A cement block accidentally falls from rest from the ledge of a 84.9-m-high building. When the block is 16.6 m above the ground, a man, 1.90 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

Answers

Answer 1

The man has at most 3.55 s to get out of the way.

What is  kinematic equations?

Kinematics can be defined as a subfield of physics developed in classical mechanics that describes the motion of points, bodies and systems of bodies without considering the forces that cause them to move.

We can solve this problem using kinematic equations.

First, we can find the time it takes for the block to fall from a height of 84.9 m using the formula:

y = 1/2 * g * t^2

where

y is the distance fallen (84.9 m)g is the acceleration due to gravity (9.81 m/s^2) t is the time

Rearranging this equation, we get:

t = sqrt(2y/g) = sqrt(2 * 84.9 m / 9.81 m/s^2) = 4.09 s (to two significant figures)

So the block will hit the ground after 4.09 s of falling.

Next, we can find the time it takes for the block to fall from a height of 16.6 m using the same formula:

t = sqrt(2y/g) = sqrt(2 * 16.6 m / 9.81 m/s^2) = 1.41 s (to two significant figures)

So the block will take 1.41 s to fall from a height of 16.6 m to the ground.

The man has to get out of the way before the block falls the remaining distance of (84.9 - 16.6) = 68.3 m. We can find the time it takes for the block to fall this distance using the same formula:

t = sqrt(2y/g) = sqrt(2 * 68.3 m / 9.81 m/s^2) = 3.55 s (to two significant figures)

Therefore, the man has at most 3.55 s to get out of the way.

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Related Questions

Describe protons.
Location:
Charge:
Mass:

Answers

The protons location is nucleus of an atom, the Mass is 1 amu and the charge is positive.

What is protons?

Protons are subatomic particles with a positive electrical charge. They are found in the nucleus of atoms and are responsible for most of the atom’s mass. Protons are one of the three main subatomic particles, along with neutrons and electrons.

Location: Proton is located in the nucleus of an atom. The nucleus is the small, dense, positively charged center of an atom. The protons, along with the neutrons, make up the nucleus of the atom.

Mass: The mass of a proton is approximately 1.007276467 u (unified atomic mass units). It is slightly heavier than a neutron, which has a mass of approximately 1 u.

Charge: A proton has a positive charge of +1 elementary charge (e). This charge is what gives the proton its repelling force to other positively charged particles and its attractive force to negatively charged particles.

Hence, a proton is a positively charged subatomic particle with a mass of 1 amu, located in the nucleus of an atom.

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Answer:

Location:

✔ nucleus

Charge:

✔ positive

Mass:

✔ one amu

Explanation:

A rotating platform with a radius of 2. 0 m makes one complete turn every 3. 0 s. The angular velocity of the platform is most nearly.

Answers

The velocity v, angular speed ω and radius r have the relationship of:
.
The angular speed (a.k.a angular frequency) ω and the period T have the relationship of:
.
So:

A wheel is formed from a hoop of mass 2.6 kg
and seven equally spaced spokes, each of mass
0.11 kg. The hoop’s radius is the length
0.73 m of each spoke.
0.11 kg
2.6 kg
0.73 m
Find the moment of inertia of the wheel
about an axis through its center and perpendicular to the plane of the wheel.
Answer in units of kg · m2
. Answer in units
of kg · m2
.

Answers

The moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel is equal to the moment of inertia of the hoop plus the moment of inertia of the spokes. The moment of inertia of the hoop is given by Ih = mr^2, where m is the mass of the hoop and r is the radius. The moment of inertia of the spokes is given by Is = 7(m/2)r^2, where m is the mass of each spoke and r is the radius. Therefore, the total moment of inertia of the wheel is given by It = Ih + Is = mr^2 + 7(m/2)r^2. Substituting in the given values gives It = 2.6(0.73)^2 + 7(0.11/2)(0.73)^2 = 1.64 kg · m^2.

The moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel is 2.09 kg·m^2.

What is the parallel axis theorem?

The moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel can be calculated by using the parallel axis theorem, which states that the moment of inertia of a rigid body about any axis is equal to the moment of inertia about a parallel axis through the center of mass plus the product of the mass and the square of the distance between the two axes.

