Select the impulse-momentum bar charts for the next problems. A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s. Both blocks stick together and move to the right. What is their speed after collision?

Answers

Answer 1

Answer with Explanation:

since the two blocks move (stick) together, the collision is inelastic, which does not conserve kinetic energy.  So do not use kinetic energy consideration.

Fortunately, in such a situation, momentum is still conserved.

Momentum of 1.0 kg block

= 1.0 * 3.0 = 3.0 kg-m/s

Momentum of second block

= 1.0 * 1.0 = 1.0 kg-m/s

Total mass after collision = 1.0+1.0 = 2.0 kg

Common velocity after collision

= total momentum / total mass

= (3.0+1.0)/2.0 = 2.0 m/s


Related Questions

A transformer is used to change: 1) power 2) frequency 3) capacitance 4) none of the above

Answers

4)none of the above.

A transformer is an electrical device  that allows to increase or decrease from one voltage level to another in an AC circuit while conserving the power.  

It is basically composed of two coils magnetically coupled in a core of ferromagnetic material increasing its permeability and effectiveness.

Answer: None of the above

Explanation:

1400 kg car has a speed of 27 m/s. If it takes 7 s to stop the car, what is the Impulse and the average force acting on the car?

Answers

Explanation:

Impulse is change in momentum:

I = Δp

I = (1400 kg) (27 m/s) - (1400 kg) (0 m/s)

I = 37800 kg m/s

Impulse is also average force times time:

I = F ΔT

37800 kg m/s = F (7 s)

F = 5400 N

A 39-kg girl is bouncing on a trampoline. During a certain interval after leaving the surface of the trampoline her kinetic energy decreases to 160 J from 490 J. How high does she rise during this interval? Neglect air resistance.

Answers

Answer:

The distance is 0.86 m.

Explanation:

Given that,

Mass = 39 kg

Initial kinetic energy, [tex]K.E_{i} = 160\ J[/tex]

Final kinetic energy, [tex]K.E_{f} = 490\ J[/tex]

We need to calculate the work done

According to work energy theorem

[tex]W = \Delta K.E[/tex]

[tex]W=K.E_{f}-K.E_{i}[/tex]...(I)

Work done is the product of the force and displacement.

[tex]W = mgh[/tex]....(II)

From equation (I) and (II)

[tex]K.E_{f}-K.E_{i}=mgh[/tex]

[tex]490-160=39\times9.8\times h[/tex]

[tex]h = 0.86\ m[/tex]

Hence, The distance is 0.86 m.

A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the particle's motion?Check all that apply.Increasing its chargeIncreasing its massIncreasing the field strengthIncreasing its speed

Answers

Answer:

Increasing its mass

Explanation:

More mass -> more inertia -> increased period of motion

Final answer:

Increasing the mass of the charged particle will increase the period of its motion in a uniform magnetic field. However, increasing the particle's charge, the field strength or the velocity would not result in an increased period.

Explanation:

The period of a charged particle's motion in a uniform magnetic field is determined by the equation T = 2πm/(qB), where T is the period, m is the particle's mass, q is its charge, and B is the strength of the magnetic field. One method to increase the period of motion is therefore to increase the particle's mass. Intuitively, this makes sense as a more massive particle is harder to turn, thus takes longer time to complete a cycle. However, increasing the charge q or the field strength B would decrease the period since they are in the denominator, while increasing the velocity would not affect the period in a uniform magnetic field, since the particle's speed is perpendicular to their motion.

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Which of the following is NOT an example of an assembly process? a) Handling b) Fitting c) Automated robot d) Orientation

Answers

Answer: I think it is D

Explanation: It is the only one that does not make sense

Final answer:

The correct answer is 'c) Automated robot' because it is a tool used within the assembly process, not a process itself like handling, fitting, or orientation.

Explanation:

The question is asking which of the provided options is not an example of an assembly process. An assembly process involves the steps required to put together parts to make a complete product. The options given are:

Handling - the act of manipulating components in preparation for assembly.

Fitting - the process of putting parts together, which is certainly part of assembly.

Automated robot - often used in assembly to perform repetitive tasks more efficiently.

Orientation - this typically refers to aligning parts in the correct position for assembly.

Based on these definitions, automated robot is not an example of an assembly process but is rather a tool that might be used in the process. Therefore, the correct answer is 'c) Automated robot'

A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vertical.
(a) How far does the mass fall (y-displacement) before reaching its lowest point?
(b) How much work is done by gravity as it falls to its lowest point?
(c) How much work is done by the string tension as it falls to its lowest point?

