A rod of m= 1.3 kg rests on two parallel rails that are L = 0.42 m apart. The rod carries a current going between the rails (bottom to top in the figure) with a magnitude 1 = 2.6 A. A uniform magnetic field of magnitude B = 0.35 T pointing upward is applied to the region, as shown in the graph. The rod moves a distance d=1.25 m. Ignore the friction on the rails. † † † Ē I Otheexpertta.com A Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails.Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails. Assume the current through the rod is constant at all times.

Answer:

[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]

Explanation:

While the satellite is in the space vehicle, it has the next potential energy

[tex]U = -\frac{GmMe}{r}[/tex]

             where G is the gravitational constant

                         m is the satellite's mass in kilograms

                        Me is the earth's mass

                        r is the orbit's radius from to the earth's center in meters

[tex]U = - \frac{6.67x10^{-11}*2721.554*5.972x10^{24} }{482803}[/tex]

[tex]U = -2.2423x10^{12} J[/tex]

The additional energy required is the difference between this energy and the energy that the satellite would have in an orbit with an altitude of 22000 mi

[tex]U = -\frac{6.67x10^{-11}*2721.554*5.792x10^{24} }{35405568}[/tex]

[tex]U = -29696124610.3 J[/tex]

Then

[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]

Answer:

Resistance, [tex]R=0.529\ \Omega[/tex]

Explanation:

Given that,

Voltage of the battery, V = 9 volts

Current produced in the circuit, I = 17 A

We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :

[tex]V=IR[/tex]

[tex]R=\dfrac{V}{I}[/tex]

[tex]R=\dfrac{9\ V}{17\ A}[/tex]

[tex]R=0.529\ \Omega[/tex]

So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.

Answer:

Explanation:

a ) The speed of light will be 3 x 10⁸ m /s

b ) Speed of light will again be  3 x 10⁸ m /s . for an observer on the ground. It is so because speed of light is independent of moving frame of reference . It is absolute . It can not be changed by changing frame of reference .

The behavior of light differs because the speed of bullet will be less in case of stationary train . For moving train , speed of bullet will be increased by amount equal to speed of train for an observer on the ground.

Answer: elastic (e = 2.43)

Explanation:

The price elasticity formulae is given below as

Elasticy of price = change in quantity demanded / change in price.

P1 =$2.60, P2 = $6.30, q1 =17 and q2 = 8

e = q2 - q1/ P2 - P1

e = 8 - 17/ 6.30 - 2.60

e = - 9 /3.7

e = - 2.43

We take the modulus of e to have a positive value. Hence e = 2.43

Since e is greater than 1, then the elasticity of demand is elastic

Final answer:

The elasticity of demand between the given prices is calculated using the midpoint formula. It results in a value of -0.865 (ignoring the minus sign), indicating that the demand is elastic.

Explanation:

To calculate the elasticity of demand, we can use the midpoint formula, which is defined as the percentage change in quantity demanded divided by the percentage change in price. The formula for computing elasticity is:

Price Elasticity of Demand (Ed) = ([(Q2 - Q1) / ((Q2 + Q1)/2)] / [(P2 - P1) / ((P2 + P1)/2)]) * 100

Using the information provided: At a price of $2.60, quantity demanded is 17, and at a price of $6.30, quantity demanded is 8. We can fill in the values:

(Q2 - Q1) is (8 - 17) = -9

(Q2 + Q1)/2 is (8 + 17)/2 = 12.5

(P2 - P1) is ($6.30 - $2.60) = $3.70

(P2 + P1)/2 is ($6.30 + $2.60)/2 = $4.45

Now putting all these into the formula:

Ed = [(-9 / 12.5) / (3.70 / 4.45)] * 100

The percentage change in quantity is -72%, and the percentage change in price is 83.15%. Therefore:

Ed = (-0.72 / 0.8315) * 100

Ed = -0.865 (ignoring the minus sign)

This value is greater than 1, which indicates that the demand is elastic between these two prices; meaning, the quantity demanded is quite responsive to price changes.

