A 0.5 kg ball is dropped from rest at a point 1.2m above the floor. The ball rebounds straight upward to a height of 0.7m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor.

Answers

Answer 1

Answer:

4.281 kgm/s upward

Explanation:

Impulse:This can be defined product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = Change in momentum.

I = mΔv....................... Equation 1.

Where m = mass of the ball, Δv = change in velocity of the ball  

and Δv = v -u

Where u = velocity of the ball before it hit the floor, v = velocity of the ball after if hit the floor

I = m(v-u) -------------- Equation 2

But

the initial kinetic energy of the ball = potential energy at the initial height (1.2 m above)

1/2mu² = mgh₁

Where h₁ = initial height. or height of the ball before collision

making u the subject of the equation,

u = √(2gh₁)........................ Equation 3

Where h₁ = 1.2 m g = 9.81 m/s²

Substitute into equation 3

u = √(2×1.2×9.81)

u =√(23.544)

u = -4.852 m/s.

Note: u is negative because the ball was moving downward at the first instance.

Similarly,

v = √(2gh₂)............................. Equation 3

h₂ = height of the ball after collision

Given: h₂ = 0.7 m, g = 9.81 m/s²

Substitute into equation

v = √(2×9.81×0.7)

v = √13.734

v = 3.71 m/s.

Also given: m = 0.5 kg,

Substituting into equation 2

I = 0.5(3.71-(4.852)

I = 0.5(8.562)

I = 4.281 kgm/s. Upward.

Thus the impulse = 4.281 kgm/s upward


Related Questions

An elevator packed with people has a mass of 1800 kg.
A) The elevator accelerates upward (in the positive direction) from rest at a rate of 1.5 m/s2 for 2.2 s. Calculate the tension in the cable supporting the elevator in newtons.
B) The elevator continues upward at constant velocity for 9 s. What is the tension in the cable during this time in newtons?
C) The elevator experiences a negative acceleration at a rate of 0.65 m/s2 for 2 s. What is the tension in the cable, in Newtons, during this period of negative accleration?
D) How far has the elevator moved above its original starting point in meters?

Answers

Answer:

a) 20.34 N

b)17.64 N

c)16.47 N

d)38.63m

Explanation:

a) F=m*a. Earth gravity is 9.8m/s2. If the elevator goes upwards, tension on the rope will be higher. To find the tension we need to all accelerates. Total accelerates effects the elevator is a=9.8+1.5=11.3 m/s2. Then;

[tex]F=1.8*11.3=20.34[/tex]

the tension is 20.34N

b) There is no accelerate in this situation therefore the tension is:

F=1.8*9.8=17.64 N

c) In this situation elevator goes down. We need to subtract gravity from elevator acceleration. a=9.8-0.65=9.15m/s2

F=1.8*9.15=16.47 N

d) Total distance is:

[tex]x=0.5*a*t^2+v*t+v*t-0.5*a*t^2\\=0.5*1.5*2.2^2+9*(1.5*2.2)+2*(1.5*2.2)-0.5*0.65*2^2\\=38.63 m[/tex]

A child has been diagnosed with an autoimmune disease that predominantly affects epithelial tissue. Histologic examination should include which body parts as potential sites of involvement? Select all that apply.a) Heartb) Bone marrowc) Bladderd) Epidermise) Blood vessel lining

Answers

Answer:C) BLADDER

D) EPIDERMIS (E) BLOOD VESSEL LINING

Explanation: Autoimmune diseases are diseases caused by the body fighting against its self. In autoimmune diseases the body's defense system tends to attack the body cells, tissues, Organs and the entire system mistaking it for a foreign body such as Viruses or Bacteria,Fungi etc.

Epithelial tissues are tissues which form the surface covering of most of the hollow Organs,all body Organs and are present inside the glands. The body parts that will be involved are the BLADDER, EPIDERMIS AND BLOOD VESSEL LINING.

_____ can be thrown, caught, and used to provide resistance for a variety of movements, in a variety of planes of motion, and at a variety of velocities.

Answers

Answer:

b. Medicine balls

Explanation:

A medicine ball (also called an exercise ball, or a health ball) is a weighted ball about the shoulder diameter (approx. 13.7 inches), often used for recovery and strength training.... In 1705 similar large balls had been used in Persia.So you can use your medicine ball as a projectile to boost the strength of your throws.