Answers

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

[tex]\theta=65.4^{\circ}[/tex]

The projection of the length of the pendulum along the vertical direction is

[tex]L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm[/tex]

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

[tex]\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m[/tex]

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

[tex]\Delta U = mg \Delta y[/tex]

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

[tex]\Delta y = 0.211 m[/tex] is the vertical displacement of the pendulum

So, the work done by gravity is

[tex]W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J[/tex]

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force

d is the displacement

[tex]\theta[/tex] is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

[tex]\theta=90^{\circ}, cos \theta = 0[/tex]

and so the work done is zero.

Final answer:

The pendulum falls 0.212 m, the work done by gravity is 0.349 J, and the work done by string tension is 0 J.

Explanation:

The first part of the question asks for the vertical displacement (y-displacement) of the pendulum. The length of the pendulum is the hypotenuse of a right triangle, and the vertical displacement is the adjacent side, so we can use the cosine function to solve: y = L*cos(θ). Plugging in the given values: y = 0.361 m * cos(65.4 degrees) = 0.149 m. So the fall is the length of the pendulum minus this displacement: 0.361 m - 0.149 m = 0.212 m.

The second part of the question asks for the work done by gravity. The work done by gravity is equal to the weight of the pendulum times the vertical distance it falls (Work = m*g*y), or 0.168 kg * 9.8 m/s² * 0.212 m = 0.349 J.

The final part of the question asks for the work done by string tension. The tension force always acts perpendicular to the direction of displacement, meaning it does no work on the pendulum, as work is defined as force times the displacement in the direction of the force. Therefore, the work done by the tension in the string is 0 J.

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An 80.0 g sample of a gas was heated from 25 ∘C25 ∘C to 225 ∘C.225 ∘C. During this process, 346 J of work was done by the system and its internal energy increased by 6085 J.6085 J. What is the specific heat of the gas?

Answers

Final answer:

The specific heat of the gas is 0.381 J/g°C.

Explanation:

The specific heat of a substance is the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius. To find the specific heat of the gas, we can use the equation:



Q = mass * specific heat * change in temperature



In this case, the mass of the gas is 80.0g, the change in temperature is 225°C - 25°C = 200°C, and the energy absorbed by the gas is 6085J. Plugging these values into the equation, we can solve for the specific heat:



6085J = 80.0g * specific heat * 200°C



Specific heat = 6085J / (80.0g * 200°C) = 0.381 J/g°C

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Final answer:

The specific heat of the given gas can be calculated using the formula c = Q / (m * ΔT). By substitifying Q = 6431J (sum of work done by the system and increase in internal energy), m = 80g (mass of gas), and ΔT = 200°C (change in temperature), we find that the specific heat of the gas is roughly 0.40 J/g°C

Explanation:

The specific heat of a substance is the heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. For the given gas, we first need to calculate the heat absorbed which is the sum of the work done by the system and the increase in its internal energy, which equals 346J + 6085J = 6431J. The following information is given, mass m = 80.0g and the change in temperature ΔT = 200°C (225 °C - 25°C).

The formula for specific heat is:
Q = m*c*ΔT
where:
Q is the heat energy absorbed (or released),
m is the mass of the substance,
c is the specific heat capacity, and
ΔT is the change in temperature.
Solving this equation for the specific heat (c), we find:
c = Q / (m * ΔT).

So, substituting the given values into this formula, we have:
c = 6431J /  (80.0g * 200°C = 0.40 J/g °C which will be the specific heat of the gas.

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A 1.55 kg falcon catches a 0.395 kg dove from behind in midair. What is their velocity after impact if the falcon's velocity is initially 29.5 m/s and the dove's velocity is 7.15 m/s in the same direction?

Answers

Answer:

The velocity after impact 24.96 m/s

Explanation:

Given that,

Mass of falcon = 1.55 kg

Mass of dove = 0.395 kg

Velocity of falcon = 29.5 m/s

Velocity of dove = 7.15 m/s

We need to calculate the velocity after impact

Using conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]

Where,

u = initial velocity

v = final velocity

[tex]m_{1}[/tex] = mass of falcon

[tex]m_{2}[/tex] = mass of dove

Put the value in the equation (I)

[tex]1.55\times29.5+0.395\times7.15=(1.55+0.395)v[/tex]

[tex]v=\dfrac{1.55\times29.5+0.395\times7.15}{(1.55+0.395)}[/tex]

[tex]v=24.96\ m/s[/tex]

Hence, The velocity after impact 24.96 m/s

Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 1900 kg/s steadily. Determine the power generation potential

Answers

Answer:

P = 2234400 Watt

Explanation:

