A helium nucleus contains two protons and two neutrons. The mass of the helium nucleus is greater than the combined masses of two protons and two neutrons because binding energy has been added. True False

Answer:

[tex]\frac{dQ}{dt} = 966 W[/tex]

Explanation:

As we know that the rate of heat transfer due to temperature difference is given by the formula

[tex]\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}[/tex]

here we know that

[tex]K = 0.69 W/m-K[/tex]

A = 4 m x 7 m

thickness = 30 cm

temperature difference is given as

[tex]\Delta T = 20 - 5 = 15 ^oC[/tex]

now we have

[tex]\frac{dQ}{dt} = \frac{(0.69W/m-K)(28 m^2)(15)}{0.30}[/tex]

[tex]\frac{dQ}{dt} = 966 W[/tex]

Answer:

The ball remain in air for 3.06 seconds.

Maximum height reached = 22.94 m

Explanation:

We have equation of motion v = u + at

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0 = 15 -9.81 t

        t = 1.53 s

Time of flight = 2 x 1.53 = 3.06 s

The ball remain in air for 3.06 seconds.

We also have equation of motion v² = u² + 2as

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0² = 15² - 2 x 9.81 x s

        s = 22.94 m

Maximum height reached = 22.94 m

Answer:

[tex]I(t)=4[1-e^{-2t}][/tex]

Explanation:

For a LR circuit as shown the current at any time t in the circuit is given by

[tex]I(t)=\frac{V}{R}[1-e^{\frac{-Rt}{L}}][/tex]

where

'V' is the voltage

'R' is resistance in the circuit

'L' is the inductance of the circuit

't' is time after circuit is turned on

Applying the given values we get

[tex]I(t)=\frac{32}{8}[1-e^{\frac{-8t}{4}}]\\\\I(t)=4[1-e^{-2t}][/tex]

Answer:

Elastic collision is that collision in which two objects are collides and separate out without and deformation in shape and size

example: if a block connected with ideal spring will collide with another block in which spring will compress during the collision and then again come to its natural state.

Inelastic collision in that in which two objects will collide and then stick with each other after collision and there is permanent deformation in the objects.

example : if two cars will collide and stuck together after collision and move together with same speed

So here in both examples and both type of collision there is no external force during collision of two objects so here total momentum of the system will remains conserved.

But since the shape will change in inelastic collision so kinetic energy is not conserved in inelastic collision while in elastic collision kinetic energy is conserved

Final answer:

An example of an elastic collision is two colliding billiard balls, with both momentum and kinetic energy conserved, while an inelastic collision, such as a car crash, conserves momentum but not kinetic energy.

Explanation:

An example of an elastic collision is a collision between two billiard balls where they bounce off of each other without any deformation or generation of heat. In this case, both momentum and kinetic energy are conserved. Momentum conservation means the total momentum before the collision is equal to the total momentum after the collision. Similarly, kinetic energy conservation implies that the total kinetic energy before the collision is the same as the total kinetic energy after the collision.

In contrast, an example of an inelastic collision is a car crash, where the cars collide and deform, often sticking together. In this type of collision, while again momentum is conserved, kinetic energy is not conserved because some of it has been transformed into other forms of energy, such as heat or sound. The reduction in internal kinetic energy is a hallmark of inelastic collisions.

The conservation of momentum and kinetic energy together allow us to calculate the final velocities of objects in one-dimensional, two-body collisions. However, in inelastic collisions, energy is not conserved due to the conversion of kinetic energy to other energy forms.

Answer:

Power dissipated in 3 ohms resistor is 5.32 watts                

Explanation:

Resistor 1, R₁ = 3 ohms

Resistor 2, R₂ = 6 ohms

Resistor 3, R₃ = 4 ohms

Voltage source, V = 12 V

We need to find the power dissipated in the 3 ohms resistor. Firstly, we will find the equivalent resistance of R₁ and R₂.

