A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at 100 rpm. What is its rotational kinetic energy?

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 8.5 m/s

     Final velocity, v = 0 m/s    - At maximum height

     Time, t = ?

     Acceleration , a = -9.81 m/s²

     Substituting

                      v = u + at  

                      0 = 8.5 + -9.81 x t

                      t = 0.867 s

  It takes 0.867 seconds to get to the top of its motion

Answer:

0.87 s

Explanation:

initial velocity, u = 8.5 m/s

Let it takes time t to reach to maximum height. At maximum height the velocity is zero, so, v = 0

Use first equation of motion

v = u - gt

where, g be the acceleration due to gravity

0 = 8.5 - 9.8 t

t = 0.87 s

Thus, the time taken to reach at top is 0.87 s.

Answer:

Explanation:

Check the attachment for solution

Answer: The final speed of the incoming ball is approximately 200m/s

Explanation:

Using the law of conservation of momentum.

m1(u1) + m2u2 = m1v1 + m2v2

And also law of conservation of kinetic energy for elastic heads on collision we can derive the formula for elastic heads on collision which is given below:

For elastic heads on collision.

v1 = [( m1 - m2)/(m1+m2)] u1 ......1

v2 = [(2m1)/(m1+m2)]u1 ......2

Where,

m1 and m2 are the mass of the incoming and stationary ball respectively.

u1 and u2 are the initial speed of the incoming and stationary ball respectively.

v1 and v2 are the final speed of the incoming and stationary ball respectively.

a) to determine the final speed of the incoming ball using equation 1

v1 = [( m1 - m2)/(m1+m2)]u1

Since m1 >> m2

m1 - m2 ~= m1 and m1 +m2 ~= m1

So, equation 1 becomes

v1 ~= [m1/m1]u1

v1 ~= u1

Since u1 = 200m/s

v1 ~= 200m/s

Additional tips: using equation 2 we can derive the approximate final speed of the stationary ball following the same assumptions. If well solved v2 = 2u1 = 400m/s

Answer:

The kinetic energy when the film vault landed is 12744000J.

Explanation:

The kinetic energy is defined as:

[tex]k_{e} = \frac{1}{2}mv^{2}[/tex]  (1)

Where m is the mass and v is the velocity.

By means of equation 1, the kinetic energy of the film vault when it landed can be determined

[tex]k_{e} = \frac{1}{2}(1770kg)(120m/s)^{2}[/tex]  

But [tex]1 J = kg.m^{2}/s^{2}[/tex]

[tex]k_{e} = 12744000J[/tex]

Hence, the kinetic energy when the film vault landed is 12744000J.

The kinetic energy of the film vault when it landed was 1,919,200 Joules.

The kinetic energy of the film vault can be calculated

The kinetic energy of the film vault can be calculated using the formula: KE = 0.5 * mass * velocity^2. Given that the mass of the film vault is 1770 kg and its velocity is 120 m/s, we can substitute these values into the formula:

KE = 0.5 * 1770 kg * (120 m/s)^2 = 0.5 * 1770 * 14400 = 1,919,200 J

Therefore, the kinetic energy of the film vault when it landed was 1,919,200 Joules. using the formula: b mass ity^2. Given that the mass of the film vault is 1770 kg and its velocity is 120 m/s, we can substitute these values into the formula:

KE = 0.5 * 1770 kg * (120 m/s)^2 = 0.5 * 1770 * 14400 = 1,919,200 J

Therefore, the kinetic energy of the film vault when it landed was 1,919,200 Joules.

https://brainly.com/question/26472013

#SPJ3

Answer:

46 days

Explanation:

It is a very logical question which does not require any mathematical calculation rather requires thinking.

It is said in the question every day, the patch doubles in size. if it takes 48 days for the patch to cover the entire lake.

Let total area of the lake be 100 m^2. The at the end of  48th day, it would have covered 100 m^2. So, at the end of 47th day it would have covered only 50 m^2, since every day, the patch doubles in size. Similarly on 46th day it would have covered 25 m^2, which is quarter of 100 m^2.

So ,it would take 46 days for the patch to cover quarter the lake.

