1. What is the wave speed of a wave that has a frequency of 1500 Hz and a wavelength of 0.50 m?

Answers

Answer 1

Answer:

750m/s

Explanation:

v = fλ = (1500 Hz)(0.5 m) = 750m/s

Answer 2

Answer:

0.5 m and 1500 Hz divided by 750 m/s equals 750 m/s.

Explanation:

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Related Questions

a mass is vibrating on a string. its frequency describes ( )

Answers

Answer:

The frequency of a mass vibrating on a string describes the number of complete cycles of vibration that occur per unit of time, usually measured in hertz (Hz).

Explanation:

Ok so my question is “ blank heat is a measure of how much energy it takes to raise the temperature of a substance.” Help PLEASEE I’m stuck!

Answers

Specific heat is a measure of how much energy it takes to raise the temperature of a substance.

What is temperature ?

Temperature is defined as the measurement of degree of amount of hotness or coldness of a body.

Here,

Specific heat is a measure of how much energy it takes to raise the temperature of a substance. More clearly, specific heat is the amount heat required to raise the temperature of unit mass through one degree.

If an amount of heat Q is given to a body of mass m and ΔT is the rise in temperature. Then specific heat capacity,

      C = Q/mΔT

Its unit is Jkg⁻¹K⁻¹

Specific heat of a substance is a constant but it changes slightly with change in temperature.

The rise in temperature is small for body having large specific heat.

Hence,

Specific heat is a measure of how much energy it takes to raise the temperature of a substance.

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Kyle is blowing leaves with a leaf blower. He lifts the 1600 newton leaf blower at a
distance of 6 meters. What are the joules of work being put out?

PLS HELP <3

Answers

Answer:

To calculate the work being done, we need to use the formula:

work = force x distance x cos(theta)

where force is in newtons, distance is in meters, and theta is the angle between the direction of the force and the direction of the movement.

In this case, the force is the weight of the leaf blower, which is 1600 N, the distance is 6 meters, and the angle between the force and the movement is 0 degrees (since Kyle is lifting the leaf blower straight up). So we have:

work = 1600 N x 6 m x cos(0°)

work = 9600 J

Therefore, the joules of work being put out by Kyle are 9600 J.

Explanation:

A car accelerates uniformly from rest to a speed of 40.0 mi/h in 13.0 s.
(a) Find the distance the car travels during this time.
m

(b) Find the constant acceleration of the car.
m/s2

Answers

Answer:

b

Explanation:

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3. A shaft of 100 mm diameter rotates at 120 rad/s in a bearing 150 mm long. If the dial clearance is 0-2 mm and the absolute viscosity of the lubricant is 20 kg/ms find the power loss in the bearing.

Answers

The power loss in the bearing is 1.36 W shaft of 100 mm diameter rotates at 120 rad/s in a bearing 150 mm long.

What is viscosity?

The viscosity of a fluid is a measure of its resistance to flow. A fluid with a high viscosity resists movement because its molecular structure causes a lot of internal friction.

The viscosity of a fluid is a measure of its resistance to flow. A fluid with a high viscosity resists movement because its molecular structure causes a lot of internal friction.

To calculate the power loss in the bearing, we can use the formula:

P = F × V

To calculate the frictional force, we can use the formula:

F = μ × A × P

Where μ is the coefficient of friction, A is the area of the bearing, and P is the pressure of the lubricant.

First, we need to calculate the pressure of the lubricant:

P = μ × Viscosity × (V/D)

Where D is the diameter of the shaft.

P = (0.02) × 20 × (120/0.1) = 480 N/[tex]m^2[/tex]

Next, we can calculate the area of the bearing:

A = π × (D/2)^2 × L

Where L is the length of the bearing.