Power generation is given as rate of work done

Here the turbine generator is located at 120 m below the free surface

So here rate of work done is given as rate of potential energy

[tex]Power = \frac{dw}{dt}[/tex]

[tex]Power = \frac{mgh}{t}[/tex]

so we have

[tex]Power = (\frac{dm}{dt})gh[/tex]

now we have

[tex]Power = 1900(9.8)(120) = 2234400 Watt[/tex]

so the power generation potential will be 2234400 Watt

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6 m, y = 8.5 m, and has velocity ~vo = (9 m/s) ˆı + (−2.5 m/s) ˆ . The acceleration is given by ~a = (4.5 m/s 2 ) ˆı + (3 m/s 2 ) ˆ . What is the x component of velocity after 3.5 s? Answer in units of m/s.

Answers

The x component of velocity after 3.5 seconds is 24.75 m/s, calculated using the equation v = v0 + at with the given initial velocity and constant acceleration.

To calculate the x component of velocity after 3.5 seconds for a particle under constant acceleration, we use the equation:

v = v0 + at

Here v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time elapsed. Given v0 = 9 m/s in the x direction, and acceleration a = 4.5 m/[tex]s^2[/tex] in the x direction, and time t = 3.5 s, the calculation is:

v = 9 m/s + (4.5 x (3.5))

The x component of velocity after 3.5 seconds is:

v = 9 m/s + 15.75 m/s

v = 24.75 m/s

what is periodicity?

Answers

Hello There!

"Periodicity" are types of trends that are seen in element properties.

This is the same thing as periodic trends. These are patterns that are present in the periodic table. These trends show different properties of elements and how characteristics increase or decrease.

What is the difference between center of gravity and center of mass?

Answers

The center of mass is the mean position in an object. Then there’s the center of gravity which is the point where gravity appears to act.

A 320.9 ng sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 47 days the amount of radioactive substance had decreased to 40.11 ng. How many half‑lives of the unknown radioactive substance have occurred?

Answers

Answer:

3 half life of the unknown radioactive substance have occurred.

Explanation:

Mass of sample = 320.9 ng

Mass after 1 half life = 0.5 x 320.9 = 160.45 ng

Mass after 2 half life = 0.5 x 160.45 = 80.225 ng

Mass after 3 half life = 0.5 x 80.225 =40.11 ng

So 3 half life of the unknown radioactive substance have occurred.

Final answer:

The unknown radioactive substance went through approximately 3 half-lives in the 47 days period.

Explanation:

The subject of your question is related to radioactive decay, particularly the concept of a half-life, which is a term in physics used to describe the time it takes for half the atoms in a sample to decay. In your case, the initial mass of your substance was 320.9 ng and it decreased to 40.11 ng after 47 days. To figure out how many half-lives have occurred, we need to understand that with each half-life, the quantity of the substance halve its original mass.

A useful method to answer your question is to divide the final amount by the starting amount and then take the logarithm base 2 of the result. This will give us the number of times the amount halved, which is the number of half-lives. In your case, the calculation is like this: number of half lives = log2(320.9 ng / 40.11 ng) = approx. 3 half-lives.

This means that in 47 days, approximately 3 half‑lives of the unknown radioactive substance have occurred, based on the given data. It's important to note that because this is an approximate result, the true half-life of the substance might be slightly smaller or larger.

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The brightest star in the night sky in the northern hemisphere is Sirius. Its distance from Earth is estimated to be 8.7 light years. A light year is the distance light travels in one year. Light travels at a speed of 3.00 × 108 m/s. Calculate the distance from Earth to Sirius in miles. (1 mi = 5280 ft) g

Answers

Answer:[tex]5.11(10)^{13}miles[/tex]

Explanation:

A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "

In other words: It is the distance that the light travels in a year.  

This unit is equivalent to [tex]5.879(10)^{12}miles[/tex], which mathematically is expressed as:

[tex]1Ly=5.879(10)^{12}miles[/tex]

Doing the conversion:

[tex]8,7Ly.\frac{5.879(10)^{12}miles}{1Ly}=5.11(10)^{13}miles[/tex]  This is the distance from Earth to Sirius in miles.

Final answer:

The distance from Earth to Sirius is approximately 50 trillion miles.