Explanation:

The given statement is absolutely true.  this is because magnitude of a vector is always non negative, it can not be zero unless its a zero vector. So, in the given question, final position vector of a moving object has a smaller magnitude than the initial position vector, so, magnitude is neither zero nor negative. Hence, it has a positive magnitude.

Final answer:

The displacement of an object is the final position vector minus the initial position vector and is a vector with both magnitude and direction. The magnitude of displacement does not have to be positive if the final position vector has smaller magnitude than the initial; it depends on the direction relative to the chosen coordinate system.

Explanation:

When discussing the displacement of an object, it is important to understand that it refers to the change in the object's position and is defined as the final position vector minus the initial position vector. Displacement, being a vector quantity, has both magnitude and direction. If the magnitude of the final position vector is smaller than the magnitude of the initial position vector, this does not necessitate that the magnitude of the displacement is positive. In fact, the magnitude of the displacement vector could be either positive or negative based on the assigned coordinate system.

In a one-dimensional coordinate system, the direction of motion can be designated as positive or negative. For example, if rightward motion is deemed positive, then a leftward motion would be negative. Similarly, in vertical motion, upward movement is often taken as positive and downward as negative, though these conventions can be reversed based on the scenario and convenience. The direction of the displacement vector essentially depends on the chosen orientation of the positive direction.

Therefore, if the final position vector has a smaller magnitude than the initial position vector, and if the positive direction is assigned towards the final position, the displacement vector magnitude could indeed be negative, not necessarily positive, as it signifies a change in position opposite to the assigned positive direction.

Answer:

Height will be equal to 0.4489 m

Explanation:

We have given mass of the toy m = 0.50 kg

Initial kinetic energy K = 2.2 J

We have to fond the height of the hill over which car roast

When car will roast the hill its kinetic energy will be converted into potential energy and at maximum height all kinetic energy will be converted into potential energy

So at maximum height [tex]mgh=K[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

So 0.50×9.8×h = 2.2

h = 0.4489 m

The amplitude where two pulses cross on a string rigidly attached to a post is momentarily zero due to destructive interference. However, when the end is free to move and two pulses cross, they superimpose constructively, resulting in a combined amplitude of 86 cm.

When Anthony sends a series of pulses of amplitude 43 cm down a string tied to a post:

(a) If the string is rigidly attached to the post, the reflected pulse inverts. Therefore, when two pulses of the same amplitude cross, they interfere destructively at the instant they coincide and the amplitude at the point of crossing is momentarily zero. However, since they're not lasting interactions, the amplitude of each pulse remains 43 cm before and after crossing.

(b) If the end is free to slide up and down, the reflected pulse does not invert and remains in phase with the incident pulse. Now, as two pulses of the same amplitude (43 cm) cross, they temporarily superimpose constructively, resulting in a combined amplitude of 86 cm, which is the sum of their individual amplitudes.

The phenomenon described here is a demonstration of wave interference, specifically in the context of reflected transverse waves on a string. It shows the distinct difference between reflections from fixed end (inverting) and free end (non-inverting) boundaries.

For a rigidly attached end, the amplitude is 0 cm due to destructive interference, while for a free boundary, it is 86 cm due to constructive interference.

When dealing with wave pulses on a string, the amplitude of the resulting wave at a point where two pulses are crossing depends on the boundary conditions at the end of the string.

When the string is rigidly attached, the reflected pulse inverts upon reflection. This means that the original pulse and the reflected pulse will be out of phase by 180 degrees when they meet. If the amplitude of each pulse is 43 cm, the resulting amplitude at the point of crossing will be zero (43 cm - 43 cm = 0 cm) due to destructive interference.

When the end at which reflection occurs is free to move up and down (a free boundary condition), the reflected pulse remains in phase with the incident pulse. Therefore, the amplitudes add constructively. Hence, at the crossing point, the resulting amplitude will be the sum of the individual amplitudes: 43 cm + 43 cm = 86 cm.

In summary, the conditions at the end of the string determine whether the interference is constructive or destructive, affecting the amplitude at the crossing point of two pulses.