[tex]\dfrac{1}{R'}=\dfrac{1}{R_1}+\dfrac{1}{R_2}[/tex]

[tex]\dfrac{1}{R'}=\dfrac{1}{3}+\dfrac{1}{6}[/tex]

R' = 2 ohms

Now R' is connected in series with R₃. Their equivalent is given by :

[tex]R_{eq}=R'+R_3[/tex]

[tex]R_{eq}=2+4[/tex]

[tex]R_{eq}=6\ ohms[/tex]

Total current flowing through the circuit, [tex]I=\dfrac{12}{6}=2\ A[/tex]

Voltage across R', [tex]V'=IR'=2\times 2=4\ V[/tex]

The voltage across R₁ and R₂ is 4 volts as they are connected in parallel. So, current across 3 ohm resistor is,

[tex]I=\dfrac{4}{3}=1.33\ A[/tex]

Power dissipated is given by, P = I × V

[tex]P=1.33\ A\times 4\ \Omega[/tex]

P = 5.32 watts

So, 5.32 watt of power is dissipated  in 3 ohms resistor. Hence, this is the required solution.

Answer:

122241.02 J

Explanation:

Work Done = 230000 J

[tex]T_H=\text {High Temperature Reservoir}=830\ K\\T_L=\text {Low Temperature Reservoir}=288\ K\\\text{In case of Carnot Cycle}\\\text{Efficiency}=\eta\\\eta=1-\frac{T_L}{T_H}\\\Rightarrow \eta =1-\frac{288}{830}\\\Rightarrow \eta=0.65\\[/tex]

[tex]\eta=\frac{\text{Work}}{\text{Heat}}\\\Rightarrow \eta=\frac{W}{Q_H}\\\\\Rightarrow 0.65=\frac{230000}{Q_H}\\\Rightarrow Q_H=\frac{230000}{0.65}\\\Rightarrow Q_H=352214.02[/tex]

[tex]Q_L=Q_H-W\\\Rightarrow Q_L=352214.02-230000\\\Rightarrow Q_L=122241.02\ J[/tex]

∴Heat exhausted into the air to produce 230,000 J of work is 122241.02 Joule

Correct Answer:

122000 J

Answer:

Final kinetic energy, [tex]KE_f=5\ J[/tex]  

Explanation:

It is given that,

Kinetic energy of toy car, [tex]KE_i = 8\ J[/tex]

Frictional force, F = 0.5 N

Distance, d = 6 m

We need to find the kinetic energy after this frictional force has acted on it. We know that frictional force is an opposing force. The work done by the toy car is given by :

[tex]W=F\times d[/tex]

[tex]W=-0.5\ N\times 6\ m[/tex]

W = -3 J

We know that the work done is equal to the change in kinetic energy. Let [tex]KE_f[/tex] is the final kinetic energy.

[tex]W=KE_f-KE_i[/tex]

[tex]-3=KE_f-8[/tex]

[tex]KE_f=5\ J[/tex]

So, the final kinetic energy is the toy car is 5 J. Hence, this is the required solution.

The period of the vertical block-spring system on the moon is approximately 2.45 seconds.

The period of a vertical block-spring system is determined by the square root of the ratio of the mass of the block to the spring constant. Since the mass and spring constant remain the same, the period will only be affected by the acceleration due to gravity.

On the moon, where the acceleration due to gravity is about 1/6 that of Earth, the period of the system will be sqrt(6), or approximately 2.45 seconds.

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Answer:

v = 2.43 m/s

Explanation:

As we know that block is placed at rest at distance

d = 2 m

so here the centripetal force on the block to move in the circle is due to static friction force

now when block is just going to slide over the disc

then the friction force is maximum static friction which is given as

[tex]\mu_s mg = \frac{mv^2}{R}[/tex]

now we have

[tex]v = \sqrt{\mu_s Rg}[/tex]

now plug in the values in the above equation

[tex]v = \sqrt{0.3(2)(9.81)}[/tex]

[tex]v = 2.43 m/s[/tex]

To determine the maximum speed at which the block can attain before it begins to slip, calculate the force of static friction using the given coefficient and weight of the block. Then, use the equation fs = m(a+g) to determine the maximum speed.

To determine the maximum speed at which the block can attain before it begins to slip, we need to calculate the force of static friction. The formula for static friction is fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is the weight of the block, which is mg. The maximum force of static friction is therefore μs(mg). Since the block is on a horizontal surface, the normal force is equal to the weight of the block, and the maximum force of static friction is μs(mg). We can calculate μs using the given coefficient of static friction and determine the maximum speed using the equation fs = m(a+g), where fs is the force of static friction, m is the mass of the block, a is the acceleration, and g is the acceleration due to gravity.