Here in the Question,

First, we need to find the moment of inertia of the hoop about an axis through its center and perpendicular to the plane of the hoop. This is a well-known result from basic mechanics and is given by:

I_hoop = 1/2 * m_hoop * r^2

where m_hoop is the mass of the hoop and r is its radius. Substituting the given values, we get:

I_hoop = 1/2 * 2.6 kg * (0.73 m)^2 = 1.26 kg·m^2

Next, we need to find the moment of inertia of a single spoke about an axis through its center and perpendicular to its length. This is also a well-known result from basic mechanics and is given by:

I_spoke = 1/12 * m_spoke * L^2

where m_spoke is the mass of the spoke and L is its length. Since the spokes are equally spaced around the hoop, we can consider a single spoke and multiply its moment of inertia by 7 to account for all the spokes. Substituting the given values, we get:

I_spoke = 1/12 * 0.11 kg * (0.73 m)^2 = 0.005 kg·m^2

Therefore, the moment of inertia of all the spokes combined is:

I_spokes = 7 * I_spoke = 0.035 kg·m^2

Finally, we can use the parallel axis theorem to find the moment of inertia of the entire wheel about an axis through its center and perpendicular to the plane of the wheel:

I_wheel = I_hoop + I_spokes + 7 * m_spoke * r^2

Substituting the given values, we get:

I_wheel = 1.26 kg·m^2 + 0.035 kg·m^2 + 7 * 0.11 kg * (0.73 m)^2 = 2.09 kg·m^2

Therefore, The wheel's moment of inertia about an axis passing through its center and perpendicular to its plane is 2.09 kg·m^2.

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In the experiment that student needs to know the mass of four different objects: a shoe, piece of paper,book, and a tissue.which of the following units of measurement should the student use

Answers

The unit of measurement the student should use is the gram. The correct option is c.

What are the measurements units?

The International System of Units (SI) uses the gram (formerly gram; SI unit symbol g) as the unit of mass that corresponds to one-thousandth of a kilogram.

A student must know the masses of four different objects for an experiment: a shoe, a piece of paper, a book, and a tissue. The learner should utilize grams as their measurement units.

Therefore, the correct option is c. gram is used to measure the mass of objects and things.

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The question is incomplete. The missing options are given below:

milligram

ounce

gram

kilogram

Calculate the total work done on a mass m as it moves from position x1 = 0 m to x2 = 40 m
F1 = 5 N F2 = 6 N angle between them 50 degree
F3 = 2 N F4 = 2 N Angle Between them 20 Degree

Answers

The total work done on the mass as it moves from x1 = 0 m to x2 = 40 m is approximately 515.17 J.

What is Work Done?

Work is a physical quantity that describes the amount of energy transferred when a force acts on an object and causes it to move. When a force acts on an object and causes it to move in the direction of the force, work is said to be done on the object. Mathematically, work is defined as the dot product of force and displacement:

Work = Force x Displacement x cos(theta)

To calculate the total work done on the mass as it moves from position x1 to x2, we need to find the net work done by all the forces on the mass. The net work done by a force is given by the formula:

W = F * d * cos(theta)

where W is the work done, F is the force, d is the displacement of the mass, and theta is the angle between the force and the displacement.

First, we can calculate the work done by each force separately and then add them up to find the total work done.

Work done by F1:

W1 = F1 * (x2 - x1) * cos(0) = 5 N * 40 m * cos(0) = 200 J

Work done by F2:

W2 = F2 * (x2 - x1) * cos(50°) = 6 N * 40 m * cos(50°) ≈ 165.41 J

Work done by F3:

W3 = F3 * (x2 - x1) * cos(20°) = 2 N * 40 m * cos(20°) ≈ 74.88 J

Work done by F4:

W4 = F4 * (x2 - x1) * cos(20°) = 2 N * 40 m * cos(20°) ≈ 74.88 J

The total work done on the mass is the sum of the work done by each force:

W_total = W1 + W2 + W3 + W4 ≈ 515.17 J

Therefore, the total work done on the mass as it moves from x1 = 0 m to x2 = 40 m is approximately 515.17 J.

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You will now use the observations you have made so far to predict what the strength of the field will be at point (0, 50).A. Using the right-hand rule, which direction is the magnetic field at (0, 50)?B. Since the magnetic field of the Earth and the induced field are at right angles, you can use the Pythagorean Theorem to determine the strength of the combined field.

Answers

In A. part, the magnetic field at (0, 50) is in west direction. In B. part, the strength of the field at (0,50) is 2.06 G.

A. The current is flowing up for west as shown by the front view figure at the bottom of the gadget. Your fingers will curve to the west if you wrap your right hand around the wire with your thumb up. Put a compass at (0,50) to check the direction as well. It indicates west.

B. By using the Pythagorean Theorem to determine the strength of the combined field, the strength of the field at (0,50) is 2.06 G.

The earth's magnetic field strength= 0.50 G

The induced current magnetic field strength= 2.0

B is given by=

[tex]\sqrt{0.50^{2} - 2.00^{2} }\\ =\sqrt{0.25-4.00}\\ =2.06[/tex]

Hence, we can also check by putting the probe on (0,50) and the probe reads 2.06 G.

Therefore, the strength of the field at (0,50) is 2.06 G.