Remember, we ignore the negative sign and use the absolute value when discussing elasticity, even though it is understood that price and quantity demanded move in opposite directions.

Final answer:

The correct statement about a wheel with four forces of the same magnitude but different directions acting tangentially at its rim is that both the net force on the wheel and the net torque about the CM are zero.

Explanation:

When four forces of the same magnitude but differing directions act tangentially at the rim of a uniform wheel free to spin about its center, the net force acting on the wheel may be zero if the forces cancel each other out. However, the direction and point of application of these forces are critical in determining the net torque. If the forces are applied in such a way that they create rotational effects that cancel each other, the net torque about the center of mass (CM) will also be zero. Therefore, the correct statement about the wheel is:

D. The net force on the wheel and the net torque on the wheel about the CM are zero.

To put it simply, the forces are arranged in a way that causes them to balance out, leading to no linear acceleration (net force is zero), and they are positioned symmetrically around the center of mass, producing no rotational acceleration (net torque is zero).

Answer:7.92 N

Explanation:

Given

mass of book [tex]m=1.83\ kg[/tex]

coefficient of static friction [tex]\mu _s=0.442[/tex]

coefficient of kinetic friction [tex]\mu _k=0.240[/tex]

To move the book, one need to overcome  the static friction  

Static friction [tex]F_s=\mu _sN[/tex]

[tex]F_s=\mu _s\times 1.83\times 9.8[/tex]

[tex]F_s=0.442\times 1.83\times 9.8[/tex]

[tex]F_s=7.92\ N[/tex]

After overcoming the Static friction , Force needed to move the block is

[tex]F_k=\mu _kN[/tex]

[tex]F_k=0.240\times 1.83\times 9.8[/tex]

[tex]F_k=4.30\ N[/tex]

Final answer:

The force needed to begin moving a 1.83 kg book on a flat desk, given a coefficient of static friction of 0.442, is approximately 7.90 N. This is calculated using the book's weight and the static friction formula fs(max) = μs * N.

Explanation:

To calculate the force needed to begin moving the book, we need to use the coefficient of static friction and the book's weight. The formula to find the maximum static frictional force (fs max) that must be overcome to start moving the object is fs(max) = μs * N, where μs is the coefficient of static friction and N is the normal force.

The book's weight is given by W = m * g, where m is the mass of the book (1.83 kg) and g is the acceleration due to gravity (9.81 m/s²). The weight of the book is equivalent to the normal force (N) exerted by the desk on the book since the book is resting on a flat surface and there are no other vertical forces acting on it. This simplifies the normal force to N = W = m * g.

Using the given coefficient of static friction (0.442) and the calculated normal force, the force needed to begin moving the book is fs(max) = 0.442 * (1.83 kg * 9.81 m/s²) ≈ 7.90 N. Therefore, a force slightly greater than 7.90 N is required to overcome static friction and start moving the book.

Answer:

Mechanical would have been conserved if only the force of gravity (the weight of the object does work on the system). The tension force does work also on the system but negative work instead. The net force acting of the system is zero since the upward tension in the string suspending the object is equal to the weight of the object but acting in the opposite direction. As a result they cancel out. In the equation above the effect of the tension force on the object has been neglected or not taken into consideration. For the mechanical energy E to be conserved, the work done by this tension force must be included into the equation. Otherwise it would seem as though energy has been generated in some manner that is equal in magnitude to the work done by the tension force.

The conserved form of the equation is given by

E = K + Ug + Wother.

In this case Wother = work done by the tension force.

In that form the total mechanical energy is conserved.

Final answer:

The correct answer is:  An external force is doing work on the system.

Explanation:

When the object is lowered at constant velocity, an external force (in this case, the force exerted by the person lowering the object) is doing work against gravity to keep the velocity constant. This work done by the external force results in a change in the object's gravitational potential energy (∆Ug = -mgh).