Answer:

For  the pendulum , yes and for the spring ,No

Explanation:

We will examine the two cases,

For pendulum;

[tex]T =2\pi \sqrt{\frac{l}{g} }[/tex]

we see that for a pendulum, the period depends on the gravitational acceleration of the region it is. so the period for earth will differ compared to that of the moon.

For a spring;

[tex]T=2\pi \sqrt{\frac{m}{k} }[/tex]

form the equation above is obvious that the period of a spring in vibration is independent of gravitational acceleration of the region it is.

The mass of the bob has no effect on the motion of a pendulum; the period of vibration is determined by the pendulum's length and the acceleration due to gravity.

The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum's length) and by the acceleration due to gravity.

Answer:

b. is the process in which two smaller nuclei combine to form a larger nucleus

Answer: The balance of care is sympathetic and loving

Explanation: According to Erikson's theory, when the balance of care is sympathetic and loving the psychological conflict of the first year is resolved on the positive side.

According to Erikson's theory of psychosocial development, the psychological conflict of the first year is resolved on the positive side when the infant develops a sense of trust in their caregivers.

According to Erik Erikson's theory of psychosocial development, when an infant develops a sense of trust in their caregivers during the first year of life, the psychological conflict of that stage is resolved on the positive side. Erickson named this first stage trust versus mistrust. In this stage, caregivers who are responsive and sensitive to their infant's needs help the child to develop a sense of trust; thus, the child will see the world as a safe, predictable place. If the caregivers are unresponsive and do not meet the infant's needs, feelings of anxiety, fear, and mistrust emerge; the child may see the world as unpredictable.

Successful completion of each developmental task at various stages results in a sense of competence and a healthy personality, according to Erikson. Failure to master these tasks may lead to feelings of inadequacy and affect the formation of a positive self-concept.

https://brainly.com/question/30761435

#SPJ3

Answer: a. F doubled

b. F reduced by one-quarter i.e

1/4*(F)

c. 1/9*(F)

d. F increased by a factor of 4 i.e 4*F

e. F reduces 3/4*(F)

Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is

F = k*(qA*qB)/d²

a. If qA is doubled therefore the force is doubled since they are directly proportional.

b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4

Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.

c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9

d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4

e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,

3/4*F

Coulomb's Law explains how changes in charge and distance affect the electric force between two charges.

Coulomb's Law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Let's analyze the given scenarios:

https://brainly.com/question/20935307

#SPJ3

Answer:

2 N/C direction of the force

-2N/C opposite to the direction of the force

Explanation:

E = Electric field

q = Charge

Electrical force is given by

[tex]F=Eq\\\Rightarrow E=\dfrac{F}{q}\\\Rightarrow E=\dfrac{40}{20}\\\Rightarrow E=2\ N/C[/tex]

The magnitude of the force is 2 N/C

The force acting on the charge is positive so the direction of the electric field is positive.

[tex]F=Eq\\\Rightarrow E=\dfrac{F}{q}\\\Rightarrow E=\dfrac{20}{-10}\\\Rightarrow E=-2\ N/C[/tex]

The magnitude of the force is 2 N/C

The direction of the electric field is negative and opposite to the direction of the force as the charge is negative.

To determine the magnitude and direction of an electric field, use the formula E = F/Q. For a positively charged object, the electric field has the same direction as the force. For a negatively charged object, the electric field has the opposite direction of the force.

To determine the magnitude and direction of an electric field given the charge of an object and the force exerted on it, you can use the formula: E = F/Q, where E is the electric field, F is the force, and Q is the charge.

For a positively charged object (+20.0μC) with a force of 40.0μN due east: The electric field E = F/Q = (40.0μN)/(20.0μC) = 2.0 N/C. The direction of the electric field is in the same direction as the force, due east.

For a negatively charged object (-10.0μC) with a force of 20.0μN due west: The electric field E = F/Q = (20.0μN)/(-10.0μC) = -2.0 N/C. The direction of the electric field is in the opposite direction of the force, due east, as per the convention that electric fields are directed away from positive charges or towards negative charges.