When a .22-caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure causes the force that pushes the bullet through the barrel. The barrel has a length of 0.61 m and an opening whose radius is 2.8 x 10²³ m. A bullet (mass = 2.6 x 10²³ kg) has a speed of 370 m/s after passing through this barrel. Ignore friction and determine the average pressure of the expanding gas.

Answers

Answer:

[tex]1.18454\times 10^{-19}\ Pa[/tex]

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

m = Mass of bullet = [tex]2.6\times 10^{23}\ kg[/tex]

r = Radius of barrel = [tex]2.8\times 10^{23}\ m[/tex]

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{370^2-0^2}{2\times 0.61}\\\Rightarrow a=112213.11475\ m/s^2[/tex]

Pressure is given by

[tex]P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{ma}{\pi r^2}\\\Rightarrow P=\dfrac{2.6\times 10^{23}\times 112213.11475}{\pi (2.8\times 10^{23})^2}\\\Rightarrow P=1.18454\times 10^{-19}\ Pa[/tex]

The pressure of the expanding gas is [tex]1.18454\times 10^{-19}\ Pa[/tex]

What type of nuclear radiation is emitted when carbon-14 decays

Answers

Answer:

Beta radiation

Explanation:

Beta radiation is a radioactive phenomenon of nuclear decay in which an unstable atom or isotop, by transforming a neutron into a proton, or by transforming a proton into a neutron, becomes stable. For example, the decay of carbon 14 produces beta radiation.

A rock is thrown upward with a velocity of 26 meters per second from the top of a 43 meter high cliff, and it misses the cliff on the way back down. When will the rock be 6 meters

Answers

Final answer:

To solve the problem, we use the kinematic equation for distance considering the initial velocity, acceleration due to gravity, and the distance from the top of the cliff to 6m above the ground. The equation gives two possible times: one when the rock is on the way up, and one when it's on the way down.

Explanation:

In this physics question, we are dealing with kinematic equations which describe the motion of the rock. The equation we will use for this question is d = vit + 0.5gt^2, where 'd' represents total distance, 'vi' is initial velocity, 'g' is acceleration due to gravity and 't' is time.

Here, d = 43m - 6m = 37m (distance from the top of the cliff to 6m above the ground), vi = 26m/s (initial velocity of the rock when thrown upwards), and g = 9.8m/s^2 (acceleration due to gravity, but since the rock is thrown downwards, we take it as positive).

Now plug these values into the equation, you have two possible times- one for the rock on its way up and one for on its way down.

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A helicopter, initially hovering 40 feet above the ground, begins to gain altitude at a rate of 21 feet per second. Which of the following functions represents the helicopter’s altitude above the ground y, in feet, t seconds after the helicopter begins to gain altitude?

Answers

Answer: y = 40 + 21t

Explanation:

Apply the equation of distance covered.

d = vt + C

Where d is the distance covered

v = velocity , t = time

C = constant = initial distance covered

For the case above....

d = y

y(t) = vt + C

But y(0) =40 = C

C = 40ft

velocity v = 21 ft/s

Therefore, the equation of the altitude is given by;

y(t) = 21t +40

y = 40 + 21t

Final answer:

The function representing the helicopter's altitude as it gains height is y = 21t + 40. The helicopter gains altitude at a rate of 21 feet per second and it starts at 40 feet above the ground.

Explanation:

The function that represents this situation is a linear function because the change in the helicopter's altitude is constant over time.

The formula for a linear function is y = mx + b, where m is the slope (rate of change), b is the y-intercept (initial value), y is the dependent variable (altitude in this case), and x is the independent variable (time in this case).

In this scenario, we're given that the helicopter starts 40 feet above the ground (our y-intercept, b) and it's gaining altitude at 21 feet per second (our slope, m).

Therefore, the function to represent the helicopter’s altitude, y, after t seconds can be written as y = 21t + 40.

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What do we expect to happen to Earth’s air temperature if the amount of energy from the sun that reaches the Earth’s surface is reduced by half?

Answers

Explanation:

The Natural Greenhouse Effect However, the amount that directly escapes to space is only about 12 percent of incoming solar energy. The remaining fraction—a net 5-6 percent of incoming solar energy—is transferred to the atmosphere when greenhouse gas molecules absorb thermal infrared energy radiated by the surface.