Since work is being done on the system externally, the total mechanical energy of the system (the sum of kinetic and potential energies) is not conserved.

However, in the scenario described, an external force is doing work on the system. Work is being done to counteract gravity and maintain a constant velocity. This work done by the external force results in a change in the object's gravitational potential energy (∆Ug = -mgh), where "m" is the mass of the object, "g" is the acceleration due to gravity, and "h" is the vertical distance through which the object is lowered.

As a result, the total mechanical energy of the system (the sum of kinetic and potential energies) is not conserved. The work done by the external force manifests as a change in the object's gravitational potential energy, causing a decrease in potential energy as the object is lowered. This decrease in potential energy is exactly balanced by the work done by the external force, so the total mechanical energy of the system remains constant, despite the fact that kinetic energy remains constant throughout the process.

Answer:

V3 = 7.802 m/s

Explanation:

m1 = 640 Kg, M2 = 816 kg, V1 = 63.9 Km/h = 17.75 m/s, V2 =0 m/s

Let V3 is the combine velocity after collision.

According to the law of conservation of momentum

m1 v1 + m2 v2 = (m1 + m2) v3

⇒ V3 =( m1 v1 + m2 v2 ) / (m1 + M2)

V3 =  ( 640 Kg × 17.75 m/s + 816 kg × 0m/s) / (640 Kg + 816 kg)

V3 = 7.802 m/s

Answer:

28.088 km/h or 7.802 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')....................... Equation 1

Where m = mass of the automobile, m' = mass of the car, u = initial velocity of the car, u' = initial velocity of the car, V = velocity of the two car after collision

Make V the the subject of the equation

V = (mu+m'u')/(m+m')................ Equation 2

Given: m = 640 kg, m' = 816 kg, u = 63.9 km/h, u' = 0 m/s (parked).

Substitute into equation 2

V = (640×63.9+816×0)/(640+816)

V = 40896/1456

V = 28.088 km/h = 28.088(1000/3600) m/s = 7.802 m/s

Hence the speed of the two cars after they collide = 28.088 km/h or 7.802 m/s

Answer:

Explanation:

Total charge passed

= 1.25 x 60 x 60 x 5.1 C

= 22950 C

Equivalent mass of copper

= 63.5 / 2

= 31.75 g

96500 coulomb is required to obtain 31.75 g of copper

22950 C will release (31.75 / 96500) x 22950

= 7.55 g of copper .

Answer and Explanation:

The answer is attached below

The velocity of both the objects after the collision is [tex]\frac{25}{11} m/s[/tex].

(a) the relative velocity of the masses is zero.

(b) the compression of the spring is 0.185m.

The given case is an example of an inelastic collision in which two objects after the collision move together. The momentum of the system is conserved, therefore,

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

here, m₁ = 5kg , m₂ = 6kg

u₁ = 5 m/s , initial velocity of mass m₁, and

u₂= 0, initial speed of the mass m₂

[tex]5\times5+0=(5+6)v\\\\v=\frac{25}{11}m/s[/tex]

(a) When the spring is fully compressed both the masses move with the same velocity, therefore the relative speed of the masses is zero.

(b) from the law of conservation of energy:

the initial kinetic energy of the masses is converted into final kinetic energies and the potential energy of the spring:

[tex]\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u^2_2=\frac{1}{2} (m_1+m_2)v^2+\frac{1}{2}kx^2[/tex]

where x is the compression of the spring

[tex]\frac{1}{2}\times5\times5^2+0=\frac{1}{2}(5+6)\times(\frac{25}{11})^2=\frac{1}{2}\times2000x^2\\\\x=0.185m[/tex]

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Answer:

Explanation:

This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.

Applying extra gas to a car moving at a constant velocity does not violate Newton's 1st Law as the additional force is required to counteract other forces, such as air resistance and road friction, allowing the car to maintain its velocity. This is, in fact, a confirmation of Newton's 1st Law.