Explanation:

To calculate the distance from Earth to Sirius in miles, we need to convert the light year measurement to miles. We know that 1 light year is the distance light travels in one year, which we can calculate by multiplying the speed of light by the number of seconds in a year. Here's the step-by-step calculation:

Speed of light = 3.00 × 10^8 m/s

Number of seconds in a year = 365 days * 24 hours * 60 minutes * 60 seconds = 31,536,000 seconds

Distance in meters = speed of light * number of seconds in a year = 3.00 × 10^8 m/s * 31,536,000 seconds

Distance in miles = distance in meters / (5280 ft/mi * 12 in/ft)

Plugging in the values into the equation, we get:

Distance in miles = (3.00 × 10^8 m/s * 31,536,000 seconds) / (5280 ft/mi * 12 in/ft)

Simplifying the equation, the distance from Earth to Sirius is approximately 50 trillion miles.

A 1.5-kg object has a velocity of 5j m/s at t = 0. It is accelerated at a constant rate for five seconds after which it has a velocity of (6i + 12j ) m/s. What is the magnitude of the resultant force acting on the object during this time interval?

Answers

Answer:

2.76 N

Explanation:

m = mass of the object = 1.5 kg

v₀ = initial velocity at t = 0, = 0 i + 5 j

v = final velocity of the object at t = 5, = 6 i + 12 j

t = time interval = 5 sec

a = acceleration of the object = ?

Acceleration of the object is given as

[tex]a = \frac{v - v_{o}}{t}[/tex]

inserting the values

a = ((6 i + 12 j) - (0 i + 5 j))/5

a = (6 i + 7 j)/5

a = 1.2 i + 1.4 j

magnitude of the acceleration is given as

|a| = √((1.2)² + (1.4)²)

|a| = 1.84 m/s²

magnitude of the resultant force is given as

|F| = m |a|

|F| = (1.5) (1.84)

|F| = 2.76 N

Final answer:

To find the magnitude of the resultant force, we use Newton's second law of motion. Evaluating the acceleration at 2.0s gives a magnitude of 24.8 m/s^2. Using the formula F = ma, the magnitude of the resultant force is 37.2 N.

Explanation:

To find the magnitude of the resultant force acting on the object during the time interval, we need to use Newton's second law of motion, which states that the force is equal to the mass of the object multiplied by its acceleration.

First, we need to find the acceleration of the object. We can use the formula:

a(t) = 5.0i + 2.0tj - 6.0t^2 km/s^2

By evaluating a(2.0 s), we get a magnitude of 24.8 m/s^2.

Now, we can use Newton's second law:

F = ma

Substituting the values, we get:

F = 1.5 kg * 24.8 m/s^2

F = 37.2 N

Therefore, the magnitude of the resultant force acting on the object during this time interval is 37.2 N.

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A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 16 cm and the capacitance is 113 pF. What is the radius of the outer sphere? (Give your answer in decimal using cm as unit)

Answers

Answer:

19 cm

Explanation:

inner radius, a = 16 cm, C = 113 p F = 113 x 10^-12 F

Let b cm be the radius of outer sphere.

The formula of capacitance of spherical capacitor is given by

C = 4π∈0 a b / (b - a)

a b / ( b - a) = C / 4π∈0

a b / ( b - a ) = 113 x 10^-12 x 9 x 10^9

a b / ( b - a ) = 1.017

16 x 10^-2 x b x 10^-2 = 1.017( b - 16) x 10^-2

0.16 b = 1.017 b - 16.272

0.857 b = 16.272

b = 19 cm

a mass suspended by a spring stretches an additional 4 ccm when an additional 10 gram mass is attached to it. what is the value of the spring constant k, in si units

Answers

Answer:

2.45 N/m

Explanation:

We can find the spring constant by using Hooke's law:

[tex]F=kx[/tex]

where

F is the force applied to the spring

k is the spring constant

x is the stretching of the spring

Here, the force applied is the weight of the mass hanged on the spring. The mass is

m = 10 g = 0.010 kg

So the weight is

[tex]F=mg=(0.010 kg)(9.8 m/s^2)=0.098 N[/tex]

while the stretching is

x = 4 cm = 0.04 m

So the spring constant is

[tex]k=\frac{F}{x}=\frac{0.098 N}{0.04 m}=2.45 N/m[/tex]

A car is traveling at 53.0 mi/h on a horizontal highway. (a) If the coefficient of static friction between road and tires on a rainy day is 0.104, what is the minimum distance in which the car will stop?

Answers

Answer:

902 ft

Explanation:

First convert mi/h to ft/s.

53.0 mi/h × (5280 ft / mi) × (1 h / 3600 s) = 77.7 ft/s

Sum of the forces on the car in the y direction:

∑F = ma

N - W = 0

N = mg

Sum of the forces on the car in the x direction:

∑F = ma

-F = ma

-Nμ = ma

Substituting;

-mgμ = ma

-gμ = a

Acceleration is constant, so:

v² = v₀² + 2a(x - x₀)

(0 ft/s)² = (77.7 ft/s)² + 2(-32.2 ft/s² × 0.104)(x - 0)

x = 902 ft

The minimum stopping distance is 902 ft.