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A ball is thrown downward with a large starting velocity. Will this accelerate more rapidly than one that is just dropped at the same time?

Answers

No, a ball thrown downward with a large starting velocity will not accelerate more rapidly than one that is just dropped at the same time, assuming that both are experiencing the same gravitational field.

Both objects will experience the same b, which is approximately 9.8 m/s^2 near the surface of the Earth. The initial velocity of the thrown ball will only affect its initial speed, but it will not change the acceleration due to gravity.

Therefore, both objects will accelerate at the same rate and will fall at the same speed. However, the thrown ball will cover a greater distance than the dropped ball before hitting the ground, as it has an initial velocity in addition to the acceleration due to gravity.

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The specific heat of copper is 387 J/kg C. The temperature of a 0.35-kg sample of copper decreases from 74.0 °C to 21.0 °C. How much heat flows out of
the copper sample during this temperature drop?

Answers

The amount of heat that flows out of the copper sample during this temperature drop is approximately 4,953.75 J.

What is the amount of heat flowing out?

The amount of heat that flows out of the copper sample can be calculated using the formula:

Q = mcΔT

where;

Q is the amount of heat transferred, m is the mass of the copper sample, c is the specific heat of copper, and ΔT is the change in temperature of the sample.

Plugging in the given values, we get:

Q = (0.35 kg) x (387 J/kg C) x (74.0 °C - 21.0 °C)

Q = 4,953.75 J

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ASAP I WILL GIVE BRAINLEST!!

A 3.0 kg object swings back and forth as a simple pendulum with a small amplitude. The potential energy U of the object as a function of distance x from its equilibrium position is shown above. This particular object has a total energy E of 0.4 J.

b) What is the farthest the object moves along the x-axis in the positive direction? Explain your reasoning.
c) Determine the object’s kinetic energy when its displacement is -8 cm.
d) What is the object’s speed at x = 0?

Answers

10cm is the farthest the object moves along the x-axis in the positive direction .0.15J is the object’s kinetic energy when its displacement is -8 cm. 0.316m/s is the object’s speed at x = 0.

Define kinetic energy.

Kinetic energy, which may be seen in the movement of an item or subatomic particle, is the energy of motion. Kinetic energy is present in every particle and moving object. Examples of kinetic energy in action include a person walking, a baseball soaring through the air, a piece of food falling from a table, and a charged particle in an electric field.

Given,

Total energy is 0.4J

m is 3kg

The farthest the object moves along the x-axis in the positive direction would be as potential energy is 0.4J. So from given diagram, displacement will be 10cm

If displacement is -8cm , P.E from diagram will be 0.25J

According to energy conservation formula ,

ME ⇒ U+KE

KE ⇒ ME-U ⇒ 0.4-0.25 ⇒ 0.15J

At x ⇒ 0,

KE ⇒ 1/2 mv^2

0.15 ⇒ 0.5*3*v^2

v ⇒ 0.316m/s

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You are asked to relocate a sunken boat to be used as an artificial reef. The boat's mass is 4,000Kg and the volume of its solid parts is 3m3. Assume the ocean water density is 1.02g/ml. Approximately, what is the minimal volume of an air bag (in liters) that should be attached to it so it can be lifted under water to be moved to its new location (the air bag weighs 10Kg)?

Answers

To lift the boat of 4,000Kg under water, the minimal volume of an air bag required is approximately 38,235.29 liters. This is calculated by using the Archimedes' principle, which states that an object submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object. Thus, the force required to lift the boat is equal to 4,000Kg, while the weight of the air bag is 10Kg. Therefore, the buoyant force required to lift the boat is equal to 3,990Kg. The volume of fluid displaced is equal to the volume of the solid part of the boat plus the volume of the air bag. Thus, the volume of the air bag required is equal to 3,990Kg/1.02g/ml, which is approximately 38,235.29 liters.

TRUE OR FALSE:  Increasing the time of impact during a collision increases the force experienced by the object. * True False ​

Answers

Answer:

False

Explanation:

I'm pretty sure increasing the time of impact actually decreases the force because it is being spread out.

Answer:

False.

Explanation:

Extending the time results in decreasing the force and thus minimizing the effect of the force in the collision.

Let E-> =3i^+1j^ and F -> =1i^−3j^. (Note: the -> means it is above the letter!)
A. Find the magnitude of E -> .
B. Find the magnitude of F ->.
C. Find the magnitude of G -> = E-> +F -> .
D. Find the magnitude of H-> =−E -> −2F -> .

Answers

A. the magnitude of E ->  is √10.

B.  the magnitude of F -> is √10

C. the magnitude of G -> = E-> +F -> is √20

D. the magnitude of H-> =−E -> −2F ->  is √50.

What is magnitude?