How star properties affect star formation? Provided following are the spectral types of four different main-sequence stars. Rank the stars based on the strength of the radiation pressure that pushes outward as they are forming, from highest pressure to lowest pressure.A) O9.B) A5.C) G2.D) M6.

Answers

Answer:

-M6

-G2

-A5

-O9

Explanation:

What do you call the gravitational attraction between you and the earth?

A.weight

B. mass

Answers

Answer:

A. Weight

Explanation:

Mass And Weight

Mass is a fundamental property of objects and is defined as a measure of the amount of matter in the object. The symbol for mass is m and its SI unit is the kilogram (kg). At normal speeds, i.e. much smaller than the speed of light, the mass is considered as a constant property.

The weight W is defined as the force of gravity on the object, that is, the gravitational attraction between the object and the Earth. Its SI unit is the Newton (Nw or N).

A car is driving around a circular track at constant speed. At one instant, the car is driving northward and sometime later the car is driving westward. What is the direction of the car’s average change in velocity during this time interval?

Answers

Answer:

Southwest

Explanation:

when the car is moving around a circular track at constant speed, velocity changes due to change in the direction. The direction at any instant is given by tangent drawn at the point on the circular path.

At instant 1: direction is northwards

At instant 2: direction is westwards

Let the constant speed be v. Then, at instant 1, the velocity would be:

[tex]v_1=v \hat j[/tex]

At instant 2, the velocity would be:

[tex]v_2 = v (-\hat i)[/tex]

Change in velocity = [tex]v_2-v_1 = v(-\hat i - \hat j)[/tex]

The direction of car's average change would be southwest.

Final answer:

The car's average change in velocity has a northwest direction.

Explanation:

The direction of the car's average change in velocity is determined by the direction of the velocity vectors at the two instances. In this case, when the car is driving northward, its velocity vector points towards the north. Later, when the car is driving westward, its velocity vector points towards the west. Therefore, the car's average change in velocity will have a direction that is a combination of north and west, which is the northwest direction.

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An astronaut is in an all-metal chamber outside the space station when a solar storm results in the deposit of a large positive charge on the station. Which statement is correct?

a. The astronaut must abandon the chamber immediately to avoid being electrocuted.
b. The astronaut will be safe only if she is wearing a spacesuit made of non-conducting materials.
c. The astronaut does not need to worry: the charge will remain on the outside surface.
d. The astronaut must abandon the chamber if the electric field on the outside surface becomes greater than the breakdown field of air.
e. The astronaut must abandon the chamber immediately because the electric field inside the chamber is non- uniform.

Answers

Answer:

c. The astronaut does not need to worry: the charge will remain on the outside surface.

Explanation:

The all-metal chamber (conductor) receives an external electromagnetic field, so it is positively charged in the direction the electromagnetic field is going, and negatively charged in the opposite direction. Since the chamber is polarized, it generates an electric field equal in magnitude, but opposite the external electromagnetic field, then the net electric field inside the conductor is 0.

When all boxes (conductors) receive environmental electromagnetic frequencies, they become electrostatically attracted inside the plane of the magnetic wave and charged negatively in a reverse way.

Since the chambers are polarized, they generate an electric field of similar size and opposite to the outside magnetic wave, hence the net electric field within the wire is 0.Whenever a solar flare deposits large amounts of positive energy on the spaceship, one astronaut is in an all-metal room outside the station. An astronaut does not need to be concerned because the charge should remain from the outside surface.

Therefore, the answer is " Option c".

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__________ brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction. Control systems apply the brakes momentarily to one of the drive wheels whenever the wheel speed sensors indicate a wheel is going faster than the others during acceleration.

Answers

"Anti-Lock" brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.

Explanation:

The safety anti-skid braking system is known as "anti-lock braking system" having huge application on land vehicles like one, two and multiple wheeler vehicles and aircraft. During braking, it avoids wheels to get locked by building tractive contacts to the road's surface.

This seems to be an automated system work on the principles of techniques - threshold and cadence braking. The wheel velocity sensors are utilized by ABS to find whether one or more than one wheels chose to get lock while braking.