A = π × (0.1/2)^2 × 0.15 = 0.001178 [tex]m^2[/tex]

Now we can calculate the frictional force:

F = μ × A × P = (0.02) × 0.001178 × 480 = 0.0113 N

Finally, we can calculate the power loss:

P = F × V = 0.0113 × 120 = 1.36 W

Therefore, the power loss in the bearing is 1.36 W.

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What type of wave is:
a) Light
b) Sound

Answers

Answer:

Explanation:

a) transverse

b)longitudinal waves

1. A fairground ride consists of a large vertical drum that spins so
fast that everyone inside it stays pinned against the wall when
the floor drops away. The diameter of the drum is 10 m. Assume
that the coefficient of static friction between the drum and the
rider's clothes is 0.15.
a) What is the minimum speed required for the riders so that
they stay pinned against the inside of the drum when the
floor drops away?
b) What is the angular velocity of the drum at this speed?

Answers

a) Minimum speed required for the riders so that they stay pinned against the inside of the drum is 18m/s.

b) Angular velocity of the drum at this speed having diameter of the drum is 10 m is 3.6 rad/s

What is Friction ?

Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force.

frictional force is directly proportional to the Normal(N).

i.e. [tex]F_{fri}[/tex] ∝ N

[tex]F_{fri}[/tex] = μN

where μ is called as coefficient of the friction. It is a dimensionless quantity.

When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.

Given,

Diameter of the drum D = 10m , Radius r = 5m

Coefficient of static friction = 0.15

a) To stay everyone pinned against the wall of drum. Frictional Force must be equal to weight mg which are opposite to each other.

μN = mg ........1)

Centrifugal acceleration = Normal

mv²÷r = N

With this equation 1 becomes

μ[tex]\frac{mv^{2} }{r}[/tex] = mg

v² = rg÷μ

v² = 5m*9.8m/s ÷ 0.15

v² = 326.6

v= [tex]\sqrt{326.6}[/tex] = 18.07 ~ 18 m/s

Hence minimum speed required for the riders is 18m/s.

b) for angular velocity of the drum, V=rω

ω = v÷r

ω = 18m/s ÷ 5m

ω = 3.6 rad/s

hence angular velocity of the drum at 18m/s speed is 3.6 rad/s

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If you stretched a rubber band so that it had 100-J of potential energy, with how much kinetic energy will the rock leave the slingshot?

Answers

Answer:

less than 100-J

Explanation:

The potential energy held in the stretched rubber band is turned into kinetic energy of the rock when it is released, assuming that the rubber band is used to launch a rock from a slingshot.

The mass of the rock and the effectiveness of the slingshot in transmitting the energy from the rubber band to the rock are two elements that affect how much kinetic energy the rock will have. To estimate the kinetic energy, though, we may make certain generalizations.

Assume that no energy is lost as a result of friction or air resistance and that the entire potential energy held in the rubber band is transformed into the kinetic energy of the rock. In this hypothetical situation, the potential energy of the stretched rubber band would be equal to the kinetic energy of the rock.

As a result, the rock will have 100 J of kinetic energy when it exits the slingshot if the rubber band contains 100 J of potential energy. The actual kinetic energy of the rock would be less than 100 J since some energy will be wasted owing to things like friction.

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What is the power involved in lifting a 20- kg object 1.0m in 1.0 s?

Answers

Answer:

Explanation:

Power=work done/time

work done=force* displacement

force=mass*acceleration due to gravity

Therefore, power= mass*acceleration due to gravity*displacement/time

20*9.8*1/1=196watts

A person can jump a horizontal distance of 1.31 m on the Earth. The acceleration of gravity is 12.3 m/s^2.

a) How far could he jump on the Moon, where the free-fall acceleration is 0.258g Answer in units of m.

b) Repeat for Mars, where the acceleration due to gravity is 0.293g. Answer in units of m.