Even while driving at a constant velocity on a straight highway, one must occasionally apply a bit of extra gas, which means applying an extra external force.

However, this does not violate Newton's 1st Law, or the law of inertia, which states that an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

The reason for this is the existence of other forces acting on the car that work against the forward motion. These can include air resistance (or drag), friction from the road, and uphill forces if the road is inclined.

The extra force (gas) provided helps overcome these forces, allowing the car to maintain a constant velocity. So, this activity is actually a confirmation of Newton's first Law, not a violation.

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Explanation:

Formula for path difference is as follows.

             x = 2tn

and,     refractive index (n) = [tex]\frac{\lambda}{2t}[/tex]

Thickness is calculated as follows.

           Thickness (t) = [tex]\frac{volume}{area}[/tex]

       Area = [tex]\pi r^{2}[/tex]

                = [tex]\pi \times (\frac{d}{2})^{2}[/tex]

                = [tex]\frac{0.49 \times 10^{-6}}{3.14 \times 0.01 m}[/tex]

                = [tex]1.56 \times 10^{-8}[/tex] m

Now, the refractive index will be calculated as follows.

For drop,      n = [tex]\frac{\lambda}{2t}[/tex]

For B drop,    n = [tex]\frac{\lambda}{26t}[/tex]

So,     n = [tex]\frac{640 \times 10^{-9}}{26 \times 1.56 \times 10^{-8}}[/tex]

            = [tex]\frac{640 \times 10^{-9}}{40.56 \times 10^{-8}}[/tex]

            = 1.5

Thus, we can conclude that index of refraction of the oil is 1.5.

Answer:

10ns

Explanation:

Suppose the EM wave travels at light speed [tex]c = 3\times10^8 m/s[/tex]. A change in travel distance of the order of 3 m would result of a change in travel time of

[tex] \Delta t = \frac{\Delta s}{c} = \frac{3}{3\times10^8} = 10^{-8} s = 10 ns[/tex]

To detect a change in travel distance of about 3 m, a GPS receiver must measure a change in travel time of approximately 10 nanoseconds. This is calculated using the speed of light and the time = distance/speed formula.

To determine the associated travel time that must be measured for a change in travel distance of 3 meters, we need to know the speed of electromagnetic waves, which is the speed of light, typically denoted 'c'. The speed of light is approximately 3.00 x 108 meters per second (m/s).

Now we can calculate the time it takes for light to travel a given distance using the formula time (t) = distance (d) / speed (s). Substituting for our given distance (3 m) and the speed of light, we get t = 3 m / 3.00 x 108 m/s. This results in a time of approximately 1 x 10-8 seconds or 10 nanoseconds (ns).

Therefore, to detect a change in travel distance of about 3 m, the GPS receiver must be able to measure a change in travel time of roughly 10 nanoseconds.

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Kinetic energy lost in collision is 10 J.

Explanation:

Given,

Mass, [tex]m_{1}[/tex] = 4 kg

Speed, [tex]v_{1}[/tex] = 5 m/s

[tex]m_{2}[/tex] = 1 kg

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

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

[tex]\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2[/tex]

By plugging in the values we get,

[tex]KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\[/tex]

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

[tex]KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)[/tex]

[tex]KE = 40J + KE(lost)[/tex]

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

Answer:

The lethal voltage for the electrician under those conditions is 126.5 V.

Explanation:

To discover what is the lethal voltage to the electrician we need to find out what is the voltage that produces 55 mA = 0.055 A when across a resistance of 2300 Ohms (Electrician's body resistancy). For that we'll use Ohm's Law wich is expressed by the following equation:

V = i*R

Where V is the voltage we want to find out, i is the current wich is lethal to the electrician and R is his body resistance. By applying the given values we have:

V = 0.055*2300 = 126.5 V.