The magnitude or size of a mathematical object is described as a property which determines whether the object is larger or smaller than other objects of the same kind.

A. The magnitude of the vector E -> is given by the formula:

|E -> | = √(3^2 + 1^2) = √(9 + 1) = √10

So, the magnitude of E -> is √10.

B. The magnitude of the vector F -> is given by the formula:

|F -> | = √(1^2 + (-3)^2) = √(1 + 9) = √10

So, the magnitude of F -> is √10.

C. The magnitude of the vector G -> is given by the formula:

G -> = E-> + F ->

|G -> | = √((3 + 1)^2 + (1 - 3)^2) = √(4^2 + (-2)^2) = √(16 + 4) = √20

So, the magnitude of G -> is √20.

D. The magnitude of the vector H -> is given by the formula:

H -> = -E-> - 2F->

|H -> | = √((-3 - 2 * 1)^2 + (-1 - 2 * -3)^2) = √((-5)^2 + 5^2) = √50

So, the magnitude of H -> is √50.

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Croquet balls must have a mass of .50 kg. A red croquet ball moving at 10 m/s strikes an at rest green croquet ball head-on. If the red ball stops after hitting the green ball, what will be the final speed of the green ball

Answers

The final velocity of the green ball is 5.0 m/s.

What will be the final speed of the green ball?

The final speed of the green ball can be determined using the law of conservation of momentum.

The momentum of the system (red and green ball) before the collision is equal to the momentum of the system after the collision, assuming there are no external forces acting on the system.

Before the collision, the momentum of the red ball is given by:

p1 = m1v1 = 0.50 kg   x 10 m/s = 5.0 kg m/s

After the collision, the momentum of the green ball is given by:

p2 = m2  x v2

Using the law of conservation of momentum, we have:

p1 + p2 = (m1 + m2) v1

5.0 kg m/s + p2 = (0.50 kg + 0.50 kg) x 10 m/s

5.0 kg m/s + p2 = 1.0 kg * 10 m/s

5.0 kg m/s + p2 = 10.0 kg m/s

p2 = 10.0 kg m/s - 5.0 kg m/s = 5.0 kg m/s

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A motorcycle moves according to the velocity-versus-time graph shown in the figure
Find the average acceleration of the motorcycle during segment A of the motion.
(part a)
Find the average acceleration of the motorcycle during segment B of the motion.
(part b)
Find the average acceleration of the motorcycle during segment C of the motion.
(part c)

Answers

Average acceleration A = 2 m/s² Average acceleration B = 0 m/s² Average acceleration C = -0.5 m/s²

What is acceleration?

The rate of change in velocity over time is called acceleration. The unit of measurement for this vector quantity is meters per second squared (m/s²). Acceleration can be either positive (speeding up) or negative (speeding down). It can also be referred to in terms of direction, such as acceleration to the left or right.

a) V=10m/s

u=0m/s

t= 5 second

a = (v-u)/t

   = (10-0)/5

   = 2 m/s²

b) The body moves with constant velocity 10m/s , so acceleration is 0m/s

c The velocity is falling , so the body is,

a= (v-u)/t= (5-10)/10

  = -5/10

  = -0.5 m/s²

Therefore, the Average acceleration of A, B and C are 2 m/s², 0m/s and -0.5 m/s²

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A 0.0272 kg arrow is fired with a velocity of 87.3 m/s from a 0.0245 kg bow. Determine the equivalent impulse value imparted on the arrow. (hint: you can also use newton's third law of motion)* A.2.37 Ns B.3.11 Ns C.113.5 Ns D.214.2 Ns​

Answers

Answer:

Explanation:

C. 113.5 Ns

What is the mathematical expression for the quantization of energy in a system described by the Schrödinger equation and how does it relate to the concept of quantized energy levels in an atom?

*Answer = 50 Points

Answers

Answer:

The mathematical expression for the quantization of energy in a system described by the Schrödinger equation is given by the eigenvalue equation:

HΨ = EΨ

Where H is the Hamiltonian operator, Ψ is the wave function, and E is the eigenvalue that represents the quantized energy of the system.

The concept of quantized energy levels in an atom is related to the quantization of energy in the Schrödinger equation. In quantum mechanics, atoms can only exist in certain energy levels or states, which are determined by the solutions to the Schrödinger equation. These energy levels are quantized, meaning that the energy can only take on specific values, and not any value in between. This results in the characteristic spectra of atomic systems, where the electrons in an atom can only transition from one energy level to another by absorbing or emitting a photon with an energy that corresponds to the difference in energy between the two levels.

In summary, the quantization of energy in a system described by the Schrödinger equation is the foundation for the concept of quantized energy levels in atoms, which has important implications for our understanding of the behavior of atoms and the properties of materials.

