A ball with a mass of 4 kg is initially traveling at 2 m/s and has a 5 N force applied for 3 s. What is the initial momentum of the ball?


Answer 1


The initial momentum of the ball is 8 kg-m/s.


Given that,

Mass of the ball is 4 kg

Initial speed of the ball is 2 m/s

Force applied to the ball is 5 N for 3 seconds

It is required to find the initial momentum of the ball. Initial momentum means that the product of mass and initial velocity of the ball. It is given as :

p_i=mu\n\np_i=4\ kg* 2\ m/s\n\np_i=8\ kg-m/s

So, the initial momentum of the ball is 8 kg-m/s.

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A toy car that is 0.12 m long is used to model the actions of an actual car thatis 6 m long. Which ratio shows the relationship between the sizes of the
model and the actual car?
A: 5:1
B: 1:50
C: 50:1
D: 1:5



it is B 1:50


just did it on apex

Final answer:

The ratio that shows the relationship between the sizes of the model car and the actual car is 1:50. This is because the actual car is 50 times longer than the model car.


The relationship between the sizes of the model car and the actual car is represented by a ratio. To find this ratio, we can divide the length of the actual car by the length of the model car. So, 6 m / 0.12 m = 50. This means that the actual car is 50 times longer than the model car, or in other words, the model car is 1/50th the size of the actual car. Therefore, the correct ratio is 1:50.

Learn more about Ratios here:



An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materials have this space/size? (d) Assuming that we can measure the velocity to an accuracy of 10%. Use the Heisenberg uncertainty principle to calculate the uncertainty in the position.



P = 2.91*10^{-24} kg m/s

\lambda = 2.73 *10^(-10) m

size of atom hat lie in range of 1 to 5 Angstrom

\Delta x = 0.2272 Angstrom



p = mv

where m is mass of electron

so momentum p can be calculated as

p = 9.11*10^{-31} *3.2*10^{6}

P = 2.91*10^{-24} kg m/s

b) wavelength

\lambda = (h)/(mv)

where h is plank constant

so\lambda = (6.626*10^(-34))/(2.91*10^(-24))

\lambda = 2.73 *10^(-10) m

c) size of atom hat lie in range of 1 to 5 Angstrom

d) from the information given in the question we have

(\Delta v)/(v) = 0.1

\Delta v = 0.1 v

we know that

\Delta p *\Delta x = (h)/(4\pi)

m \Delta v \Delta x =(h)/(4\pi)

\Delta x = (h)/(m \Delta v)

\Delta x  = (2.272)/(0.1)                      [\Delta v = 0.1 v]

\Delta x = 0.2272 Angstrom

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.



Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

u_(1) = u_(g) = 2553.6 kJ/kg

v_(1) = v_(g) = 0.4625 m^(3)/kg

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, v_(2) = v_(g) = 0.4625 m^(3)/kg

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at v_(2) = v_(g) = 0.4625 m^(3)/kg and temperature T_(2) = 360^(o)C so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

 u_(2) = u_{\text{at 5 bar, 400^(o)C}} + (\frac{v_(2) - v_{\text{at 5 bar, 400^(o)C}}}{v_{\text{at 7 bar, 400^(o)C - v_(at 5 bar, 400^(o)C)}}})(u_{at 7 bar, 400^(o)C - u_(at 5 bar, 400^(o)C)})

     u_(2) = 2963.2 + ((0.4625 - 0.6173)/(0.4397 - 0.6173))(2960.9 - 2963.2)

                   = 2963.2 - 2.005

                   = 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

      Q - W = \Delta U + \Delta K.E + \Delta P.E ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

            \Delta K.E = \Delta P.E = 0

Now, equation will be as follows.

           Q - W = \Delta U + \Delta K.E + \Delta P.E

           Q - 0 = \Delta U + 0 + 0

           Q = \Delta U

Now, we will obtain the heat transfer per unit mass as follows.

          (Q)/(m) = \Delta u

         (Q)/(m) = u_(2) - u_(1)

                      = (2961.195 - 2553.6)

                      = 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

Final answer:

The heat transfer is 227.4 kJ per kg of water.


Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. To determine the heat transfer in kJ per kg of water, we need to calculate the heat absorbed by the water as it reaches 360°C.