Answers

Answer:

a) On the Moon, where the acceleration due to gravity is 0.258g:

First, we need to find the acceleration due to gravity on the Moon:

g_Moon = 0.258g_Earth

g_Moon = 0.258(12.3 m/s^2)

g_Moon = 3.17 m/s^2

Now we can use the range formula for projectile motion to find the distance he could jump:

R = (v^2/g) sin(2θ)

Assuming the same initial velocity and angle of jump, we can rearrange the formula to solve for R:

R = (v^2/g) sin(^2/g_Earth) sin(2θ) * (g_Moon/g_Earth)

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.17 m/s^2) / (12.3 m/s^2)

R = 0.191 m

Therefore, he could jump approximately 0.191 m on the Moon.

!

a) On the Moon, where the acceleration due to gravity is 0.258g:

First, we need to find the acceleration due to gravity on the Moon:

g_Moon = 0.258g_Earth

g_Moon = 0.258(12.3 m/s^2)

g_Moon = 3.17 m/s^2

Now we can use the range formula for projectile motion to find the distance he could jump:

R = (v^2/g) sin(2θ)

Assuming the same initial velocity and angle of jump, we can rearrange the formula to solve for R:

R = (v^2/g) sin(

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^2/g_Earth) sin(2θ) * (g_Moon/g_Earth)

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.17 m/s^2) / (12.3 m/s^2)

R = 0.191 m

Therefore, he could jump approximately 0.191 m on the Moon.

b) On Mars, where the acceleration due to gravity is 0.293g:

Similarly, we need to find the acceleration due to gravity on Mars:

g_Mars = 0.293g_Earth

g_Mars = 0.293(12.3 m/s^2)

g_Mars = 3.61 m/s^2

Using the same formula and rearrangement as in part a, we can find the distance he could jump on Mars:

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.61 m/s^2) / (12.3 m/s^2)

R = 0.223 m

Therefore, he could jump approximately 0.223 m on Mars.

# Calculate the radius of water drop which look just remain suspendat in an electric-ff- eld of 300 viem and charged with one election-​

Answers

The radius of a water droplet suspended in an electric field can be calculated using the following formula:

r = (3qE/4πρg)^(1/2)

where r is the radius of the droplet, q is the charge on the droplet, E is the strength of the electric field, ρ is the density of the droplet material (assumed to be that of water, which is 1000 kg/m^3), and g is the acceleration due to gravity.

What is the radius of water drop which look just remain suspendat in an electric-ff- eld of 300 viem and charged with one election?

In this case, q = -1.602 × 10^-19 C (the charge on an electron), E = 300 V/m, ρ = 1000 kg/m^3, and g = 9.81 m/s^2.

Plugging in these values, we get:

r = (3(-1.602 × 10^-19 C)(300 V/m)/(4π(1000 kg/m^3)(9.81 m/s^2)))^(1/2)

= 1.83 × 10^-5 m

Therefore, the radius of the water droplet is approximately 1.83 × 10^-5 meters.

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A 500 g object is dropped from a height of 2 meters. What is its kinetic energy just before it hits the ground?

Answers

Answer: 9.8 J

Explanation:

Since the gravitational potential energy of the object is mgh or mass*acceleration due to gravity*initial height, its [tex]U_{g}[/tex] is 9.8 J. Due to the Law of Conservation of Energy, its kinetic energy will also be 9.8 J. This can be seen in the equation [tex]KE_{i}+ PE_{i}= KE_{f} + PE_{f}[/tex]. Since there is no initial kinetic energy and no final potential energy, its initial potential energy is equal to its final kinetic energy.

Answer:

9.81 joules.

Step by step solved:

The kinetic energy (KE) of an object is given by the formula:

KE = 1/2 * m * v^2

where m is the mass of the object and v is its velocity.

When an object is dropped from rest, it gains speed as it falls due to the force of gravity. The velocity of a falling object can be calculated using the formula:

v = √(2gh)

where g is the acceleration due to gravity (9.81 m/s^2) and h is the height from which the object is dropped.