Plug in the given values: ms = 0.3, m = (mass of the block), a = 2 m/s², and g = 9.8 m/s².

Then, solve for the maximum speed by rearranging the equation to isolate v: v = sqrt((fs - m*g)/m).

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Answer:

acceleration is 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k  ) m/s

Explanation:

given data

uniform electric field E  = 54.0 ĵ V/m

uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T

velocity v  = 170 î m/s.

to find out

acceleration

solution

we know magnetic force for proton is

i.e = e (velocity × uniform magnetic field)

magnetic force = e (170 î × (0.200 î + 0.300 ĵ + 0.400 k) )

magnetic force = e (- 68 ĵ + 51 k  )   ..................1

and now for electric force for proton i.e

= uniform electric field × e ĵ

electric force = e (54 ĵ )   ............2

so net force will be  add magnetic force + electric force

from equation 1 and 2

e (- 68 ĵ + 51 k  )  + e (54 ĵ )

e (- 14 ĵ + 51 k  )

so the acceleration (a)  for proton will be

net force = mass × acceleration

a = e (- 14 ĵ + 51 k  )  / 1.6 [tex]10^{-19}[/tex] / 1.67 × [tex]10^{-27}[/tex]

acceleration = 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k  ) m/s

The acceleration of the proton can be determined using the equation F = qvBsinθ. Plug in the given values to calculate the acceleration.

The acceleration of the proton can be determined using the equation F = qvBsinθ, where F is the force, q is the charge of the proton, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Since the question does not provide the angle, we assume that θ = 0. Therefore, the acceleration of the proton is given by:

a = (q * v * B) / m

where m is the mass of the proton. Plugging in the given values, we can calculate the acceleration.

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Answer:

[tex]F = 56.75 N[/tex]

Explanation:

As per the free body diagram of the box we can say that the applied force must be greater than or equal to the sum of friction force and component of the weight of the box along the hill

So we can say that

[tex]F_{net} = F cos35 - (mgsin20 + F_f)[/tex]

on the other side the force perpendicular to the plane must be balanced so that it remains in equilibrium in that direction

so we can say that

[tex]F_n + Fsin35 = mgcos20[/tex]

now we will have

[tex]F_f = \mu F_n[/tex]

[tex]F_f = (0.20)(mg cos20 - Fsin35)[/tex]

now we have

[tex]0 = Fcos35 - mg sin20 - (0.20)(mg cos20 - Fsin35)[/tex]

[tex]F(cos35 + 0.20 sin35) = mg sin20 + 0.20 mgcos20[/tex]

[tex]F = \frac{mg sin20 + 0.20 mgcos20}{(cos35 + 0.20 sin35)}[/tex]

[tex]F = \frac{100(sin20 + 0.20 cos20)}{(cos35 + 0.20 sin35)}[/tex]

[tex]F = 56.75 N[/tex]

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, [tex]L=50\ mH=0.05\ H[/tex]

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

[tex]X_L=2\pi fL[/tex]

[tex]X_L=2\pi \times 400\ Hz\times 0.05\ H[/tex]

[tex]X_L=125.66\ \Omega[/tex]

[tex]X_L=125.7\ \Omega[/tex]

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

To calculate the inductive reactance for a 50 mH inductor with a 400 Hz source, apply the formula XL = 2πfL, resulting in an inductive reactance of 125.6 ohms.

To determine the inductive reactance of a 50 mH inductor across a 15 volt, 400 Hz source, we use the formula for inductive reactance, which is:

L = 2πfL

where:

XL is the inductive reactance

f is the frequency of the AC source (400 Hz in this case)

L is the inductance of the coil (50 mH or 0.050 H)

Substituting the given values:

XL = 2π x 400 Hz x 0.050 H

XL = 125.6 Ω

Therefore, the inductive reactance is 125.6 ohms (Ω).

Final answer:

The required force is approximately 21.63 newtons.

Explanation:

The question involves determining the average force that a shooter must exert to keep a gun from moving while it fires bullets. To find this force, we can use the concept of impulse, which relates force, time, and change in momentum.

