In a hydraulic lift, the radius of the small and large pistons are

A 57 g ice cube can slide without friction up and down a 33 ∘ slope. The ice cube is pressed against a spring at the bottom of t

beks73 [17]

The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy.

0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h 

0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h 

∆h = 0.51 m = 51 cm 

This is the height gained 
Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm 

In the second case, the stored spring energy is converted into height energy AND frictional heat energy. 

The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. 

The Frictional energy converted is F . d 

F ( the frictional force ) = µ . N 

N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 

Total energy converted 

0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) 

Solve for d 

d = 0.528 = 53 cm

5 0

3 years ago

A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2

stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

$s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2$

For the bottom of window (position B)

$s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2$

$\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673$

We also have

$t_B-t_A=0.27$

Solving

$t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s$

So after 1.11 seconds ball reaches at top of window,

We have equation of motion v = u + at

$v_A=0+9.81\times 1.11=10.89m/s$

Speed at which the ball passes the window’s top = 10.89 m/s

7 0

3 years ago

70 pointss yall !!! helpp

Ilya [14]

A: the type of plant

B: how tall the plant is

5 0

3 years ago

Read 2 more answers

A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c

natita [175]

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

$mg = \rho V_{displaced} g$

$V_{displaced} = \frac{m}{\rho}$

$V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3$

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

$V\rho_w = m_{ice}$

$V = \frac{0.200}{1000} = 2\times 10^{-4} m^3$

So here extra volume that rise in the level will be given as

$\Dleta V = V - V_{displaced}$

$\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}$

$(\pi (0.039^2) h = 0.41 \times 10^{-4}$

$h = 0.86 cm$

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

5 0

4 years ago

How does mass effect potential and kinetic energy on a roller coaster?

PilotLPTM [1.2K]

Potential energy is measured by mass * gravity * height. So, the larger the mass on a roller coaster, the more potential energy it has.

Also, the higher it is, the more potential energy it has.

6 0

3 years ago