A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and the secondary is connected to a floodlight that draws 5A, what is the power output? Please show ALL of your work.

Answers

Answer:

2000 W

Explanation:

First of all, we need to find the output voltage in the transformer, by using the transformer equation:

[tex]\frac{V_1}{N_1}=\frac{V_2}{N_2}[/tex]

where here we have

V1 = 200 V is the voltage in the primary coil

V2 is the voltage in the secondary coil

N1 = 250 is the number of turns in the primary coil

N2 = 500 is the number of turns in the secondary coil

Solving for V2,

[tex]V_2 = N_2 \frac{V_1}{N_1}=(500) \frac{200 V}{250}=400 V[/tex]

Now we can find the power output, which is given by

P = VI

where

V = 400 V is the output voltage

I = 5 A is the output current

Substituting,

P = (400 V)(5 A) = 2,000 W

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. (a) Determine the time necessary for the projectile to reach the ground below. (b) Determine the distance from the base of the building that the projectile lands. (c) Determine the horizontal and vertical components of the velocity just before the projectile reaches the ground.

Answers

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

[tex]y = h + \frac{1}{2}at^2[/tex]

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

[tex]0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s[/tex]

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

[tex]v_x = 23.4 m/s[/tex]

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

[tex]d=v_x t = (23.4 m/s)(3.35 s)=78.4 m[/tex]

(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

[tex]v_x = 23.4 m/s[/tex]

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

[tex]v_y = u_y +at[/tex]

where

[tex]u_y = 0[/tex] is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

[tex]v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s[/tex]

and the negative sign means it points downward.

The projectile shot from the roof of the building travel in the projectile motion to reach to the ground.

(a) The time necessary for the projectile to reach the ground below is 3.35 seconds.(b) The distance from the base of the building that the projectile lands is 78.4 meters.(c) The horizontal and vertical components of the velocity just before the projectile reaches the ground is 23.4 and 32.8 m/s downward respectively.

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

For the motion of horizontal and vertical direction we use following equation in projectile motion.

[tex]y=y_o+\dfrac{1}{2}gt^2[/tex]

Here, (g) is the gravity and (t) is time.

Here, in the given problem, the projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall.

(a) The time necessary for the projectile to reach the ground below-

As the height of the projectile before the shot is 55 meters and the the height of the projectile is 0 meter when it reaches to the ground. Thus from the above equation,

[tex]0=55+\dfrac{1}{2}\times9.81\times(t)^2\\t=3.35\rm s[/tex]

Thus the time required for the projectile to reach the ground below is 3.35 seconds.

(b) The distance from the base of the building that the projectile lands-

The distance traveled by the object is the product of velocity and time taken. Thus, distance from the base of the building that the projectile lands is,

[tex]d=23.4\times3.35\\d=78.4\rm m[/tex]

(c) The horizontal and vertical components of the velocity just before the projectile reaches the ground-

Horizontal component of velocity is equal to the velocity of projectile shot just before it reaches to the ground. Thus, horizontal component of velocity is,

[tex]v_x=23.4\rm m/s[/tex]

Now the time taken is 3.35 seconds. Thus the vertical component of velocity can be found with the formula of first equation of motion with  0 initial velocity as,

[tex]v_y=0+(-9.81)\times(3.35)\\v_y=-32.8\rm m/s[/tex]

The projectile shot from the roof of the building travel in the projectile motion to reach to the ground.

(a) The time necessary for the projectile to reach the ground below is 3.35 seconds.(b) The distance from the base of the building that the projectile lands is 78.4 meters.(c) The horizontal and vertical components of the velocity just before the projectile reaches the ground is 23.4 and 32.8 m/s downward respectively.

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Suppose a 50-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.25 m2. It is stretched to have no area in 0.1s. What is the direction and magnitude of the induced emf if the uniform magnetic field has a strength of 1.5T?