The gun fires 115 bullets per minute, which is 115/60 bullets per second. Multiplying the number of bullets per second by the mass and speed of each bullet gives the total momentum per second, which is the same as the force when mass is in kg and speed is in m/s.

First, we find the momentum of each bullet:

Bullet speed (v) = 342 m/s

F ≈ 21.63 N

The shooter must exert an average force of approximately 21.63 newtons to keep the gun from moving.

Answer:

c).  [tex]a = 0.60 m/s^2[/tex]

Explanation:

As we know that weight of the automobile is given here

so weight = mass times gravity

[tex]W = mg[/tex]

[tex]9800 = m(9.8)[/tex]

[tex]m = 1000 kg[/tex]

now from Newton's law

[tex]F = ma[/tex]

[tex]600 = 1000 a[/tex]

[tex]a = \frac{600}{1000}[/tex]

[tex]a = 0.60 m/s^2[/tex]

The question is about finding the acceleration of a car with a given weight and applied force. To do this, the weight of the car is first converted to mass, then using Newton's Second Law (F = ma), we can determine the acceleration. The calculated acceleration of the car is 0.6 m/s^2.

The subject matter of this question is related to physics, specifically Newton's Second Law of Motion. Noting that the force, mass, and acceleration of the automobile are connected by the formula F = ma (force equals mass times acceleration), the given problem can be solved. First, we need to convert the weight of the automobile (9800 N) into mass, as weight is the product of mass and gravity (9.8 m /s^2). Therefore, the mass of the car is 1000 kg.

Using this converted mass and the applied force (600 N) in the formula, we can calculate the acceleration. The calculation goes as follows: 600N = 1000 kg * a, where a is the unknown acceleration we're trying to find. When you solve for 'a', you get a = 0.6 m/s^2. Hence, the acceleration of the 9800-N automobile when a forward force of 600N is applied, neglecting friction, is option c) 0.6 m/s^2.

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Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

[tex]V_{ab}[/tex] = i R

[tex]V_{ab}[/tex] = (0.1832) (65)

[tex]V_{ab}[/tex] = 11.91 Volts

(c)

Power dissipated in the resister R is given as

[tex]P_{R}[/tex] = i²R

[tex]P_{R}[/tex] = (0.1832)²(65)

[tex]P_{R}[/tex] = 2.18 Watt

Power dissipated in the internal resistance is given as

[tex]P_{r}[/tex] = i²r

[tex]P_{r}[/tex] = (0.1832)²(0.5)

[tex]P_{r}[/tex] = 0.0168 Watt

To compute the values in the question, one can utilize Ohm’s law and the power dissipation formula. The current in the circuit is 0.183 A, the terminal voltage of the battery is 11.908 V, the power dissipated in the 65.0 Ω resistor is 2.18 W, and the energy wasted in the battery's internal resistance is 0.0167 W.

To calculate the current in the circuit, we can use Ohm's law which states that the current is equal to the voltage divided by the resistance. This gives us: I = emf/(R+r) where I is the current, emf is the electromotive force of the battery, R is the load resistance and r is the internal resistance of the battery.

Substituting the values given in the question, we get: I = 12.0 V/(65.0 Ω + 0.5 Ω) = 0.183 A. Therefore, the current flowing through the circuit is 0.183 Amperes.

Now that we have the current, we can calculate the terminal voltage of the battery. The terminal voltage (Vab) is given by the equation: Vab = emf - Ir. Substituting the values, we get: Vab = 12.0V - (0.183A * 0.5Ω) = 11.908 V. Therefore, the terminal voltage of the battery is 11.908 Volts.

Lastly, we can calculate the power dissipated in the resistor R and in the battery’s internal resistance r using the formula: Power = I²R. For the resistor, power dissipation = (0.183A)² * 65.0Ω = 2.18 Watts, and for the battery's internal resistance, power dissipation = (0.183A)² * 0.5Ω = 0.0167 Watts.

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Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

Answer:

A

Explanation:

Iron and gadlinium are both very easily made into magnetic substances.  Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

Answer:

Option B is the correct answer.

Explanation:

Period of simple pendulum is given by the expression [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex], where l is the length of pendulum and g is the acceleration due to gravity value.

So if we have a simple pendulum with length l we can find its period. Using the above formula we can calculate  acceleration due to gravity value of that place.