Answer:

Anti- Lock

Explanation:

Anti- lock brake systems(ABS) is a safety system used in hindering the wheels from locking up when the brake is applied.

Braking release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.

Control systems apply the brakes momentarily to one of the drive wheels whenever the wheel speed sensors indicate a wheel is going faster than the others during acceleration.

A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass of the planet from this information. ___ kg

Answers

Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet =  3.56×10²⁶ Kg.

The mass of the planet is required.

The mass of the planet is [tex]3.56\times 10^{26}\ \text{kg}[/tex]

M = Mass of planet

r = Radius of orbit = [tex]7\times 10^7\ \text{m}[/tex]

t = Time period = [tex]6\times 60\times 60+38\times 60=23880\ \text{s}[/tex]

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

Mass is given by

[tex]M=\dfrac{4\pi^2r^3}{t^2G}\\\Rightarrow M=\dfrac{4\pi^2\times (7\times 10^7)^3}{23880^2\times 6.674\times 10^{-11}}\\\Rightarrow M=3.56\times 10^{26}\ \text{kg}[/tex]

The mass of the planet is [tex]3.56\times 10^{26}\ \text{kg}[/tex]

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What is it called when a solid turns directly into a gas

Answers

Answer:

Sublimation

Explanation:

Sublimation is the process of conversion of solids directly into vapor on the supply of heat. This occurs because the vapor pressure of the sublimate becomes higher than the atmospheric pressure and the melting has not yet happened because of insufficient heat. In other words, the substance cannot exist in its liquid form at the given temperature and the atmospheric pressure.

Example: Dry ice, camphor

A 11.6-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 65.3 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 3.50 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey's arm.

Answers

Answer:

217.611 N

331.407 N

Explanation:

m = Mass of monkey = 11.6 kg

v = Velocity of the monkey = 3.5 m/s

r = Radius = 65.3 cm

g = Acceleration due to gravity = 9.81 m/s²

Centripetal force is given by

[tex]F_c=\dfrac{mv^2}{r}\\\Rightarrow F_c=\dfrac{11.6\times 3.5^2}{0.653}\\\Rightarrow F_c=217.611\ N[/tex]

The magnitude of the centripetal force is 217.611 N

Balancing the forces we get

[tex]F_c=T-mg\\\Rightarrow T=F_c+mg\\\Rightarrow T=217.611+11.6\times 9.81\\\Rightarrow T=331.407\ N[/tex]

The tension in the monkey's arm is 331.407 N

Satellite 1 revolves around a planet at the altitude equal to one-half the radius of the planet. The period of revolution of satellite 1 is . What is the period of revolution of an identical satellite 2 that revolves around the same planet at the altitude equal to the radius of the planet?

Answers

Answer:

To calculate the period of satellite orbiting around a planet, we use Kepler's third law;

Square of T = [(4π)/(G*m)] * R^3.

Therefore,

T = sqrt{[(4π)/(G*m)]*R^3}.

T is the period, m is mass orbiting satellite, G is gravitational constant, R is the radius of of the planet, r is the radius of the orbiting satellite.

For Satellite 1, r is one-half of the planet, that is r = (3/2) * R

For satellite 2, r = R

Explanation:

To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should you choose to throw a rubber ball or a beanbag of equal size and weight?

Answers

Answer:

Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.

For Rubber Ball

Upon elastic collision it will reverses the direction and move with velocity equal or less then original

change in momentum = P

[tex]P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\ P = -2mv_{i}[/tex]

For Beanbag

value of impulse will large if velocity is zero.

[tex]v_{f}=0\\ P = -mv_{i}[/tex]

Explanation:

The rubber ball will impact a greater force on the heavy bowling pin because its kinetic energy will be conserved while the beanbag will impact lesser force due to loss of kinetic energy.

A collision between two objects can be elastic or inelastic.

In elastic collision both momentum and kinetic energy are conserved.In inelastic collision only momentum is conserved.

The impulse experienced by each throwing object is equal to change in the momentum of the object.

[tex]J = \Delta P[/tex]

The rubber ball will impact a greater force on the heavy bowling pin because its kinetic energy will be conserved while the beanbag will impact lesser force due to loss of kinetic energy.