The lethal voltage for the electrician under those conditions is 126.5 V.

Answer:

126.5 V

Explanation:

Using Ohm's Law,

V = IR............................. Equation 1

Where V = Voltage, I = current, R = resistance.

Note: The current needed to bring about the fatal voltage is equal to the current that will cause human being to be electrocuted.

Given: I = 55 mA = 55/1000 = 0.055 A, R = 2300 Ω

Substitute into equation 1

V = 0.055×2300

V = 126.5 V.

Hence the fatal voltage = 126.5 V

Answer:

58.6 N

Explanation:

We are given that

[tex]q_1=-7.97\mu C=-7.97\times 10^{-6} C[/tex]

[tex]q_2=3.55\mu C=3.55\times 10^{-6} C[/tex]

Using [tex]1\mu C=10^{-6} C[/tex]

[tex]r=6.59 cm=6.59\times 10^{-2} m[/tex]

[tex]1 cm=10^{-2} m[/tex]

The magnitude of force that one particle exerts on the other

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Where [tex]k=9\times 10^9[/tex]

Substitute the values

[tex]F=\frac{9\times 10^9\times 7.97\times 10^{-6}\times 3.55\times 10^{-6}}{(6.59\times 10^{-2})^2}[/tex]

F=58.6 N

Answer: [tex]||\vec A|| \approx 9.95 m,||\vec B|| \approx 8.19 m[/tex]

Explanation:

Magnitudes can be calculated by using Pythagorean theorem:

[tex]||\vec A|| =\sqrt{(5m)^{2}+(-7m)^{2}+(5m)^{2}}\\||\vec A|| \approx 9.95 m\\||\vec B|| =\sqrt{(3m)^{2}+(7m)^{2}+(-3m)^{2}}\\||\vec B|| \approx 8.19 m\\[/tex]

Answer:

V = 144mi/h

Explanation:

Please see the attachment below.

Answer:

42.4 Npa

Explanation:

Explanation is attached in the picture below

Explanation:

Below is an attachment containing the solution.

To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.

To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for electric force is:

F = k * (q1 * q2) / r^2

Where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges. In this case, the charge of the Moon is 27.3% of the charge of the Earth, and the radius of the Moon is also 27.3% of the radius of the Earth. Using these values, we can calculate the electric force.

Let's assume the charge of the Earth is q1 and the charge of the Moon is q2. Since the charge of the Moon is 27.3% as large as the charge of the Earth, we can write q2 = 0.273 * q1. Similarly, the radius of the Moon is 27.3% of the radius of the Earth, so we can write r = 0.273 * R, where R is the radius of the Earth. Plugging these values into Coulomb's law formula:

F = k * (q1 * (0.273 * q1)) / (0.273 * R)^2

Simplifying the equation, we get:

F = k * (q1^2 * 0.273) / (0.273^2 * R^2)

The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.

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Answer:

91.017N

Explanation:

Parameters

L=4.67m, m=0.192kg, t = 0.794s, The pulse makes four trips down and back along the cord, we have 4 +4 =8 trips( to and fro)

so N= no of trips = 8, From  Wave speed(V) = N *L/t , we have :

V= 8*4.67/0.794 = 47.0529 m/s.

We compute the cords mass per length,  Let it be P

P = M/L = 0.192/4.67 = 0.04111 kg/m

From  T = P * V^2  where T = Tension, we have

T = 0.04111 * (47.0529)^2

T =  91.017N.

The tension in the cord is 91.017N

Answer:

so 40 ways for distinguishable

4 ways for bosons

1 way for fermions

Explanation:

The explanation is attached below

Answer:

Therefore, the force exerted by the magnetic field on the particle goes along the positive y- direction

Explanation:

The equation for the force exerted on a particle due to the magnetic field is equal to:

F = q*(v * B)

If we replace vi for v, +q for q and -Bk for B, we have:

F = +q*(vi*(-Bk)) = -q*v*B*(i * k)

From this equation we have that the vector i is in direction -x and k in direction -z

We compute the cross product of two unit vectors:

i * k = -j, where j is the vector along -y direction

Replacing we have:

F = -q*v*B*(-j) = -q*v*B*j

Therefore, the force exerted by the magnetic field on the particle goes along the positive y- direction

Answer:

a. Perpendicular.