So using simple pendulum we can measure acceleration due to gravity value.

Option B is the correct answer.

Answer:

105.8 degree from + X axis

Explanation:

V1 = 46 N at 30 degree

V2 = 63 N at 151 degree

Write the vector form

V1 = 46 (Cos 30 i + Sin 30 j) = 39.84 i + 23 j

V2 = 63 (Cos 151 i + Sin 151 j) = - 55 i + 30.54 j

The resultant of V1 and V2 is given by

V = V1 + V2 = 39.84 i + 23 j - 55 i + 30.54 j = - 15.16 i + 53.54 j

The angle made by resultant of V1 and V2 with X axis is

tan∅ = 53.54 / (- 15.16) = - 3.53

∅ = 105.8 degree from + X axis.

Answer:

Distance ran by basketball player = 4.5 m

Explanation:

We have equation of motion, v = u+at

Substituting

           6 = 0 + a x 1.5

           a = 4m/s²

Now we need to find distance traveled, we have the equation of motion

           v² = u² + 2as

Substituting

           6² = 0² + 2 x 4 x s

           8 s = 36

              s = 4.5 m

Distance ran by basketball player = 4.5 m

Answer:

1.27 m

Explanation:

Distance = 192 m

number of rotations = 48

Distance traveled in one rotation = 2 x π x r

Where, r be the radius of wheel.

so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r

It is equal to the distance traveled.

192 = 48 x 2 x 3.14 x r

r = 0.637 m

diameter of wheel = 2 x radius of wheel = 2 x 0.637 = 1.27 m

Answer:

1.29 meters

Explanation:

In order to calculate this you first have to calculate the circumference of the wheel, as you know to travel 192 meters the wheel did 48 revolutions, that means that it took 48 circumferences of the wheel to cover 192 meters, we first have to divide the 192 meters by the circumference:

Circumference:192 m/48=4m

So to know the diameter we just need to divide the circumference by pi:

diameter=circumference/pi

Diameter=4 meters/3.14

Diameter= 1.2861 meters.

Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

         v² = u² + 2as

         v² = 44.50² + 2 x (-9.81) x 71.32

         v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

        v = u + at

         24.10 = 44.50 - 9.81 t , t = 2.07 s

         -24.10 = 44.50 - 9.81 t , t = 6.99 s      

Since the second ball throws after 2.3 seconds  we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

           s = ut + 0.5 at²

           71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

           u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.  

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.  

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

Answer:

a) [tex]a = 0.9 m/s^2[/tex]

Explanation:

As per Newton's 2nd law we know that net force on an object of mass "m" is given by the formula

[tex]F = ma[/tex]

so net force is the product of mass and acceleration

here we know that

mass = 5 kg

net force on the particle is 4.5 N

so from above equation we have

[tex]4.5 = 5 \times a[/tex]

so the acceleration is given as

[tex]a = \frac{4.5}{5} = 0.9 m/s^2[/tex]

Answer:

There  will be no voltage across  secondary coil because DC voltage source is used in primary coil so there is no electromotive force induced in secondary coil.

Explanation:

In this question we have given

AA battery is used to supplies a constant voltage of 1.5 volts to primary coil of  transformer. In this case, voltage source is DC source which is providing constant voltage and for a transformer to work it is necessary to use an AC source.

Therefore, no EMF will induce in the secondary coil

and we know that in a transformer,

[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1} }=\frac{N_{2} }{N_{1} }[/tex]...........(1)

hence from above equation it is clear that,

[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1}}[/tex]............(2)

Here ,

[tex]E_{2}=0[/tex]

Put [tex]E_{2}=0[/tex] in equation (2)

We got

[tex]V_{2}=0[/tex]

There  will be no voltage across  secondary coil because there is no electromotive force induced in secondary coil.

Answer:

The change in momentum of an object is 1.6 kg-m/s.  

Explanation:

It is given that,

Mass of the object, m = 0.3 kg

Force acting on it, F = 8 N

Time taken, t = 0.2 s

We need to find the change in momentum of an object. The change in momentum is equal to the impulse imparted on an object i.e.