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Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well.Suppose that the moment of inertia of a skater with arms out and one leg extended is 2.9 kgm2 and for arms and legs in is 0.90 kgm2 . If she starts out spinning at 4.5rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?

Answers

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is 1.4 rev/s

Explanation:

given information:

moment inertia of arm and leg when in, I₁ = 0.9 kgm²

moment inertia of arm and leg when extended, I₂ = 2.9 kgm²

angular speed when in, ω₁ = 4.5 rev/s

so, her angular speed (in rev/s) when her arms and one leg open outward is

L₁ = L₂

I₁ω₁ = I₂ω₂

ω₂ = I₁ω₁/I₂

     = 0.9 x 4.5/2,9

     = 1.4 rev/s

Using conversions and data, determine the number of hydrogen atoms required to obtain 8.0 kg of hydrogen. A hydrogen atom has a mass of 1.0 u.

Answers

Answer:

The answer to your question is 4.82 x 10²⁷ atoms of Hydrogen

Explanation:

Data

final mass = 8.0 kg

mass of hydrogen = 1.0 u

number of hydrogens = ?

Process

1.- Use proportions to solve this problem

                  1 u -------------------------  1.66 x 10⁻²⁷ kg

                 x u -------------------------  8 kg

                 x = (8 x 1) / 1.66 x 10⁻²⁷

                 x = 8 / 1.66 x 10⁻²⁷

                x = 4.82 x 10 ²⁷ u or 4.82 x 10²⁷ atoms of Hydrogen

As you jump across a small stream, does a horizontal force keep you moving forward? If so, what is that force? [Note—this question seems to be deliberately designed to trick you! Is there really a horizontal force being applied once you’ve left the ground?]

Answers

There is no horizontal force

Explanation:

When you jump across the stream, your motion is the same of the motion of a projectile, which consists of two independent motions:

A uniform motion (constant velocity) along the horizontal directionA uniformly accelerated motion (with acceleration equal to the acceleration of gravity) in the vertical direction

We notice that as you are in the air, there is only one force acting on your body: the force of gravity, whose direction is downward, and causes your body to accelerate downward with an acceleration equal to

[tex]g=9.8 m/s^2[/tex]

However, there is no force acting on you in the horizontal direction: therefore, your acceleration in this direction is zero, and so your horizontal velocity is constant (that's why you keep moving forward).

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Which are true?
1) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.
2) Electric field lines near positive point charges radiate outward.
3) The electric force acting on a point charge is proportional to the magnitude of the point charge.
4) Electric field lines near negative point charges circle clockwise.
5) The electric field created by a point charge is constant throughout space.

Answers

Answer:

Following are the correct options.

1. In a uniform electric field, the field lines are straight, parallel and uniformly spaced. ( As the field strength does not change in the uniform electric field that is why the field lines are parallel, equally spaced etc).

2. Electric field lines near positive point charges radiate outward. (Due to repulsive forces field lines radiate in outward direction)

3.The electric force acting on a point charge is proportional to the magnitude of the point charge. ( As columb law states that the electrostatic forces between the point charges is proportional to the product of their magnitude and inversely proportional to square of distance between them).

Explanation:

Final answer:

In a uniform electric field, electric field lines are straight, parallel, and uniformly spaced while near positive point charge it emits outwards. Also, the electric force on a point charge is proportional to its magnitude. But, lines near negative point charges radiate inward, not circle clockwise. The electric field created by a point charge isn't constant across space, but diminishes with distance.

Explanation:

Of the five statements you provided:

In a uniform electric field, the field lines are indeed straight, parallel, and uniformly spaced. Electric field lines near positive point charges do radiate outward. The electric force acting on a point charge is proportional to the magnitude of the point charge.However, electric field lines near negative point charges don't circle clockwise, they radiate inward towards the negative charge.Additionally, the electric field created by a point charge isn't constant throughout space; it diminishes with distance from the point charge according to Coulomb's Law.

Therefore, statements 1, 2 and 3 are true, while 4 and 5 are false.

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A 92-kg fullback moving south with a speed of 5.8 m/s is tackled by a 110-kg lineman running north with a speed of 3.6 m/s. Assuming momentum conservation, determine the speed and direction of the two players immediately after the tackle.Give the direction as an angle, in degrees, south of west.