Unit 3
The Flow of Energy in Systems
EVIDENCE NOTEBOOK
11. How could the surfaces and structures in an urban area be related to the
urban heat island effect? Record your evidence

Answers

Answer:

The surfaces and structures in an urban area can have a significant influence on the urban heat island effect. For example, paved surfaces like roads and buildings absorb more heat than surfaces covered in vegetation, and structures like high-rise buildings can trap and reflect heat, resulting in higher temperatures in the urban area. Additionally, urban areas usually have less vegetation than their rural counterparts, meaning there is less vegetation to absorb heat from the sun and provide shade, further contributing to the urban heat island effect. Evidence of this can be seen in research at louisvilleky.gov/government/sustainability/urban-heat-island-project.

. Find the net resistance of 113 such strands if they are placed side by side to form a cable of the same length as a single strand.

Answers

The net resistance of 113 strands placed side by side is 1/113th of the resistance of a single strand.

Assuming that each strand has the same resistance, the net resistance of 113 strands placed side by side can be found by calculating the equivalent resistance of a parallel combination of 113 resistors. The formula for calculating the equivalent resistance of a parallel combination of resistors is:

1/R = 1/R1 + 1/R2 + ... + 1/Rn

where R is the equivalent resistance, and R1, R2, ..., Rn are the resistances of the individual components.

In this case, we have 113 strands, so n = 113. Since the strands are placed side by side, they are in parallel, so we can use the above formula to find the equivalent resistance:

1/R = 1/R1 + 1/R2 + ... + 1/R113

R = 1 / (1/R1 + 1/R2 + ... + 1/R113)

Since we don't know the resistance of a single strand, we cannot calculate the exact value of the net resistance. However, if we assume that each strand has the same resistance, we can use the formula for the equivalent resistance of n equal resistors in parallel:

1/R = n / R1

R = R1 / n

Substituting n = 113, we get:

R = R1 / 113

This means that the net resistance of 113 strands placed side by side is 1/113th of the resistance of a single strand.Assuming that each strand has the same resistance, the net resistance of 113 strands placed side by side can be found by calculating the equivalent resistance of a parallel combination of 113 resistors. The formula for calculating the equivalent resistance of a parallel combination of resistors is:

1/R = 1/R1 + 1/R2 + ... + 1/Rn

where R is the equivalent resistance, and R1, R2, ..., Rn are the resistances of the individual components.

In this case, we have 113 strands, so n = 113. Since the strands are placed side by side, they are in parallel, so we can use the above formula to find the equivalent resistance:

1/R = 1/R1 + 1/R2 + ... + 1/R1₁₃

R = 1 / (1/R₁ + 1/R₂ + ... + 1/R1₁₃)

Since we don't know the resistance of a single strand, we cannot calculate the exact value of the net resistance. However, if we assume that each strand has the same resistance, we can use the formula for the equivalent resistance of n equal resistors in parallel:

1/R = n / R₁

R = R1 / n

Substituting n = 113, we get:

R = R₁ / 113

This means that the net resistance of 113 strands placed side by side is 1/113th of the resistance of a single strand.

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A small glass bead has been charged 3 60nCA small metal ball bearing 2.60 cm above the bead feels a 1.8 * 10 2 * N downward electric force

Answers

Answer:

The electric field strength at the location of the metal ball bearing is 1.8 * 10^2 N/C.


Two balls are dropped to the ground from different heights. One is dropped 1.5 s after the other, but they
both strike the ground at the same time, 5.0 s after the first was dropped. (a) What is the difference in the
heights from which they were dropped? (b) From what height was the first ball dropped? Ans. (a) 63 m;
(b) 0.12 km

Answers

Answer:

Height difference: approximately [tex]63\; {\rm m}[/tex].

The first ball was dropped from a height of approximately [tex]123\; {\rm m}[/tex].

(Assumptions: both balls were released from rest, air friction is negligible, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

Under the assumptions, both ball would accelerate at a constant [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].

Let [tex]t[/tex] denote the time since the first ball was released.

Height of the first ball at time [tex]t[/tex] can be modelled with the SUVAT equation [tex]h(t) = (1/2)\, a\, t^{2} + u\, t + h_{0}[/tex], where [tex]u[/tex] is the initial velocity. However, since [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] by assumption, this equation simplifies to [tex]h(t) = (1/2)\, a\, t^{2} + h_{0}[/tex].

Since this ball reached the ground after [tex]t = 5.0\; {\rm s}[/tex], [tex]h(5.0) = 0\; {\rm m}[/tex]. In other words:

[tex]\begin{aligned}\frac{1}{2}\, (-9.81)\, (5.0)^{2} + h_{0} = 0\end{aligned}[/tex].