Learn more about Electric Field here:

https://brainly.com/question/33547143

#SPJ3

Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction[tex],\mu_k=0.11[/tex]

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

[tex]F=\mu_k N[/tex]

Where N= Normal=mg

[tex]F=0.11\times 634=69.74 N[/tex]

Hence, the force is needed  to pull the sled across the snow at constant speed=69.74 N

Answer:

(A) t_H = Vo×Sin (theta)/g

(B) t,_R = (2VoSin(theta))/g

(C) H = Vo²Sin²(theta) / 2g

(D) Range = Vo²Sin(2×theta) / g.

Explanation:

The detailed solution to the problem can be found in the attachment below

To find the time it takes for the projectile to reach its maximum height, use the equation t_H = v_0 sin(θ) / g. To find the time the projectile hits the ground, use the equation t_R = 2v_0 sin(θ) / g. The maximum height attained by the projectile can be found using the equation H = (v_0^2 sin^2(θ)) / (2g). The total distance traveled in the x-direction (range) can be found using the equation R = (v_0^2 sin(2θ)) / g.

To find the time it takes for the projectile to reach its maximum height, we can use the equation for vertical motion:

t_H = v_0 sin(\theta) / g

To find the time the projectile hits the ground, we can use the equation for vertical motion:

t_R = 2v_0 sin(\theta) / g

The maximum height attained by the projectile can be found using the equation:

H = (v_0^2 sin^2(\theta)) / (2g)

The total distance traveled in the x-direction (range) can be found using the equation:

R = (v_0^2 sin(2\theta)) / g

Answer:

Explanation:

Here is the full question:

Two neutron stars are separated by a distance of 1.0 × 1010 m. They each have a mass of 1.0 × 1030 kg and a radius of 1.0 × 105 m. They are initially at rest with respect to each other. As measured from that rest frame, how fast are they moving when (a) their separation has decreased to one-half its initial value and (b) they are about to collide?

solution:

G is the gravitational constant, the value is,

[tex]G=6.67\times 10^{-11}\frac{m^3}{kg.s^2}[/tex]

For half distance is,

[tex]U'=2U\\\\-dU=\frac{-2Gm^2}{R}\\\\dKE=-dU[/tex]

a)

If the sepperation confined to one-half its initial value,

The velocity is,

[tex]K=\frac{Gm^2}{2R}\\\\\frac{1}{2}mv^2=\frac{Gm^2}{2R}\\\\v=\sqrt{\frac{Gm}{R}}\\\\=\sqrt{\frac{(6.67\times 10^{-11}\frac{m^3}{kg.s^2})(10^{30}kg)}{(10^{10}m)}}\\\\=81,670m/s\\\\v=8.2\times 10^4m/s[/tex]

b)

[tex]dU=Gm^2(\frac{1}{R}-\frac{1}{2r})\\\\dKE=Gm^2(\frac{1}{2r}-\frac{1}{R})=mv^2\\\\\therefore dKE=-dU\\\\v=\sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}\\\\\sqrt{(6.67\times 10^{-11}\frac{m^3}{kg.s^2})(10^{30}kg))(\frac{1}{2(10^5m)}-\frac{1}{(10^{10}m)})}\\\\v=1.8\times 10^7m/s[/tex]

This observation is not consistent with a Sun-centered model, because in this model "the rise and set of all objects depends only on Earth’s rotation".

Answer: Option D

Explanation:

Copernican heliocentrism, the astronomical model's title which was created by Nicolaus Copernicus and released in 1543. This model placed the Sun in motionless position near the Universe's center, with Earth and the planets of other which orbiting it in a circular orbits, altered by epicycles, and also at constant speeds.

Out of several planets in a determined order the Earth orbits around a stationary sun having three motions like annual revolution, daily rotation, and its axis is tilted annually. The planets' retrograde motion is understood by motion from the Earth.

Answer:

2.57832 atm

Explanation:

[tex]P_1[/tex] = Initial pressure = 2.1 atm

[tex]P_2[/tex] = Fianl pressure

[tex]T_1[/tex] = Initial Temperature = (21+273.15) K

[tex]T_2[/tex] = Final Temperature = (88+273.15) K

From Gay-Lussacs law we have

[tex]\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\Rightarrow P_2=\dfrac{P_1\times T_2}{T_1}\\\Rightarrow P_2=\dfrac{2.1\times (88+273.15)}{21+273.15}\\\Rightarrow P_2=2.57832\ atm[/tex]

The final pressure would be 2.57832 atm

Answer:

2.58 atm

Explanation:

P1 = 2.1 atm

T1 = 21 °C = 294 K

T2 = 88 °C = 361 K

P2 = ?