Explanation:

The relation between angular velocity vector and linear velocity vector is described by a cross product:

[tex]\vec v = \vec \omega \times \vec r[/tex]

Where [tex]\vec r[/tex] is perpendicular to the rotation axis and [tex]\vec \omega[/tex] is parallel to the same axis. By definition of cross product, [tex]\vec v[/tex] is a vector which is perpendicular to both vectors. Therefore, linear velocity vector is perpendicular to angular velocity vector. The correct answer is A.

The angular velocity vector, which points along the rotation axis, is perpendicular (Option A) to the linear velocity vector, which is tangent to the rotation path.

The angular velocity of a point on a rotating object is a vector that points along the axis of rotation. This vector's direction is determined with respect to the right-hand rule: if the fingers on your right-hand curl from the x-axis toward the y-axis, your thumb points in the direction of the positive z-axis. For an angular velocity pointing along the positive z-axis, the rotation is counterclockwise, whereas for an angular velocity pointing along the negative z-axis, the rotation is clockwise.

On the other hand, the linear velocity of the same point is a vector that is tangent to the path of rotation. It represents the instantaneous linear speed of the point as it moves along the path.

Therefore, given the definitions and directions of both vectors, the angular velocity vector of a point on a rotating object is perpendicular to the linear velocity vector of that point.

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Answer:

Pithball electroscope is used to determine if any object has static charge.

It consists of one or two small balls of a lightweight non-conductive substance. When this material is moved near an object having static charge, polarization will be induced in the atoms of pithballs which will either attract or repel the object depending on the nature of the charge in it.

It will not move in case it is brought near to a neutral object.

Similarly, the pith ball will move when it will brought into the magnetic field of the magnet as it will also induce polarization within the atoms of the pithballs.

Answer:

(a) v = 15m/a

(b) No they won't feast because the rock can only rise to a height of 11.5m which is less than 15m.

Explanation:

Please see the attachment below for film solution.

The rock does not reach the fruit because the maximum height it achieves is 11.48 meters, which is below the required 15 meters. Therefore, the contestants will go hungry.

Let's solve the problem step by step. First, we'll determine the speed with which the rock leaves the spring, and then we'll check if this speed is enough to reach the fruit hanging in the tree.

Part (a): Speed of the rock leaving the spring

Using the principle of conservation of energy, the potential energy stored in the spring when compressed will be converted into the kinetic energy of the rock when the spring is released.

The potential energy stored in the spring [tex]\( E_{\text{spring}} \)[/tex] is given by:

[tex]\[E_{\text{spring}} = \frac{1}{2} k x^2\][/tex]

Substituting the given values:

[tex]\[E_{\text{spring}} = \frac{1}{2} \times 1000 \, \text{N/m} \times (0.30 \, \text{m})^2\][/tex]

[tex]\[E_{\text{spring}} = \frac{1}{2} \times 1000 \times 0.09\][/tex]

[tex]\[E_{\text{spring}} = 45 \, \text{J}\][/tex]

This energy will be converted into kinetic energy [tex]\( E_{\text{kinetic}} \)[/tex] of the rock:

[tex]\[E_{\text{kinetic}} = \frac{1}{2} m v^2\][/tex]

Equating the spring potential energy to the kinetic energy:

[tex]\[45 \, \text{J} = \frac{1}{2} \times 0.4 \, \text{kg} \times v^2\][/tex]

Solving for ( v ):

[tex]\[45 = 0.2 \times v^2\][/tex]

[tex]\[v^2 = \frac{45}{0.2}\][/tex]