Simplify and solve for the initial height of this ball, [tex]h_{0}[/tex]:

[tex]\begin{aligned}h_{0} &= -\frac{1}{2}\, (-9.81)\, (5.0)^{2} \\ &\approx 123\; {\rm m}\end{aligned}[/tex].

In other words, the first ball was dropped from a height of approximately [tex]123\; {\rm m}[/tex].

Similarly, the height of the second ball may be modelled as [tex]h(t) = (1/2)\, a\, t^{2} + h_{0}[/tex].

Since this ball reached the ground [tex]t = (5.0 - 1.5)\; {\rm s} = 3.5\; {\rm s}[/tex] after being released, [tex]h(3.5) = 0\; {\rm m}[/tex]. The initial height of this ball would be:

[tex]\begin{aligned}h_{0} &= -\frac{1}{2}\, (-9.81)\, (3.5)^{2} \\ &\approx (-60)\; {\rm m}\end{aligned}[/tex].

Subtract the initial height of the second ball from that of the first ball to find the difference in initial height:

[tex](123 - 60) \; {\rm m} \approx 63\; {\rm m}[/tex].

In which collision(s) is momentum conserved?
A.Elastic
B.Inelastic
C.Perfectly inelastic​

Answers

Momentum is conserved in both elastic and perfectly inelastic collisions.

In an elastic collision, the total momentum of the colliding objects is conserved before and after the collision. This means that the sum of the momentum of the objects before the collision is equal to the sum of the momentum of the objects after the collision.

In a perfectly inelastic collision, the two objects stick together after the collision, forming a single object with new momentum. In this case, the total momentum of the system is also conserved.

However, in an inelastic collision, momentum is not conserved, as some of the momenta are transformed into other forms of energy, such as heat or sound. This means that the total momentum of the objects before the collision is not equal to the total momentum of the objects after the collision.

Answer:

Explanation:

A

1. we know that the total amount of heat that flows out of the sample and into the water at a specific time is given byLaTeX: Q\:=\:m_sc_s\left(T_{s,i}-T_s\right)Q=mscs(Ts,i−Ts), whereLaTeX: T_sTs is the temperature of the sample at a specific time and, again,LaTeX: T_{s,i}Ts,i is the initial temperature of the sample (at time 0). To simplify the math, we may neglect the heat leak term here to say that this is roughly the same amount of heat the flows into the water, soLaTeX: Q=m_wc_w\left(T_w-T_{w,i}\right)Q=mwcw(Tw−Tw,i), whereLaTeX: T_wTw is the temperature of the water at this same specific time andLaTeX: T_{w,i}Tw,i is the initial temperature of the water.
In the lab, we will measure both the sample and water temperatures as a function of time, but the important quantity is the difference between these temperatures since this is what drives the heat flow between the center of the sample and the water. Using the above equations (solving for the temperatures of the sample and the water bath at a particular time), we can find the relationship between the total amount of heat flow and the difference in the temperatures of the center of the sample and water at some moment in time. This yields _________________________________.
sample and water at some moment in time. This yields _________________________________.
Group of answer choices

Answers

Option D: the link between the total heat flow and the temperature difference between the sample's Centre and the water at a specific time.

[tex]Q\:=\:m_sc_s\left(T_{s,i}-T_s\right)[/tex]

[tex]T_s\right =(T_{s,i}-T_s\right))[/tex]

[tex]Q=m_wc_w\left(T_w-T_{w,i}\right)[/tex]

[tex]Q=m_wc_w\left(T_w-T_{w,i}\right)[/tex]

[tex]T_{diff} =(T_{s}-T_w\right))[/tex]

        = [tex]T_{s,i} -\frac{Q}{m_{s}C_{s}} -(T_{w,i}\right +\frac{Q}{m_{s}C_{s}} )[/tex]

        =[tex](T_{s,i} - T_{w,i} )-Q(\frac{1}{m_{s}C_{s}} +\frac{1}{m_{w}C_{w}})[/tex]

Specific time refers to a precise moment in time, often denoted by a particular time and date. It can be expressed in different ways depending on the context, such as using a 24-hour clock or the AM/PM system. Specific time is essential for scheduling events, meetings, and appointments, and for coordinating activities across different time zones. It is also crucial for time-sensitive activities such as transportation, where schedules must be coordinated down to the minute. The concept of specific time is used in many fields, including science, technology, business, and everyday life. In modern times, technologies such as smartphones and computers have made it easier than ever to track and coordinate specific times across the globe.