Answers

Answer:

EMF = 187.5 volts

Explanation:

As per Faraday's law of electromagnetic induction we know that rate of change in the flux will induce EMF

so we can say

[tex]EMF = -\frac{d\phi}{dt}[/tex]

[tex]EMF = \frac{\phi_i - \phi_f}{\Delta t}[/tex]

now we will have

[tex]EMF = \frac{NBA - 0}{\Delta t}[/tex]

[tex]EMF = \frac{50(1.5)(0.25)}{0.10}[/tex]

[tex]EMF = 187.5 Volts[/tex]

So the induced EMF in the coil will be 187.5 Volts

Question Part Points Submissions Used A car is traveling at 53.0 km/h on a flat highway. (a) If the coefficient of friction between road and tires on a rainy day is 0.115, what is the minimum distance in which the car will stop? Incorrect: Your answer is incorrect. m (b) What is the stopping distance when the surface is dry and the coefficient of friction is 0.575? m

Answers

(a) 95.9 m

The initial velocity of the car is

[tex]u=53.0 km/h = 14.7 m/s[/tex]

The car moves by uniformly accelerated motion, so we can use the SUVAT equation:

[tex]v^2 - u^2 = 2ad[/tex]

where

v = 0 is the final velocity

d is the stopping distance of the car

a is the acceleration of the car

The force of friction against the car is

[tex]F_f = - \mu mg[/tex]

where

[tex]\mu=0.115[/tex] is the coefficient of friction

m is the mass of the car

[tex]g = 9.8 m/s^2[/tex] is the acceleration due to gravity

According to Newton's second law, the acceleration is

[tex]a=\frac{F}{m}=\frac{-\mu mg}{m}=-\mu g[/tex]

Substituting into the previous equation:

[tex]v^2 - u^2 = -2\mu g d[/tex]

and solving for d:

[tex]d=\frac{v^2 -u^2}{-2\mu g}=\frac{0-(14.7 m/s)^2}{-2(0.115)(9.8 m/s^2)}=95.9 m[/tex]

(b) 19.1 m

This time, the coefficient of friction is

[tex]\mu = 0.575[/tex]

So the acceleration due to friction is:

[tex]a=-\mu g = -(0.575)(9.8 m/s^2)=-5.64 m/s^2[/tex]

And substituting into the SUVAT equation:

[tex]v^2 - u^2 = 2ad[/tex]

we can find the new stopping distance:

[tex]d=\frac{v^2 -u^2}{-2a}=\frac{0-(14.7 m/s)^2}{2(-5.64 m/s^2)}=19.1 m[/tex]

How far away was a lightning strike if thunder is heard 7.20 seconds after the flash is seen? (Assume that sound traveled at 350.0m/s during the storm)

Answers

Answer: 2.52 Kilometers

Explanation:

We know that the formula to calculate speed is given by :-

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

Given: The speed of sound during the storm : [tex]V=3350.0m/s[/tex]

The time taken by sound to travel : [tex]t= 7.20\text{ seconds}[/tex]

Then , the distance traveled by sound :-

[tex]\text{Distance}=350\times7.2=2520\text{ meters}=2.52\text{ kilometers}[/tex]

To find the distance of a lightning strike heard 7.20 seconds after the flash, multiply the sound's speed (350.0 m/s) by the time delay (7.20 s), resulting in a distance of 2520 meters.

We use the speed of sound. Since sound traveled at 350.0 m/s during the storm, we multiply the time it took to hear the thunder by the speed of sound to calculate the distance:

Distance = Speed of Sound × Time

Substituting in the given values:

Distance = 350.0 m/s × 7.20 s = 2520 meters

Therefore, the lightning struck approximately 2520 meters away.

An object is traveling such that it has a momentum of magnitude 23.3 kg.m/s and a kinetic energy of 262 J. Determine the following. (a) speed of the object in meters per second. (b) mass of the object in kilograms.

Answers

Explanation:

It is given that,

Momentum of an object, p = 23.3 kg-m/s

Kinetic energy, E = 262 J

(a) Momentum is given by, p = mv

23.3 = mv...........(1)

Kinetic energy is given by, [tex]E=\dfrac{1}{2}mv^2[/tex]

m = mass of the object

v = speed of the object

[tex]E=\dfrac{1}{2}\times (mv)\times v[/tex]

[tex]262=\dfrac{1}{2}\times 23.3\times v[/tex]

v = 22.48 m/s

(2) Momentum, p = mv

[tex]m=\dfrac{p}{v}[/tex]

[tex]m=\dfrac{23.3\ kg-m/s}{22.48\ m/s}[/tex]

m = 1.03 Kg

Hence, this is the required solution.

Final answer:

An object with momentum 23.3 kg.m/s and kinetic energy of 262J is traveling at a speed of approximately 30.21 m/s and its mass is approximately 0.771 kg based on the physics principles of kinetic energy and momentum.

Explanation:

The question involves the physics concepts of momentum and kinetic energy. We are given the momentum (p) of 23.3 kg.m/s and the kinetic energy (KE) of 262 J of an object.