[tex]\[v^2 = 225\][/tex]

[tex]\[v = \sqrt{225}\][/tex]

[tex]\[v = 15 \, \text{m/s}\][/tex]

So, the speed of the rock as it leaves the spring is [tex]\( 15 \, \text{m/s} \).[/tex]

Part (b): Checking if the rock can reach the fruit

To determine if the rock can reach the height of 15 m, we use the following kinematic equation for vertical motion:

[tex]\[h = v_0 t - \frac{1}{2} g t^2\][/tex]

First, determine the time ( t ) it takes to reach the maximum height where the vertical velocity becomes zero:

[tex]\[v = v_0 - g t\][/tex]

At the maximum height, [tex]\( v = 0 \):[/tex]

[tex]\[0 = 15 - 9.81 t\][/tex]

[tex]\[t = \frac{15}{9.81}\][/tex]

[tex]\[t \approx 1.53 \, \text{s}\][/tex]

Now, calculate the maximum height [tex]\( h_{\text{max}} \):[/tex]

[tex]\[h_{\text{max}} = v_0 t - \frac{1}{2} g t^2\][/tex]

[tex]\[h_{\text{max}} = 15 \times 1.53 - \frac{1}{2} \times 9.81 \times (1.53)^2\][/tex]

[tex]\[h_{\text{max}} = 22.95 - 11.47\][/tex]

[tex]\[h_{\text{max}} = 11.48 \, \text{m}\][/tex]

The maximum height reached by the rock is approximately 11.48 m, which is less than the 15 m needed to reach the fruit.

Answer:

2y

Explanation:

Electric field in terms of Electric potential is given as:

E = dV/dr(x, y, z)

Where r(x, y, z) = position in x, y, z plane

The y component of the Electric field will be:

Ey = -dV/dy

Given that

V = 2x - y² - cos(z)

dV/dy = -2y

=> E = - (-2y)

E = 2y

Answer:

(a) linear momentum = 8.75.  angular momentum =0. (b) 2.5m/s

Explanation:

A and B are on horizontal and so is the velocity given to A = 3.5m/s i (in i direction which is x direction ), this means that there would be motion only in x direction.

let's first calculate linear momentum.

(A)

p(total)= P_A+P_B = (2.5Kgx3.5m/s)+(1kgx0m/s) = 8.75m/s.

Angular momentum.

L=m*v x r. as the motion is only in x direction and there is no rotation in the system, there fore m*v x r = 0

(B)

Velocity of A and B come from the fact that total linear momentum is conserved.

p_final = P_A+P_B = (mA=mB)v_new = mA*V1+mB*vB.

second term on right is = 0 because B has 0 velocity.

solving for V_new and with the values of all unknown substituted in gives

V_new = (mA*VA)/(mA+mB)= 8.75/3.5= 2.5m/s

The diagram showing the 2 spheres is missing, so i have attached it.

Answer:

A) Linear momentum =(8.75 kg/m.s)i

Angular momentum =  (-0.5kg.m²/s)k

B) Va' = (1.5m/s)i and Vb' = (5 m/s)j

Explanation:

First of all, let's find the position of the mass centre;

y' = Σ(mi.yi)/mi = [2.5(0) + 1(0.2)]/(2.5+1) = 0.2/3.5 =0.057143 m

A) Linear momentum is given as;

L = m(a) x v(o) = (2.5 x 3.5)i = (8.75 kg/m.s)i

Angular momentum is given as;

HG = Vector GA  x m(a) x v(o)

Where vector GA is the position of the mass centre;

Thus;

HG = 0.057143j x (2.5 x 3.5)i = (-0.5 kg.m²/s) k

B) from conservation of linear momentum;

MaVo = Ma Va' + Mb Vb

2.5 x 3.5 = 2.5Va' + 1 Vb'

8.75 = 2.5Va' + 1 Vb' - - - - eq(1)

Also, for conservation of angular momentum ;

raMaVo = - raMa Va' + rbMb Vb'

from the diagram attached, ra + rb = 0.2

Now, ra is the same as  value as that of the centre of the mass.