In this problem, the mass of the object is 500 g or 0.5 kg, and the height from which it is dropped is 2 meters. Using the formula for velocity, we get:

v = √(2gh) = √(2 x 9.81 m/s^2 x 2 m) = √(39.24) = 6.27 m/s

Now that we know the velocity of the object just before it hits the ground, we can calculate its kinetic energy using the formula:

KE = 1/2 * m * v^2 = 1/2 * 0.5 kg * (6.27 m/s)^2 = 9.81 J

Therefore, the kinetic energy of the 500 g object just before it hits the ground is 9.81 joules.

I can’t figure it out

Answers

Answer:

Explanation: In anyway can you get a better picture just a close picture i cant read it

Four students each construct a claim about the gravitational attraction on a probe passing between Mercury and Venus when the planets are nearest in their orbits.



Student W claims that the probe is pulled harder by the planet with the greater mass when halfway between the planets.

Student X claims that the probe is pulled harder by the planet nearest the Sun when halfway between the planets.

Student Y claims that the probe is pulled harder by the planet with the greater mass anywhere between the planets.

Student Z claims that the probe is pulled harder by the planet nearest the Sun anywhere between the planets.


Which student made a claim that is best supported by evidence from the table and diagram?

Answers

Answer:

Without the table and diagram, it's difficult to provide a specific answer. However, in general, the gravitational attraction between two objects depends on their masses and the distance between them. The force of attraction is stronger when the masses are greater and when the objects are closer together.

Based on this understanding, we can evaluate the claims made by each student:

Student W claims that the probe is pulled harder by the planet with the greater mass when halfway between the planets. This claim is partially supported by the evidence, since the gravitational force is stronger when the masses are greater. However, it doesn't take into account the distance between the planets and the probe.

Student X claims that the probe is pulled harder by the planet nearest the Sun when halfway between the planets. This claim is not supported by the evidence, since the distance between the probe and each planet is not specified.

Student Y claims that the probe is pulled harder by the planet with the greater mass anywhere between the planets. This claim is partially supported by the evidence, since the gravitational force is stronger when the masses are greater. However, it doesn't take into account the distance between the planets and the probe.

Student Z claims that the probe is pulled harder by the planet nearest the Sun anywhere between the planets. This claim is not supported by the evidence, since the distance between the probe and each planet is not specified.

Overall, Student W and Student Y made claims that are partially supported by the evidence, but neither claim takes into account the distance between the planets and the probe. Therefore, it's difficult to determine which claim is best supported by the evidence without more information.

Explanation:

Can someone please teach me how to do this

I need to find the tension in the bottom rope

Answers

The solutions for questions A and B are

28.1866N T_1=0 N

What is tension?

Generally, Tension is a physical force that pulls on an object, often tending to stretch it. It is typically measured in units of force per area, such as newtons per square meter (N/m2). Tension is an important concept in mechanics, physics, engineering, and other fields.

(a) T_1=m_1(g-a)

T_1=4.81 (9.8-3.94)

T_1=28.1866N

(b) In freefall

a=g

T_1=m_1(g-g)

T_1=0 N

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What is the velocity of a 1,000.0 kg car if its kinetic energy is 200 kJ?

Answers

Answer: 20

Explanation:

What is the wavelength (in nm) of maximum intensity for a celestial body with a temperature of 50000 K?

Answers

The wavelength of maximum intensity for a celestial body with a temperature of 50000 K is approximately 577 nm.

What is wavelength?

To find the wavelength of maximum intensity for a celestial body with a temperature of 50000 K, we can use Wien's displacement law, which relates the temperature of an object to the wavelength at which it emits the most radiation. The law is given by:

λ_max = b / T

where λ_max is the wavelength of maximum intensity, b is Wien's displacement constant (2.898 × 10⁻³ m·K), and T is the temperature in Kelvin.

