A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.


Answer 1


6.5e-4 m


We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=(1)/(2) kx^(2) +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

(1)/(2) kx^(2) +PE1=PE2

PE2-PE1=(1)/(2) kx^(2)

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

mgh=(1)/(2) kx^(2)

Given m=2100kg


x=11cm=0.11 m





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Diethyl ether has a H°vap of 29.1 kJ/mol and a vapor pressure of 0.703 atm at 25.0°C. What is its vapor pressure at 60.0°C?



The vapor pressure at 60.0°C is 2.416 atm


To solve this problem, we use Clausius-Clapeyron equation

ln(P_2)/(P_1) = (-\delta H)/(R)[(1)/(T_2)-(1)/(T_1)]= (\delta H)/(R)[(1)/(T_1)-(1)/(T_2)]


Initial pressure P₁ = 0.703 atm

Initial Temperature T₁ = 25+273 = 298K

Final temperature T₂ = 60+273 = 333K

Change in enthalpy of vaporization ΔH = 29.1 KJ/mol = 29100J/mol

R is Boltzman constant = 8.314 J/K.mol

ln(P_2)/(P_1) = (29100)/(8.314)[(1)/(298)-(1)/(333)] =1.23449

(P_2)/(P_1) = e^(1.23449)(P_2)/(P_1) = 3.43663

P₂ = P₁ (3.43663) = (0.703 atm)(3.43663) = 2.416 atm

P₂ = 2.416 atm

Therefore, the vapor pressure at 60.0°C is 2.416 atm.

What is the magnitude of a vector that has the following components: x = 32 m y = -59 m




Since the x and y components are given

The vectors Magnitude = √32²+(-59)²


A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?





According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis,  and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

\int\ {E} \, dA = E*A = E*\pi  *R^(2)

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

Flux = E*π*R²

(a) Is the velocity of car A greater than, less than, or the same as thevelocity of car B?
(b) Is the initial position of car A greater than, less than, or equal to the
initial position of car B?
(c) In the time period from t = 0 tot = 1 s, is car A ahead of car B,
behind car B, or at the same position as car B?


a. ) Is the velocity of car A  less than the velocity of car B b. the initial position of car A greater than the initial position of car B  c. ahead In the time period from t = 0 tot = 1 s, is car A ahead of car B?.

what is velocity ?

Velocity is the parameter which is different from speed,  can be defined as the rate at which the position of the object is changed with respect to time, it is basically speeding the object in a specific direction in a specific rate.

Velocity is a  vector quantity which shows both magnitude  and direction  and The SI unit of velocity is meter per second (ms-1). If there is a change in magnitude or the direction of velocity of a body, then it is said to be accelerating.

Finding the final velocity is simple but few calculations and basic conceptual knowledge are needed.

For more details regarding velocity, visit




a. less than, b. greater than, c. ahead


7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, whatforce does the ball experience to accelerate from rest to 73 m/s?



3.65 x mass


Given parameters:

Time  = 20s

Initial velocity  = 0m/s

Final velocity  = 73m/s


Force the ball experience  = ?


To solve this problem, we apply the equation from newton's second law of motion:

    F  =  m (v  - u)/(t)  

m is the mass

v is the final velocity

u is the initial velocity

 t is the time taken


  F  = m ((73 - 0)/(20) )  = 3.65 x mass

Final answer:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, use Newton's second law of motion.


To calculate the force experienced by the ball to accelerate from rest to 73 m/s, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a).

Since the ball starts from rest, its initial velocity (vi) is 0 m/s. The final velocity (vf) is 73 m/s. The time (t) taken for the impact is given as 2 x 10 seconds. So, the acceleration (a) can be calculated using the formula a = (vf - vi) / t.

Substituting the given values into the equation, we have a = (73 - 0) / (2 x 10) = 3.65 m/s^2.

Now, we can find the force (F) using the formula F = m * a. If the mass of the ball is known, we can substitute it into the equation to find the force experienced by the ball.

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A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?



The time for final 15 cm of the jump equals 0.1423 seconds.


The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as



'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

0^2=u^2-2* 9.81* 0.76\n\n\therefore u=√(2* 9.81* 0.76)=3.86m/s

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

v^(2)=3.86^2-2* 9.81* 0.66\n\n\therefore v=√(3.86^2-2* 9.81* 0.66)=1.3966m/s

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as


where symbols have the usual meaning

Applying the given values we get