(a) The formula for kinetic energy is KE = 0.5 * m * v^2, where m is the mass of the object and v is its velocity. We can rearrange to find v = sqrt((2*KE) / m). The mass can be obtained from the momentum formula, p = m * v, hence m = p / v. Substituting the second equation into the first gives v = sqrt((2 * KE * v) / p), which simplifies to v = sqrt((2 * 262 J) / 23.3 kg.m/s) = 30.21 m/s.

(b) With the velocity calculated in (a), the mass of the object can now be found by rearranging the momentum formula to m = p / v = 23.3 kg.m/s / 30.21 m/s = 0.771 kg.

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A current carrying circular loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counterclockwise sense. What is the direction of the magnetic field caused by this current, inside the loop? The magnetic field A) circles the loop in a counterclockwise direction. B) points straight up. C) points straight down. D) circles the loop in a clockwise direction. E) points toward the east.

Answers

Final answer:

The magnetic field direction, when current moves in a circular loop in a counterclockwise sense, points straight up. The right-hand rule is used to determine the orientation of the magnetic field.

Explanation:

When current moves in a circular path, it generates a magnetic field. Using the right-hand rule, where your thumb points to the direction of the current and your fingers curl in the direction of the magnetic field, it can be determined that when looking from above, if the current moves in a counterclockwise sense, the magnetic field direction will be pointing straight up from the loop. Hence, the correct answer is B) points straight up.

This rule, also referred to as RHR-2 (Right Hand Rule-2), is widely applied to illustrate the magnetic fields induced around current carrying conductors. The net force on a current-carrying loop of any plane shape in a uniform magnetic field is zero, but it induces a distinctive orientation of the magnetic field in its surroundings.

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The correct answer is B) points straight up. Using the right-hand rule for a counterclockwise current in a circular loop, the magnetic field inside the loop points straight up.

To determine the direction of the magnetic field inside a current-carrying circular loop, we use the right-hand rule. In this case, the current moves counterclockwise when viewed from above, so follow these steps:

Point your right thumb in the direction of the current.

Notice how your fingers curl in the direction of the magnetic field.

Since the current flows counterclockwise, the right-hand rule shows that the magnetic field lines point straight up inside the loop. Therefore, the correct answer is B) points straight up.

A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the person and the kayak is 74 kg, what is the magnitude of the net force acting on the kayak?

Answers

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

Final answer:

The magnitude of the net force acting on the kayak, which accelerates from 0 to 0.61 m/s over a distance of 0.39 m with a mass of 74 kg, is approximately 35.25 Newtons (N).

Explanation:

A person in a kayak accelerates from 0 to 0.61 m/s over a distance of 0.39 meters. To find the magnitude of the net force acting on the kayak, we can use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. We can calculate this as:

Kinetic energy (KE) = (1/2)mv2Work done (W) = Force (F) × Distance (d)So, the net work done is also equal to the change in kinetic energy, KE = W

To find the net force, we need to first calculate the change in kinetic energy, which is the kinetic energy at 0.61 m/s minus the initial kinetic energy at 0 m/s:

ΔKE = (1/2)m(v2 - 02)ΔKE = (1/2) × 74 kg × (0.61 m/s)2ΔKE = (1/2) × 74 kg × 0.3721 m2/s2ΔKE = 13.7467 kg·m2/s2 (= 13.7467 Joules)

Since work done is equal to force times distance and to the change in kinetic energy, we can express the net force as:

F = ΔKE / dF = 13.7467 J / 0.39 mF = 35.247 N  (rounded to three decimal places)

Therefore, the magnitude of the net force acting on the kayak is approximately 35.25 Newtons (N).

If the net torque is zero, what does this imply about the clockwise and counterclockwise torques.

Answers

When the net torque is zero, there is no overall force causing the object to turn clockwise or counterclockwise.

An object is said to be in a condition of rotational equilibrium if there is no net torque operating on it. In other words, the object is keeping a constant angular velocity while not suffering any rotational acceleration. The object has a constant angular momentum when the net torque is zero.

If the net torque is zero, the item is being pulled in a direction that is equal to the sum of all the clockwise and anticlockwise torques pulling on it. To put it another way, any propensity the object has to spin in one direction is counterbalanced by an equal tendency to rotate in the opposing direction.

Hence, When the net torque is zero, there is no overall force causing the object to turn clockwise or counterclockwise.

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Final answer:

When the net torque is zero, it means that the clockwise and counterclockwise torques are balanced.

Explanation:

When the net torque is zero, it implies that the clockwise and counterclockwise torques balance each other out. In other words, the total torque in the clockwise direction is equal in magnitude but opposite in sign to the total torque in the counterclockwise direction.

