I need this asap pls

The roller-coaster car shown in fig. 6-41 (h1 = 30 m, h2 = 12 m, h3 = 20 m), is dragged up to point 1 where it is released from

maxonik [38]

<span>Since there is no friction, conservation of energy gives change in energy is zero Change in energy = 0 Change in KE + Change in PE = 0 1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0 1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0 (vf^2 - vi^2) = 2 x g x (hi - hf) Since it starts from rest vi = 0 Vf = squareroot of (2 x g x (hi - hf)) For h1, no hf Vf = squareroot of (2 x g x (hi - hf)) Vf = squareroot of (2 x 9.81 x 30) Vf = squareroot of 588.6 Vf = 24.26 For h2 Vf = squareroot of (2 x 9.81 x (30 â€“ 12)) Vf = squareroot of (9.81 x 36) Vf = squareroot of 353.16 Vf = 18.79 For h3 Vf = squareroot of (2 x 9.81 x (30 â€“ 20)) Vf = squareroot of (20 x 9.81) Vf = 18.79</span>

7 0

3 years ago

What effects does the conductor have on the potential field?

liraira [26]

Type of conductors determines rate of flow of current

8 0

2 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago

momentum A proton interacts electrically with a neutral HCl molecule located at the origin. At a certain time t, the proton’s po

arlik [135]

Answer:

$\ m/s$

Explanation:

F = Force =

m = Mass of proton = $1.7\times 10^{-27\ kg$

t = Time taken = $2\times 10^{-14}\ s$

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{}{1.7\times 10^{-27}}\\\Rightarrow a=\ m/s^2$

$v=u+at\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\\\Rightarrow v=\ m/s$

The velocity of the proton is $\ m/s$

6 0

3 years ago

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,

Andre45 [30]

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules$

Power

$P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W$

Converting to hp

$1\ W=\frac{1}{745.7}\ hp$

$\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp$

The power developed by the cheetah is 9241.6 W or 12.39318 hp

7 0

3 years ago

I need this asap pls ​

I need this asap pls