Thus, ra = 0.057143.

rb = 0.2 - ra = 0.2 - 0.057143 = 0.14286

Thus;

0.057143 x 8.75 = -(0.057143 x 2.5)Va' + (0.14286 x 1) Vb'

0.5 = - 0.14286Va' + 0.14286Vb' - - - eq 2

Solving eq 1 and 2 simultaneously, we get; Va' = 1.5m/s and Vb' = 5 m/s

Answer:

37.5 N

Explanation:

Horizontal component is represented by Fx and  is given as

Fx = F cos θ

Here,

θ = 60 degrees

F= 75 N

So,

Fx= 75 cos (60°)

==> Fx = 75 ×  0.5 =37.5 N

Final answer:

The horizontal component of the force acting parallel to the floor when a 150 N force is applied at a 60-degree angle to the horizontal is 75 N, calculated using the cosine of the angle.

Explanation:

The question involves resolving a force into its components, which is a common problem in physics. When a force is applied at an angle to the horizontal, it has both horizontal and vertical components. The horizontal (parallel) component ([tex]F_{parallel}[/tex]) is found using the cosine function of the angle Θ , which in this case is 60 degrees. The formula is [tex]F_{parallel}[/tex] = F * cos(Θ), where F is the magnitude of the force.

Given that a force of 150 N is applied at an angle of 60 degrees to the horizontal, the horizontal component of the force can be calculated as follows:

[tex]F_{parallel}[/tex] = 150 N * cos(60°)
[tex]F_{parallel}[/tex] = 150 N * 0.5
[tex]F_{parallel}[/tex] = 75 N

The horizontal component of the force acting parallel to the floor is 75 N. To find the net force acting on the box and thus the acceleration, you would subtract the frictional force from the parallel component of the applied force and then use Newton's second law, F = m * a, where F is the net force, m is the mass of the object and a is the acceleration. However, the frictional force is not needed for this particular question about the horizontal component.

Answer:

The minimum slope angle for which the skier can safely traverse the snow bridge is 47.82°

Explanation:

Given that,

Weight = 700 N

Normal force = 470 N

Suppose, Find the minimum slope angle for which the skier can safely traverse the snow bridge.

We need to calculate the minimum slope angle

Using balance equation

[tex]F_{N}=mg\cos\theta[/tex]

[tex]\theta=\cos^{-1}(\dfrac{F_{N}}{mg})[/tex]

Put the value into the formula

[tex]\theta=\cos^{-1}(\dfrac{470}{700})[/tex]

[tex]\theta=47.82^{\circ}[/tex]

Hence, The minimum slope angle for which the skier can safely traverse the snow bridge is 47.82°

When a 700 N skier crosses a snow bridge that can only support 470 N, the bridge will collapse due to the skier's weight exceeding the bridge's maximum normal force. Newton's second law can be used to understand this situation, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The relationship between weight, mass, and the gravitational force can also be used to determine that the normal force should be equal to the skier's weight for the bridge to be safe.

The maximum normal force that the snow bridge can sustain is 470 N. The backcountry skier weighs 700 N, so the bridge will collapse under their weight.

To understand why, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The skier's weight is acting downwards, while the normal force of the snow bridge is acting upwards. When the skier crosses the bridge, the normal force should be equal to their weight for the bridge to be in equilibrium. However, since the normal force is less than the skier's weight, the bridge cannot sustain the weight and will collapse.

This situation can be described using the formula: Weight = mass x gravitational force. So, we can calculate the mass of the skier by dividing their weight by the gravitational force. With a weight of 700 N and a gravitational force of 9.8 m/s², the mass is approximately 71.4 kg. Therefore, the normal force supporting the skier on the bridge should be equal to 700 N for it to be safe.

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