By using the gas laws

P / T = constant keeping the volume constant.

P1 / T1 = P2 / T2

2.10 / 294 = P2 / 361

P2 = 2.58 atm

Thus, the pressure becomes 2.58 atm.

Answer:

The weight of the truck is 22072.5 N.

Explanation:

Given that,

Width of the brick =4.3 m

Length of the brick = 8.5 m

Distance = 6.15 cm

We need to calculate the volume of the water displaced by the boat with truck

Using formula of volume

[tex]V=l\times w\times d[/tex]

Put the value into the formula

[tex]V=8.5\times4.3\times6.15\times10^{-2}[/tex]

[tex]V=2.25\ m^3[/tex]

We need to calculate the mass of the water displaced by the boat with truck

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

[tex]m=\rho\times V[/tex]

Put the value into the formula

[tex]m=1000\times2.25[/tex]

[tex]m=2250\ kg[/tex]

We need to calculate the weight of the truck

Using formula of weight

[tex]W= mg[/tex]

Put the value into the formula

[tex]W=2250\times9.81[/tex]

[tex]W=22072.5\ N[/tex]

Hence, The weight of the truck is 22072.5 N.

Answer:

The vasa vasorum, the collection of small arteries and veins supplying blood to the smooth muscle and fibroblasts within blood vessels, is found in the tunica externa.

Explanation:

In order to understand the answer above let explain the concept of tunica externa

This in Latin means outer coat and can also be called 'tunica adventitia' which also means "additional coat in Lain",it is the outermost layer of the a blood vessel covering the tunica media(This is a Latin word  translated as middle coat in English which is the middle layer of an artery or a vein it lies between the tunica externa  and the tunica initima which is the innermost part of the artery or vein).

An Artery

We can define an artery as a blood vessel that transports blood from the heart to other parts of the body.

An Vein

We can define a vein as a blood vessel that transports blood to the heart

and we can define  blood vessel as a tube that carries blood in the circulatory system.  

Answer

given,

Force on car, F = 300 N

time, t = 0.250 s

distance of the force from the pivot, r= 0.3 m

Angular momentum = ?

Now,

torque acting is calculated by multiplying force with displacement.

    [tex]\tau = F \times r [/tex]

    [tex]\tau = 300 \times 0.3 [/tex]

    [tex]\tau = 90\ N.m[/tex]

we know,

torque is equal to change in angular momentum per unit time.

[tex]\tau = \dfrac{\Delta L}{\Delta t}[/tex]

Δ L = τ Δ t

initial angular  momentum is zero

L = 90 x 0.25

L = 22.5 kg.m²/s

Angular momentum is equal to 22.5 kg.m²/s

Answer:

0.15756 m/s

Explanation:

There are 10 blades and 10 gaps

To move through one blade or gap the windmill has to rotate

[tex]\dfrac{2\pi}{20}=0.31415\ rad[/tex]

This divided by the angular velocity is gives us the time

[tex]\dfrac{0.31415}{1.1}=0.28559\ s[/tex]

When the ball moves it does in a way that the ball must travel a distance of its own diameter which is [tex]4.5\times 10^{-2}\ m[/tex]

[tex]Speed=\dfrac{Distance}{Time}[/tex]

[tex]v=\dfrac{4.5\times 10^{-2}}{0.28559}\\\Rightarrow v=0.15756\ m/s[/tex]

The minimum linear speed is 0.15756 m/s

Answer:

C) 1.8×10⁵ N/C.

Explanation:

Electric field: This can be defined as the region where electric force is experienced.

Electric Field intensity: This is defined as the force per unit charge which it exert at that point.

The S.I unit of electric field is N/C.

Mathematically, Electric Field intensity can be represented as,

E = kq/r².................... Equation 1

Where E = electric field intensity, q = charge, r = distance. k = proportionality constant.