We can convert the temperature of 50000 K to Kelvin by adding 273.15 to get:

T = 50000 K + 273.15 = 50273.15 K

Plugging in the values, we get:

λ_max = (2.898 × 10⁻³ m·K) / 50273.15 K

Simplifying, we get:

λ_max = 5.77 × 10⁻⁸ meters

To convert meters to nanometers, we can multiply by 10⁹:

λ_max = 5.77 × 10⁻⁸ meters × 10⁹ nm/m = 577 nm

Therefore, the wavelength of maximum intensity for a celestial body with a temperature of 50000 K is approximately 577 nm.

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a rock rolling down a slope from rest covers a distance of 4 m in the first second. What distance will it covers in 3 sec?​

Answers

Wouldn’t the answer for this question be 12 m because 4x3

Answer: 12 meters

Explanation:

the rate is 4 meters per second
the rock rolled for 3 seconds
4 x 3 = 12

Q12. Calculate the resistance of a microwave with 5V and a current of 300mA​

Answers

Answer:

To calculate the resistance of the microwave, we can use Ohm's Law, which states that:

resistance = voltage / current

Substituting the given values into this equation, we get:

resistance = 5V / 0.3A

resistance = 16.67 ohms

Therefore, the resistance of the microwave with 5V and a current of 300mA is 16.67 ohms.

a charged cloud system produces an electric field in the air near the earth surface. When a particle (q=-2.0x10-9)is acted on by a downward electrostatic force of 3.0x10^-6 N when placed in this field, determine the magnitude of the electric field

Answers

The magnitude of the electrical field would be 1.5x10^3 N/C.

Electrical field calculation

The electrostatic force experienced by a charged particle in an electric field is given by the formula:

F = qE

Where F is the electrostatic force, q is the charge of the particle, and E is the electric field strength.

In this case, the electrostatic force acting on the particle is 3.0x10^-6 N and the charge of the particle is -2.0x10^-9 C. So we have:

3.0x10^-6 N = (-2.0x10^-9 C)E

Solving for E, we get:

E = (3.0x10^-6 N) / (-2.0x10^-9 C)

E = -1.5x10^3 N/C

Since the electric field is a vector quantity, its magnitude is always positive. Therefore, the magnitude of the electric field in this case is:

|E| = 1.5x10^3 N/C

So the magnitude of the electric field is 1.5x10^3 N/C, directed downward.

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Urgent!!!
Two students are testing out the law of conservation of momentum, by throwing balls of clay so that they collide. One student throws a 535g ball north at 12.4 m/s. The other throws a 725g ball south at 6.4 m/s. When the two balls of clay collide and form a single unit, what is the velocity of the combined unit?

1.86 m/s south

8.74 m/s north

1.86 m/s north

8.74 m/s south

Answers

Answer:

To solve this problem, we need to use the law of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.

The momentum of an object is given by its mass multiplied by its velocity, so we can calculate the initial momentum of each ball before the collision:

- The northbound ball has a momentum of 535g * 12.4 m/s = 6644 g*m/s north

- The southbound ball has a momentum of 725g * (-6.4 m/s) = -4640 g*m/s north (note that the negative sign indicates southward direction)

Adding these momenta together, we get a total momentum of 6644 g*m/s - 4640 g*m/s = 2004 g*m/s north.

After the collision, the two clay balls stick together and move as a single unit. Let's call the mass of the combined unit "M" and its velocity "v". By conservation of momentum, we know that the total momentum of the combined unit after the collision must be the same as the total momentum before:

M * v = 2004 g*m/s north

To solve for v, we need to figure out the mass of the combined unit. This is simply the sum of the masses of the two original balls:

M = 535g + 725g = 1260g

Substituting this into the equation above, we get:

1260g * v = 2004 g*m/s north

Solving for v, we get:

v = 1.59 m/s north

Therefore, the combined unit moves 1.59 m/s north after the collision.

However, the answer choices given in the problem are in meters per second, not meters per second north/south. To convert the answer, we need to add a direction. Recall that the northbound ball had a positive velocity and the southbound ball had a negative velocity. Since the combined unit is moving northward, we know its velocity must be positive.

