How much longer will it taketo travel a distance of 6ookm at
a speed of 50kmh than it
would at a
speed of 6okmh?


Answer 1


2hr much longer


Given parameters

  Distance  = 600km

   Speed 1  = 50km/h

   Speed 2 = 60km/h


How much longer will it take to travel a distance  = ?


  Speed is the distance divided by time;

          Speed  = (distance)/(time)  


           Time taken  = (Distance)/(Speed)  

Time 1;

                        = (600)/(50)

                       = 12hr

Time 2;

                       = (600)/(60)

                       = 10hr

To find how much more time;

            Time 1 will take 12hr - 10hr, 2hr much longer to travel the distance at that rate.

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In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value:g=9.80m/s (s is squared))



a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %


Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of  g = 9.8 m/s²


The difference Δ g =  9.96 -9.72 =0.24  m/s²

The\ percentage\ difference=(0.24)/(9.72)* 100=2.46\ percentage\n


For first one :

Error\ in\ the\ percentage =(9.96)/(9.81)* 100 =101.52\ perncetage

For second  :

Error\ in\ the\ percentage =(9.72)/(9.81)* 100 =99.08\ perncetage


The mean g(mean )

g(mean )=(9.96+9.72)/(2)\ m/s^2\ng(mean)=9.84\ m/s^2

The\ percentage=(9.84)/(9.8)* 100=100.40\ percentage

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Final answer:

The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.


In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:

|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%

The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:

For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%

For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%

The percent error of the mean involves doing the above but using the mean of the experimental measurements:

|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%

Learn more about Percent Difference and Error here:


A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.How fast is the skier moving when she gets to the bottom ofthe hill?



The given data is as follows.

      Mass, m = 62 kg,       Initial speed, v_(1) = 6.90 m/s

 Length of rough patch, L = 4.50 m,      coefficient of friction, \mu_(k) = 0.3

 Height of inclined plane, h = 2.50 m

According to energy conservation equation,

     (Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction

     K.E_(2) + U_(2) = K.E_(1) + U_(1) - W_(f)

    (1)/(2)mv^(2)_(2) + U_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL

Since, final potential energy is equal to zero. Therefore, the equation will be as follows.

    (1)/(2)mv^(2)_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL    

Cancelling the common terms in the above equation, we get

     (1)/(2)v^(2)_(2) = (1)/(2)v^(2)_(1) + gh - \mu_(k)gL

                         = (1)/(2)(6.90)^(2)_(1) + 9.8 m/s^(2) * 2.50 m - (0.3 * 9.8 * 4.50 m)

                         = 36.055 - 13.23

                         = 22.825

               v_(2) = √(2 * 22.825)

                           = 6.75 m/s

Thus, we can conclude that the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.



mass, m = 62 kg

initial velocity, u = 6.9 m/s

length, l = 4.5 m

height, h = 2.5 m

coefficient of friction, μ = 0.3

Final kinetic energy + final potential energy = initial kinetic energy + initial potential energy + wok done by friction

Let the final velocity is v.

0.5 mv² + 0 = 0.5 mu² + μmgl + mgh

0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.5

0.5 v² = 23.805 + 13.23 + 24.5

v² = 123.07

v = 11.1 m/s  

Give a quantitative definition of being in contact.


Two things are said to be in contact if the smallest distance between a point in one of them and a point in the other one is zero.

Which type of energy refers to the sum of potential and kinetic energies in the particles of a substance?Omotion
O internal
O heat


Final answer:

Internal energy is the sum of potential and kinetic energies in the particles of a substance.


The type of energy that refers to the sum of potential and kinetic energies in the particles of a substance is internal energy.

Internal energy is the total energy of all the particles in a substance, including the energy associated with their motion and stored energy due to their positions or arrangements. It consists of both potential energy (energy stored in the particles' positions) and kinetic energy (energy of the particles' motion).

For example, consider a gas in a container. The internal energy of the gas would be the sum of the potential energy of the gas particles due to their positions and the kinetic energy of the particles due to their motion.

Learn more about Internal Energy here:

Light from a sodium vapor lamp (λ-589 nm) forms an interference pattern on a screen 0.91 m from a pair of slits in a double-slit experiment. The bright fringes near the center of the pattern are 0.19 cm apart. Determine the separation between the slits. Assume the small-angle approximation is valid here.



separation between the slits is 0.28 mm


given data

wave length λ = 589 nm = 589 × 10^(-9) m

distance between slits and the screen D = 0.91 m

fringes weight y = 0.19 cm = 0.19 × 10^(-2) m


we find here the spacing between the two slits i.e d

so use here formula that is

y = λD ÷ d       .........................1

put here value we get

0.19 × 10^(-2) = (589*10^(-9)*0.91)/(d)

solve we get

d = 0.28 mm

A electromagnetic wave of light has a wavelength of 500 nm. What is the energy (in Joules) of the photon representing the particle interpretation of this light?



Energy, E=4.002* 10^(-19)\ J


It is given that,

Wavelength of the photon, \lambda=500\ nm=5* 10^(-7)\ m

We need to find the photon representing the particle interpretation of this light. it is given by :


E=(6.67* 10^(-34)* 3* 10^8)/(5* 10^(-7))

E=4.002* 10^(-19)\ J

So, the energy of the photon is 4.002* 10^(-19)\ J. Hence, this is the required solution.