Using the specific heat capacity of water (4,186 J/kg°C) and the change in temperature (360°C - 100°C), we can calculate the heat transfer:

Qw = mw * cw * AT = (1 kg) * (4186 J/kg°C) * (360°C - 100°C) = 227,440 J = 227.4 kJ

Therefore, the heat transfer is 227.4 kJ per kg of water.

Heat transfer is the process by which thermal energy moves from one object or substance to another due to a difference in temperature. This fundamental phenomenon occurs through three main mechanisms: conduction, convection, and radiation. Conduction involves the direct transfer of heat through a material, such as metal. Convection is the transfer of heat through the movement of fluids (liquids or gases). Radiation is the emission of electromagnetic waves carrying heat energy. Understanding heat transfer is essential in various fields, including physics, engineering, and environmental science, as it governs temperature regulation, climate dynamics, and the functioning of countless technological devices.

Learn more about Heat transfer here:



An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0% of the speed of light. Its nuclear-powered batteries supply the energy to keep its data transmitter active continuously. The batteries have a lifetime of 15.9 years as measured in a rest frame. (a) How long do the batteries on the space probe last as measured by mission control on Earth? yr
(b) How far is the probe from Earth when its batteries fail, as measured by mission control?
(c) How far is the probe from Earth, as measured by its built-in trip odometer; when its batteries fail?



22.26 years

, 15.585 light years  , 11.13 light years



t' = t/(√(1-(v/(c*v)/c))

= 15.9/√((1-0.7*0.7))

= 22.26 years


0.7*c*22.26 years

=15.585 light years  



=11.13 light years

1. Towards the end of a 400m race, Faisal and Edward are leading and are both running at 6m/s. While Faisal is 72m from the finish line Edward is 100m from the finish line. Realising this and to beat Faisal, Edward decides to accelerate uniformly at 0.2 m/s2 until the end of the race while Faisal keeps on the same constant speed. Does Edward succeed in beating Faisal?





Faisal will finish the race in ...

  (72 m)/(6 m/s) = 12 s

In order to beat Faisal, Edward's average speed in those 12 seconds must exceed ...

  (100 m)/(12 s) = 8 1/3 m/s

To achieve that average speed, Edward's acceleration must be ...

  (8 1/3 m/s -6 m/s)/(12 s/2) = 7/18 m/s² ≈ 0.3889 m/s²

Accelerating at only 0.2 m/s², Edward will not beat Faisal.


Additional comment

When acceleration is uniform, the average speed is reached halfway through the period of acceleration.

The engine of a model airplane must both spin a propeller and push air backward to propel the airplane forward. Model the propeller as three 0.30-m-long thin rods of mass 0.040 kg each, with the rotation axis at one end.What is the moment of inertia of the propeller?
How much energy is required to rotate the propeller at 5800 rpm? Ignore the energy required to push the air.


The moment of inertia of the propeller is 0.0036 kgm² and the energy required is 663.21 J

Energy required for propeller:

Given that the mass of the propellers is m = 0.040kg,

and their length is L = 0.30m

The moment of inertia of a rod with the rotation axis at one end is given by :

I = (1)/(3)m L^2

so for 3 propellers:


I = 0.04 × 0.09

I = 0.0036 kgm²

Now, the frequency is given f = 5800 rpm

so anguar speed, ω = 5800×(2π/60)

ω = 607 rad/s

Energy required:

E = ¹/₂Iω²

E = 0.5 × 0.0036 × (607)² J

E = 663.21 J

Learn more about moment of inertia:


Solution :

Given :

Length of the propeller rods, L =0.30 m

Mass of each, M = 0.040 kg

Moment of inertia of one propeller rod is given by  

$I=(1)/(3)* M * L^2$

Therefore, total moment of inertia is

$I=3 * (1)/(3)* M * L^2$

$I=M* L^2$

$I=0.04* (0.3)^2$

  $0.0036 \ kg \ m^2$

Now energy required is given by

$E=(1)/(2)* I * \omega^2 $

where, angular speed, ω = 5800 rpm

$\omega = 5800 * (2 \pi)/(60) $

 = 607.4 rad/s

Therefore energy,

$E=(1)/(2)* 0.0036 * (607.4)^2 $

   = 664.1 J