Answers

They travel at a speed of 0.68 m/s south ([tex]90^{\circ}[/tex] south of west)

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two players must be conserved before and after they collide.

Mathematically:

[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v[/tex]  

where, taking north as positive direction:

[tex]m_1 = 92 kg[/tex] is the mass of the first player

[tex]u_1 = -5.8 m/s[/tex] is the initial velocity of the first player (south, so negative)

[tex]m_2 = 110 kg[/tex] is the mass of the second player

[tex]u_2 = 3.6 m/s[/tex] is the initial velocity of the second player (north, so positive)

[tex]v[/tex] is the final combined velocity of the two players

Solving for v, we find the final velocity of the two players combined:

[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(92)(-5.8)+(110)(3.6)}{92+110}=-0.68 m/s[/tex]

where the negative sign indicates their final direction is south.

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Final answer:

To find the final speed and direction after the tackle, we use the conservation of momentum. The fullback and lineman's initial momenta are calculated and combined to find the total initial momentum, which equals the final momentum because momentum is conserved. The direction is either south (90 degrees) or north (270 degrees) depending upon the sign of the final velocity.

Explanation:

To solve this physics problem, we will use the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant. The formula for momentum (p) is the product of mass (m) and velocity (v), or p = mv.

Before the tackle, the fullback's momentum is southward and the lineman's is northward. We treat south as the positive direction. Thus, the initial momentums of the fullback and lineman are:

Fullback momentum: pFB = 92 kg × 5.8 m/sLineman momentum: pL = 110 kg × (-3.6 m/s) (negative due to the northward direction)

The total initial momentum (pinitial) is the sum of these:

pinitial = pFB + pL

After the tackle, the players stick together and move as one mass (M = 92 kg + 110 kg), with a final velocity (V), so:

pfinal = MV

Since momentum is conserved, pinitial = pfinal, and we can solve for V:

V = pinitial / M

Once we calculate the final velocity, the direction is south if V > 0, north if V < 0. To convert the direction into an angle south of west, note that if they move southward, the angle is simply 90 degrees (directly south), while any northward movement would result in a 270-degree angle.

A 0.500 kg mass is oscillating on a spring with k=330 N/m.The total energy of its oscillation is 3.24 J. What is speed of the mass when it is 0.100 m from the EP?
(Unit=m/s)

Answers

The speed is 2.5 m/s

Explanation:

The total (mechanical) energy of the mass-spring system at any point during the motion is the sum of the kinetic energy (KE) and the potential energy (PE):

[tex]E=KE+PE=3.24 J[/tex]

and it is constant.

The kinetic energy can be written as

[tex]KE=\frac{1}{2}mv^2[/tex]

where

m = 0.500 kg is the mass

v is the speed

While the potential energy is

[tex]PE=\frac{1}{2}kx^2[/tex]

where

k = 330 N/m is the spring constant

x is the elongation

So the first equation becomes

[tex]E=\frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]

Therefore, if we substitute

x = 0.100 m

We can find the speed when the elongation is x = 0.100 m:

[tex]v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(3.24)-(330)(0.100)^2}{0.500}}=2.5 m/s[/tex]

Learn more about kinetic and potential energy:

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Correct Answer:

2.5 m/s

A car traveling at 7 m/s accelerates uniformly at 2.5 m/s to reach a speed of 12 m/s. How long does it take for this acceleration to occur?

Answers

There is one mistake in the question as unit of acceleration is not written correctly.The correct question is here

A car traveling at 7 m/s accelerates uniformly at 2.5 m/s² to reach a speed of 12 m/s. How long does it take for this acceleration to occur?

Answer:

time taken =2 seconds

Explanation:

Given Data

Initial Speed Vi= 7 m/s

Final Speed Vf=12 m/s

Acceleration a= 2.5 m/s²

To find

Time taken for this acceleration

Solution

As we know that

Final velocity=Initial velocity + acceleration×time

[tex]V_{f}=V_{i}+at\\  t=\frac{V_{f}-V_{t}}{a}\\ t=\frac{12m/s-7m/s}{2.5m/s^{2} }\\ t=2 seconds[/tex]

So car takes 2 seconds for this acceleration to occur

A cyclist maintains a constant velocity of 4.1 m/s headed away from point A. At some initial time, the cyclist is 244 m from point A. What will be his displacement from his starting position after 60 s?