A satellite weighs 104 N at ground control. What best approximates the acceleration it experiences in orbit at an altitude of twice the earth's radius ifFpa = GME2thm, where r is the distance separating the centers of mass of the satellite and the Earth? O A. 111 m/s2 O B. 2.5 m/s O C. 1.1 m/s2 ○ D. 0 m/s

Answers

Answer:

[tex]a_c = 1.1 m/s^2[/tex]

Explanation:

As we know that net force on the Satellite due to gravity will provide it centripetal force

so we can say here

[tex]F_g = F_c[/tex]

[tex]\frac{GMm}{r^2} = ma_c[/tex]

now we will have

acceleration given by the equation

[tex]a_c = \frac{GM}{r^2}[/tex]

now we have

r = R + 2R = 3R

[tex]a_c = \frac{GM}{9R^2}[/tex]

also we know that acceleration due to gravity on the surface of earth is given as

[tex]g = \frac{GM}{R^2} = 9.8 m/s^2[/tex]

so the acceleration of satellite is given as

[tex]a_c = \frac{9.8}{9} = 1.1 m/s^2[/tex]

Final answer:

The acceleration of a satellite in orbit at an altitude of twice Earth's radius is best approximated using the formula for gravitational acceleration, g = GME / r². Plugging in the values and simplifying, the acceleration is approximately 2.45 m/s², with option B (2.5 m/s²) being the closest approximation.

Explanation:

Calculating the Acceleration of a Satellite in Orbit

In order to best approximate the acceleration a satellite experiences in orbit at an altitude of twice the Earth's radius, we must apply Newton's law of universal gravitation and the formula for gravitational acceleration, g = GME / r². Given that the satellite weighs 104 N at ground control which reflects the gravitational force at Earth's surface, we use this formula to find its gravitational acceleration in orbit. The relevant equation provided, Fpa = GME2thm/r, appears to be a typo, but our main equation for gravitational acceleration does not require mass, as it cancels out during the calculation:

g = GME / r²

Since we're given that the altitude is twice the Earth's radius, the distance r from Earth's center to the satellite in orbit is 3RE (1RE for Earth's radius and 2RE for the altitude above Earth). Considering this, we can calculate:

g = (6.674 x 10-11 m³/kg s²) (5.972 x 1024 kg) / (3 x 6.371 x 106 m)²

After simplifying, we get an acceleration of approximately 2.45 m/s². Therefore, option B, 2.5 m/s², best approximates the acceleration the satellite experiences in orbit at this altitude.

An object is thrown vertically upward at 27.1 m/s. The velocity of the object 3.4 seconds later is ____ m/s. Round your answer to the nearest tenth. Do not use scientific notation. Take up as positive and down as negative.

Answers

Answer:

-6.2 m/s (downward)

Explanation:

The velocity of an object thrown vertically upward is given by:

[tex]v= u + at[/tex]

where:

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration due to gravity

t is the time

In this problem,

u = 27.1 m/s

t = 3.4 s

So, the velocity after 3.4 s is

[tex]v=27.1 m/s + (-9.8 m/s^2)(3.4 s)=-6.2 m/s[/tex]

and the negative sign means the velocity points downward.

A jogger accelerates from rest to 4.86 m/s in 2.43 s. A car accelerates from 20.6 to 32.7 m/s also in 2.43 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 2.43 s?

Answers

Explanation:

It is given that,

Initially, the jogger is at rest u₁ = 0

He accelerates from rest to 4.86 m, v₁ = 4.86 m

Time, t₁ = 2.43 s

A car accelerates from u₂ = 20.6 to v₂ = 32.7 m/s in t₂ = 2.43 s

(a) Acceleration of the jogger :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{4.86\ m/s-0}{2.43\ s}[/tex]

a₁ = 2 m/s²

(b) Acceleration of the car,

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{32.7\ m/s-20.6\ m/s}{2.43\ s}[/tex]

a₂ = 4.97 m/s²

(c) Distance covered by the car,

[tex]d_1=u_1t_1+\dfrac{1}{2}a_1t_1^2[/tex]

[tex]d_1=0+\dfrac{1}{2}\times 2\times (2.43)^2[/tex]

d₁ = 5.904 m

Distance covered by the jogger,

[tex]d_2=u_2t_2+\dfrac{1}{2}a_2t_2^2[/tex]

[tex]d_2=20.6\times 2.43+\dfrac{1}{2}\times 4.97\times (2.43)^2[/tex]

d₂ = 64.73 m

The car further travel a distance of, d = 64.73 m - 5.904 m = 58.826 m

Hence, this is the required solution.

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