I am sorry for the title. I accidentally entered the title while I was writing it. I actually didn't think it would confuse people. Sorry! (but this is in the maths forum though)

 

[undalay]

edit: nvm u got it lol

Yeah my working is messy lol, i'll explain it qualitatively.

Basically, you can work out initial x and y velocity.

Now draw a triangle

This triangle will represent the angles and velocities at time T.

We know the angle will be 30, and x will be initial x.
With this info we can work out velocity of y at T.

Now we find a general formula to work out y' as a function of T.
We substitute our initial Y and our y at time T and we can rearrange the equation to find T.

Hopefully that made sense.
If not i'll present a clearer form of working upon request.

 

[lyounamu]

A particle moves in a straight line and its displacement, x cm, from a fixed origin point after t seconds
is determined by the function: x = sin t - sin t cos t - 2t.​

​

ii. Show that the particle never comes to rest and always moves in one particular direction,
stating what this direction is.

I got V = cos t - cos2t - 2 but I don't know how I can prove that it never becomes 0. I know that it is always negative but how do I show it?

What I did was that I said "whenever cos t becomes o, (i.e. when t=pi/2 or 3pi/2), cos 2t becomes 1 or -1 making it impossible to cancle -2 out to become 0. Therefore, V never becomes 0 and since this is the case, the V is always negative". ​

 

[Aerath]

You show that velocity*acceleration is positive, and therefore, it never slows down. But how you do that...I don't know. Sorry.

 

[lyounamu]

You show that velocity*acceleration is positive, and therefore, it never slows down. But how you do that...I don't know. Sorry.

I can certainly explain but I cannot show. It's not like I can make this -n^2x where n is any integer. This is not even SHM!

Meh.

 

[Aerath]

Well, I dunno how, but if velocity is -ve, then I guess it's always moving left.