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The complete question is:

1. we know that the total amount of heat that flows out of the sample and into the water at a specific time is given by LaTeX: [tex]Q\:=\:m_sc_s\left(T_{s,i}-T_s\right)Q=mscs(Ts,i−Ts)[/tex], where LaTeX: [tex]T_sTs[/tex] is the temperature of the sample at a specific time and, again, LaTeX: [tex]T_{s,i}Ts,i[/tex]is the initial temperature of the sample (at time 0). To simplify the math, we may neglect the heat leak term here to say that this is roughly the same amount of heat the flows into the water, so LaTeX: [tex]Q=m_wc_w\left(T_w-T_{w,i}\right)Q=mwcw(Tw−Tw,i)[/tex], where LaTeX:[tex]T_wTw[/tex] is the temperature of the water at this same specific time and LaTeX:  is the initial temperature of the water.

In the lab, we will measure both the sample and water temperatures as a function of time, but the important quantity is the difference between these temperatures since this is what drives the heat flow between the center of the sample and the water. Using the above equations (solving for the temperatures of the sample and the water bath at a particular time), we can find the relationship between the total amount of heat flow and the difference in the temperatures of the center of the sample and water at some moment in time. This yields _________________________________.

sample and water at some moment in time. This yields _________________________________.

Group of answer choices

A. [tex]T_{dif}=T_{s\:}-T_w=\left(T_{s,i}-T_{w,i}\right)-\left(\frac{1}{m_sc_s}-\frac{1}{m_wc_w}\right)Q[/tex]

B [tex]T_{dif}=T_{s\:}-T_w=\left(T_{s,i}-T_{w,i}\right)+\left(\frac{1}{m_sc_s}+\frac{1}{m_wc_w}\right)Q[/tex]

C.[tex]T_{dif}=T_{s\:}-T_w=\left(T_{s,i}-T_{w,i}\right)+\left(\frac{1}{m_sc_s}-\frac{1}{m_wc_w}\right)Q[/tex]

D. [tex]T_{dif}=T_{s\:}-T_w=\left(T_{s,i}-T_{w,i}\right)-\left(\frac{1}{m_sc_s}+\frac{1}{m_wc_w}\right)Q[/tex]

5. On some highways, exit signs are numbered according to the number of kilometres the
exit is from the place where the highway originated. If you are driving south and pass
exit 35 at 2:15 pm and then you pass exit 116 at 3:09 pm, what is your velocity in m

Answers

Answer:

We can start by using the formula:velocity = distance/timeFirst, we need to calculate the distance traveled in kilometers.

To do this, we can subtract the exit numbers:

116 km - 35 km = 81 km

Next, we need to convert the time difference from hours and minutes to hours:

3:09 pm - 2:15 pm = 0.9 hours

Now we can use the formula to find the velocity:

velocity = 81 km / 0.9 hours

velocity ≈ 90 km/h

Finally, we can convert this velocity to meters per second by multiplying by 1000/3600:

velocity = 90 km/h x 1000 m/km / 3600 s/h

velocity ≈ 25 m/s

Therefore, your velocity is approximately 25 m/s.

Explanation:

A is thrown up into the air with initial vem/s. How fast is the ball moving when it first reachemeters?

Answers

Once the ball is thrown, the only force acting on it is gravity, which means that it's acceleration is -9.81 m/s² (negative means downward).

List the known and unknown quantities from the question.

u = initial velocity = 20 m/s

v = final velocity = ? m/s

g = acceleration due to gravity = -9.81 m/s²

t = time interval = ? s

s displacement = 11 m

Before calculating the time it takes for the ball to reach 11 m, the final velocity needs to be calculated using the following kinematic equation.

v² = u² + 2gs

v = √(u² + 2gs)

= √((20 m/s)² + (2x-9.81 m/s² x 11 m)) = 13.57 m/s V=

Calculate the time it takes the ball to reach 11 m using the following kinematic equation.

V = u + gt

Solve for t.

t = (v-u)/g

t (13 57 m/s - 20 m/s)/(-981 m/s²) = 0.655 s

Jumping Off. A person of mass m jumps off a ledge that is a distance h off the ground. Their legs are straight in the air, but as they land, they bend their knees at the instant they touch the ground. Treated as a particle, they move an additional distance d downward as they hit the ground before coming to a complete stop. * Find an expression for the person's acceleration after they touch the ground in terms of g, h and d. * Draw a FBD of the person while they are in contact with the ground. Find an expression for the amount of force they apply on the ground to stop in distanced in terms of m, g, h, and d. * If the person is able to apply a force equal to 5.00 times their weight with their legs, what would the required distance to stop be if they jumped from a height of 3.00 m?

Answers

The expression for the person's acceleration after they touch the ground in terms of g, h and d is a = 2gh / (2h/g + sqrt(2gh + 2gd).

What is an acceleration?

An acceleration refers to the change in velocity with respect to time in terms of speed and direction. In the case given here, assuming no air resistance, the potential energy of the person at the top of the ledge is converted into kinetic energy just before they hit the ground.