Given: q = 8.0 nC = 8×10⁻⁹ C, r = 2.0 cm = 0.02 m, k = 8.99×10⁹ Nm²/C²

Substituting into equation 1

E = (8.99×10⁹×8×10⁻⁹)/0.02²

E = 71.92/0.0004

E = 1.798×10⁵ N/C.

E ≈ 1.8×10⁵ N/C.

The right option is C) 1.8×10⁵ N/C.

Answer:

U₂=45 kJ

Explanation:

Given Data

Initial Energy U₁=19.5 kJ

Qin=30 kJ

Win=500 N.m

Qout= 5 kJ

To find

Final energy U₂

Solution

From The first law of thermodynamic

ΔE = Ein - Eout + ΔQ -ΔW

or

U₂=U₁+Qin+Win-Qout

where

U₂ is final energy

U₁ is initial energy

q = energy transferred as heat to a system

w = work done on a system

U₂=(19.5+30+500×10⁻³-5) kJ  

U₂=45 kJ

Final answer:

To calculate the final energy of a system involving heat transfer and work, first, find the net heat transfer by considering heat added and lost, and then include the work done. The first law of thermodynamics is applied to combine these energies with the initial energy to find the final energy, which is 45 kJ.

Explanation:

The question involves calculating the final energy of a system after heat transfer and work done on a closed pan of water on a stove. Given that 30 kJ of heat is transferred to the water, 5 kJ is lost to the surrounding air, and the paddle-wheel work on the system is 500 N·m (or 0.5 kJ since 1 N·m = 1 J), we can use the first law of thermodynamics to find the final energy of the system. The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

First, we calculate the net heat transfer (Q) to the system by subtracting the heat lost to the air from the heat supplied: Q = 30 kJ - 5 kJ = 25 kJ. Then, we add the work done on the system (which is positive because the work is done on the system, not by it): 25 kJ + 0.5 kJ = 25.5 kJ. This is the net energy input into the system.

Finally, to find the final energy of the system, we add this net input to the initial energy of the system: 19.5 kJ (initial energy) + 25.5 kJ (net input) = 45 kJ. Therefore, the final energy of the system is 45 kJ.

Answer:

Change in potential energy, [tex]U=1.2\times 10^8\ J[/tex]

Explanation:

Given that,

The potential difference between a storm cloud and the ground is 100 million V, [tex]V=100\ million V=100\times 10^6\ V=10^8\ V[/tex]

If a charge of 1.2 C flashes in a bolt from cloud to Earth, q = 1.2 C

We need to find the change of potential energy of the charge. The relation between the potential difference and the potential energy of the charge is given by :

[tex]U=qV[/tex]

U is the potential energy of the charge

[tex]U=1.2\times 10^8\ J[/tex]

So, the change of potential energy of the charge is [tex]U=1.2\times 10^8\ J[/tex]. Hence, this is the required solution.

According to the question,

                                                = 100×10⁶ V

                                                = 10⁸ V

The change in Potential energy:

→ [tex]U = qV[/tex]

By substituting the values, we get

      [tex]= 1.2\times 10^{8} \ J[/tex]

Thus the above response is right.

Learn more about potential difference here:

https://brainly.com/question/13300351

Answer:

[tex]\theta=44.03^{o}[/tex]

Explanation:  

Here we have an inelastic collision problem. We can use the momentum (p = mv) conservation law in each component of the displacement.

So, [tex]p_{i}=p_{f}[/tex]

X-component:

[tex]m_{1}v_{i1x}+m_{2}v_{i2x}=m_{1}v_{f1x}+m_{2}v_{f2x}[/tex] (1)

Now,

Then, using this information we can rewrite the equation (1).

[tex]m_{2}v_{i2x}=v_{fx}(m_{1}+m_{2})[/tex]

[tex]v_{fx}=\frac{m_{2}v_{i2x}}{m_{1}+m_{2}}=\frac{2597.2*164}{1720+2597.2}[/tex]

[tex]v_{fx}=98.66 km/h[/tex]

Y-component:

[tex]m_{1}v_{i1y}+m_{2}v_{i2y}=m_{1}v_{f1y}+m_{2}v_{f2y}[/tex] (2)

We can do the same but with the next conditions:

Then, using this information we can rewrite the equation (2).