Answers

Answer:

[tex]d=490\ m[/tex] is his final displacement from the point A after 60 seconds.

Explanation:

Given:

Cyclist is moving away from A.

velocity of cyclist, [tex]v=4.1\ m.s^{-1}[/tex]displacement of the cyclist from point A at the time of observation, [tex]d_i=244\ m[/tex]time after which the next observation is to be recorded, [tex]t=60\ s[/tex]

Now as the cyclist is moving away from point A his change in displacement after the mentioned time:

[tex]\Delta d=v.t[/tex]

[tex]\Delta d = 4.1 \times 60[/tex]

[tex]\Delta d=246\ m[/tex]

Now the the final displacement from point A after the mentioned time:

[tex]d=d_i+\Delta d[/tex]

[tex]d=244+246[/tex]

[tex]d=490\ m[/tex]

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. At what angle (in degrees) must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s?

Answers

Answer:

The answer to your question is angle = 18.46°

Explanation:

Data

d = 75 m

v₀ = 35 m/s

α = ?

Formula

                     [tex]d = \frac{vo^{2}sin2\alpha}{g}[/tex]

solve for sin2α

                  sin 2α = [tex]\frac{dg}{vo^{2}}[/tex]

Substitution

                 sin 2α = [tex]\frac{75(9.81)}{35^{2}}[/tex]

Simplify

                sin 2α = [tex]\frac{735.75}{1225}[/tex]

Divide

                sin 2α = 0.600

Get sin⁻¹

                      2α = 36.9°

Divide by 2

                        α = 18.5°                

A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline.Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What is the mass of the wooden block?A.) 40 gB.) 20 gC.) 10 g

Answers

Answer:

b) M=20g

Explanation:

For this exercise we must use the Archimedes principle that states that the thrust that a body receives is equal to the weight of the dislodged liquid.

         B = ρ g V

Let's use balance healing for this case

Initial.

          B - W = 0

The weight of the body can be related to its density

          W = ρ V_body g

         ρ_liq g (½ V_body) = m g

Final

   Some masses were added

          M = 20 g = 0.020 kg

        B - W - W₂ = 0

        ρ_liq  g  V_Body = m g + M g

Let's replace and write the system of equations

     ½ ρ_liq  V_body = m

         ρ  V_body = m + M

We solve the equations

        2 m = m + M

        m = M

         m = 20 g

The answer is b

When the Darwin/Wallace theory of natural selection is summarized, four central postulates emerge. Which of the following is NOT one of these four natural selection postulates

Answers

Answer:

Four central postulates that emerges from Charles Darwin and Russel Wallace are detailed below. As there is no option provided in the question hence any option beyond the meaning of these four points will not be from these four postulates.

Explanation:

The four postulates were:  

More individuals are produced each generation than can survive. Phenotypic variation exists among individuals and the variation is heritable. Those individuals with heritable traits better suited to the environment will survive. When reproductive isolation occurs new species will form.

1st:

The number of individuals produced in a generation of a specie is more than that of the ones who survive. Some of them cannot withstand the environmental conditions and die.

2nd:

Observable characteristics of an individual can vary and this variation is inherited by parental cells/genes.

3rd:

The individuals who have characteristics which are suitable to the environment exists and survives while other ones do not bear the environment and die soon.

4th:

Reproductive isolation is the evolutionary mechanism which prevents members of different species from producing offspring but whenever this occurs, most of the times, new specie is formed. The survival of this new specie again depends upon suitability of its characteristics with environment.

Which conditions will result in the smallest change in momentum? a. a large force over a long time period b. a large force over a short time period c. a small force over a long time period d. a small force over a short time period

Answers

Answer:

A small force over a long time period.                                        

Explanation:

The second law of motion gives the relationship between the external force and the change in momentum of the object. The rate of change of linear momentum is equal to the external force applied. It is given by :

[tex]F=\dfrac{\Delta p}{\Delta t}[/tex]

[tex]\Delta p=F\times \Delta t[/tex]

It is clear that for the smallest change in momentum, a small force should be applied for a long period of time. Hence, the correct option is (c) "a small force over a long time period".

Answer:

D. a small force over a short time period

C is incorrect.

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