[tex]\Delta p=J=F.\Delta t[/tex]

[tex]\Delta p=8\ N\times 0.2\ s[/tex]

[tex]\Delta p=1.6\ kg-m/s[/tex]

So, the change in momentum of an object is 1.6 kg-m/s. Hence, this is the required solution.

The change in the momentum of the object is 1.6 kg-m/s.

Change in momentum is the product of force acting on a body and time

To calculate the change in momentum of the object, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The change in the momentum of the object is 1.6 kg-m/s.

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Initial momentum of the Play-doh: 0.0600 x 2.60 = 0.156 kg/m/s

Total mass of the block and play-doh: 0.38 + 0.0600 = 0.44 kg.

Final momentum is mass x velocity = 0.44v

V = Initial momentum / mass

V = 0.156 / 0.44 = 0.3545 m/s

Work done by spring is equal to the Kinetic enrgy.

Work Done by spring = 1/2 *28.0 * distance^2 = 14 * d^2

KE = 1/2 * 0.44* 0.3545^2

set to equal each other:

14 * d^2 = 0.22 *0.12567

Solve for d:

d = √(0.22*0.12567)/14

d = 0.44 meters = 4.4cm

Answer:

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Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

[tex]V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}[/tex]

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

[tex]V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}[/tex]

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

Answer:

it will go up along the inclined plane by d = 9.62 m

Explanation:

As we know that puck is moving upward along the slide

then the net force opposite to its speed is given as

[tex]F_{net} = - mgsin\theta - \mu mgcos\theta[/tex]

so here deceleration is given as

[tex]a = -g(sin\theta + \mu cos\theta)[/tex]

now plug in all values in it

[tex]a = -9.81(sin20 + 0.2cos20)[/tex]

[tex]a = -5.2 m/s^2[/tex]

now the distance covered by the puck along the plane is given as

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 10^2 = 2(-5.2)d[/tex]

[tex]d = 9.62 m[/tex]

Answer:

555

Explanation:

The scenario is shown in the image attached below. A Right Angled Triangle is formed. We have an angle of 6 degrees, the side adjacent to angle which measures 1 mile and we need to find the side opposite to the angle.

Since 1 mile = 5280 feet, the side adjacent to the angle has a measure of 5280 feet.

Tangent ratio relates the opposite and adjacent side by following formula:

[tex]tan(\theta)=\frac{Opposite}{Adjacent}[/tex]

Using the given values, we get:

[tex]tan(6)=\frac{x}{5280}\\\\ x=tan(6) \times 5280\\\\ x = 555[/tex]

Thus, rounded to nearest foot, the height of the building is 555 feet

To find the height of the building, we can use trigonometry and the given angle of elevation. The height of the building is approximately 555 feet.

To find the height of the building, we can use trigonometry and the given angle of elevation. Let x be the height of the building. From the given information, we have the opposite side (x) and the adjacent side (1 mile = 5280 feet). The tangent function is used to relate these sides: tan(6°) = x/5280 feet. Solving for x, we get x = 5280 feet × tan(6°).

Using a calculator, we find that tan(6°) is approximately 0.1051. Multiplying this by 5280 feet gives us the height of the building in feet: 0.1051 × 5280 = 554.928 feet.

Rounding to the nearest foot, the height of the building is approximately 555 feet.

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The question is about determining the initial speed of a car involved in an elastic collision with a parked truck, using the laws of conservation of momentum.

The subject of this question is Physics, and it involves the principles of conservation of momentum and elastic collisions. We are given masses for two vehicles (a car and a truck) and the post-collision speed of one of the vehicles (the truck). To find the car's initial speed, we use the equations for an elastic collision, which ensures that both momentum and kinetic energy are conserved.

For an elastic collision, the conservation of momentum is given by:

m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'

where m1 and m2 are the masses of the car and truck respectively; v1 and v2 are the initial velocities of the car and truck respectively (since the truck is initially parked, v2 = 0); v1' and v2' are the final velocities of the car and truck respectively.

To find the final velocity v1' of the car, we must use the conservation of kinetic energy along with the above equation. However, we only need the conservation of momentum equation solved for v1 (initial speed of the car) since v2' (final velocity of the truck) is given.

After plugging in the given values (car's mass 1689 kg, truck's mass 2000 kg, truck's final velocity 17 km/h), the equation is solved to find the car's initial velocity.