[tex]m_{1}v_{i1y}=v_{fy}(m_{1}+m_{2})[/tex]

[tex]v_{fy}=\frac{m_{1}v_{i1y}}{m_{1}+m_{2}}=\frac{1720*239.44}{1720+2597.2}[/tex]

[tex]v_{fy}=95.39 km/h[/tex]

Now, as we have both components of the final velocity, we can find the angle East of North. Using trigonometric functions, we have:

[tex]tan(\theta)=\frac{v_{y}}{v_{x}}[/tex]

[tex]\theta=arctan(\frac{v_{y}}{v_{x}})=arctan(\frac{95.39}{98.66})[/tex]

[tex]\theta=44.03^{o}[/tex]

I hope it helps you!

Answer:

[tex]F_a=1470\ N[/tex]

Explanation:

Friction Force

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

[tex]\displaystyle F_a-F_{r1}-F_{r2}=m.a=0[/tex]

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

[tex]\displaystyle F_a=F_{r1}+F_{r2}.....[1][/tex]

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

[tex]\displaystyle F_{r2}-T=0[/tex]

The friction forces are computed by

[tex]\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g[/tex]

[tex]\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g[/tex]

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

[tex]\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g[/tex]

Simplifying

[tex]\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)[/tex]

Plugging in the values

[tex]\displaystyle F_{a}=0.25(9.8)[400+2(100)][/tex]

[tex]\boxed{F_a=1470\ N}[/tex]

Final answer:

To make the 100 kg block slip, a horizontal force of at least 245 N must be applied to the 400 kg lower block, calculated based on the maximum static friction force determined by the coefficient of friction and the weight of the 100 kg block.

Explanation:

To find the horizontal force F applied to the lower block that is necessary for the 100 kg block (upper block) to start slipping, we need to calculate the force required to overcome the static friction between the 100 kg block and the 400 kg block. The coefficient of friction (μ) is given as 0.25 for both contacts.

First, calculate the maximum static friction force that can act on the 100 kg block:

The weight of the 100 kg block (W) = 100 kg × 9.8 m/s² = 980 N.

Maximum static friction force (Ffriction) = μ × Normal force = 0.25 × 980 N = 245 N.

To initiate slipping, the horizontal force (F) applied must at least be equal to the maximum static friction force, which is 245 N. However, since this force will be applied to the combined system of both blocks, we must consider the total mass involved when finding the acceleration caused by the applied force.

The total mass of the system is 500 kg (400 kg + 100 kg). To move this mass with an acceleration that would cause the upper block to slip, we calculate:

F = ma, where m is the total mass and a is the acceleration. Since the force to overcome static friction is 245 N, this gives us the minimum force needed to initiate slipping when applied to the 400 kg block.

Answer:

Angle at which object is launched is 17.15°

Explanation:

We have given initial velocity at which object is launched u = 15 m/sec

It rises to a height of 1 m

So height h = 1 m

We have to find the angle of projection [tex]\Theta[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We know that maximum height is given by [tex]h=\frac{u^2sin^2\Theta }{2g}[/tex]

So [tex]1=\frac{15^2\times\ sin^2\Theta }{2\times 9.8}[/tex]

[tex]sin^2\Theta =0.0871[/tex]

[tex]sin\Theta =0.295[/tex]

[tex]\Theta =sin^{-1}0.295=17.15^{\circ}[/tex]

So the angle at which object is launched is 17.15°

Answer:

λ = 502 n m

Explanation:

given,

slit separation,  d = 0.040 mm

                             = 4 x 10⁻⁵ m

Distance between the fringes, D = 4.70 m

distance between the fringes, x = 5.9 cm

                                                     = 0.059 m

wavelength of the light = ?

the separation between the fringe is given by

[tex]x = \dfrac{n\lambda\ D}{d}[/tex]

[tex]\lambda = \dfrac{x d}{n\ D}[/tex]

where as n = 1

[tex]\lambda = \dfrac{0.059\times 4\times 10^{-5}}{1\times 4.70}[/tex]

 λ = 5.02127 x 10⁻⁷ m

λ = 502 n m

hence, the wavelength of the light is equal to λ = 502 n m