Let's consider the motion of the person after they touch the ground. We assume that the person's acceleration is constant during the time they move a distance d. Let a be the acceleration of the person after they touch the ground, and let t be the time it takes for them to come to a complete stop. Then:

⇒ d = 1/2 × a t²........... (1)

⇒ v = at........(2)

⇒ h + d = 1/2 gt² + vt......... (3)

where, v is the velocity of the person just before they touch the ground, and g is the acceleration due to gravity.

Therefore,

t = (sqrt(2gh + 2gd + v²) – v) / g

a = 2(d + h) / t² – g

Substituting v = sqrt(2gh):

a = 2gh / (2h/g + sqrt(2gh + 2gd))

Therefore, the acceleration of the person after they touch the ground is:

a = 2gh / (2h/g + sqrt(2gh + 2gd))

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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of the cliff 4.0 s later.

Answers

Answer: I am assuming you're asking for the cliff height, which is 24M

We can use the kinematic equations to find the horizontal and vertical components of the ball's velocity and its final position.

The initial velocity components are:

v_x = 33 m/s * cos(60∘) = 28.6 m/s

v_y = 33 m/s * sin(60∘) = 33 * √3/2 m/s

The displacement components can be found using the kinematic equation:

x = v_x * t = 28.6 m/s * 4.0 s = 114.4 m

y = v_y * t - 0.5 * g * t^2 = 33 * √3/2 m/s * 4.0 s - 0.5 * 9.8 m/s^2 * (4.0 s)^2 = 102.4 m - 78.4 m = 24.0 m

The ball lands on the edge of the cliff, so its height must be equal to h, or:

y = h

24.0 m = h

So, the height of the cliff is 24.0 m.

This equation is known as the ideal gas law, and it can be used to predict the behavior of many gases at relatively low pressure. From this equation, you can see that as the temperature of a gas increases,.

Answers

From the ideal gas law equation ([tex]PV = nRT[/tex]), you can see that as the temperature of a gas increases, either the pressure (P) or the volume (V) of the gas must increase in order to maintain a constant number of moles (n) and a constant gas constant (R).

This can be explained by the kinetic theory of gases, which states that the temperature of a gas is proportional to the average kinetic energy of its molecules. As the temperature increases, the molecules move faster and collide with the walls of the container more frequently and with greater force, increasing the pressure. Alternatively, the molecules can also move further apart, increasing the volume of the gas. In other words, as the temperature of a gas increases, the gas will expand and/or its pressure will increase, assuming the volume or the number of moles of gas are held constant. This relationship is important for many practical applications, such as in the design of engines, refrigeration systems, and industrial processes.

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12. A truck travelling at 110 km/h has to cover a distance of 640 km in the next 6 hours. Will the driver be on time if he can hold his current speed? (4 marks)

Answers

Answer:

calculate the expected time of arrival, we need to divide the total distance by the speed of the truck.

In this case, the expected time of arrival would be:

640 km ÷ 110 km/h = 5.82 hours

So the driver should be able to arrive at the destination within 6 hours if he can hold his current speed.

14. A boy of mass m swings back and forth on the end of a rope 6 m long that is attached to a gymnasium ceiling. If he approaches to within 2 m of the ceiling during each cycle, what is his speed as he passes through the lowest point of the swing?

Answers

The boy's speed as he passes through the lowest point of the swing is 8.8 m/s.

How to find speed?

To solve this problem, we can use the conservation of energy. At the lowest point of the swing, the boy has the maximum gravitational potential energy and no kinetic energy. At the highest point of the swing, he has the maximum kinetic energy and no gravitational potential energy.

Since the energy of the system is conserved, the sum of the gravitational potential energy and kinetic energy must remain constant throughout the swing. Let's call the speed of the boy as he passes through the lowest point of the swing "v".

At the lowest point, the gravitational potential energy is given by:

mgh = (1/2)mv²

where m is the mass of the boy, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the lowest point above the ground (6 m - 2 m = 4 m).

Solving for v:

v = √(2gh) = √(2 x 9.8 x 4) = √(77.76) = 8.8 m/s

So the boy's speed as he passes through the lowest point of the swing is approximately 8.8 m/s.

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Suppose the Earth was only half the size it is now (half the mass and half the radius), what would "g" be?

Answers

If the Earth were only half the size it is now, the acceleration due to gravity (represented by "g") at the surface would also be halved.

This is because the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.

With the Earth's mass reduced by a factor of 2 and its radius reduced by a factor of 2, the distance between an object on the surface and the Earth's center would also be reduced by a factor of 2. Thus, the net effect is that the acceleration due to gravity would be halved, resulting in a smaller value of "g" than what we currently observe on Earth.

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