Therefore, the final answer is 1.59 m/s north, which corresponds to answer choice C.

Assume that bone will fracture if a shear stress more than 9.00 × 10^7 N/m^2 is exerted. What is the maximum force that can be exerted on the femur bone in the leg if it has a minimum effective diameter of 9.00 cm ? Young’s modulus for bone is 1.8 × 10^10 N/m^2 Answer in units of kN

Answers

The area of the bone is 0.785 m². Then the maximum force that can be exerted on the femur bone if the shear stress is 9 × 10⁷ N/m² is 70650 kN.

What is young's modulus ?

The Young's modulus of a material is the ratio of its stress to strain. Where stress is the force per unit area and strain be the ratio of change in length to the original length.

given stress s = 9 × 10⁷ N/m²

diameter of the bone d = 9 cm = 0.09 m.

then area = π d²/4

a = 3.14 × (0.09 m )²/4 = 0.785 m².

Stress = maximum force/area

then Fmax  = stress × area

Fmax = 9 × 10⁷ N/m²  × 0.785 m²

         = 70650 kN.

Therefore, the maximum force that can be exerted to the bone is 70650 kN.

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If the energy released by an electron making a transition from one hydrogen atom orbit to another is 3.02 ✕ 10^−19 J, what is the wavelength (in nm) of the photon?

Answers

Answer:

The wavelength of the photon can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 ✕ 10^−34 J s), c is the speed of light (2.998 ✕ 10^8 m/s), and λ is the wavelength.

λ = hc/E = (6.626 ✕ 10^−34 J s) (2.998 ✕ 10^8 m/s) / (3.02 ✕ 10^−19 J) = 656.4 nm

Explanation:

I = 1/2MR^2 is rolling with a translational speed v along a horizontal surface without slipping. The work required to stop the disk is
A. 1/2 mv^2
B. 1/2mv^2 + 1/2Iw^2
C. 1/2Iw^2
D. 1/2mv^2 - 1/2Iw^2

Answers

Answer:

B. 1/2mv^2 + 1/2Iw^2

Explanation:

URGENT HELP NEEDED!!!
Two spheres of molten rock, both moving in the positive y-direction, collide in space. One sphere has a mass of 660 kg and is moving at a velocity of 17 m/s. The other sphere has a mass of 970 kg, and is moving at a velocity of 24 m/s. When the two spheres collide, and become one unit, what will be their combined velocity?

21.2 m/s

24.0 m/s

41.0 m/s

20.5 m/s

Answers

Answer:

20.5 m/s

Explanation:

two children weigh the same. Will this seesaw balance?

Answers

yes they would balance because they have the same weight

Explanation:

hope this helps (:

Since the two children have the same weight, they have to sit at equal distances from the pivot of the seesaw.  Then they will balance.

A hockey puck with a mass of 0.171 kg slides over the ice. The puck initially slides with a speed of 5.65 m/s, but it comes to a rough patch in the ice which slows it down to a speed of 3.05 m/s.

How much energy is dissipated as the puck slides over the rough patch?

Answers

To answer the question we have, 1.93401 J of energy is lost when the puck travels over the uneven surface.

What is the easiest way to define energy?

Energy is referred to by scientists as the capacity for work. People have figured out how to transform energy from one kind to the other before employing it to accomplish tasks, making western civilisation possible.

Hockey weighs 0.171 kg.

Starting speed: 5.65 m/s

Ultimately, the speed was 3.05 m/s.

We must determine how much kinetic energy was lost on the tough patch.

[tex]E_{d}=\frac{1}{2} v^2_{2} -v^2_1[/tex]

Where, m = mass

v₁ = Initial velocity

v₂ = Final velocity

Put the value into the formula

[tex]E_{d}=\frac{1}{2}[/tex] × 0.171 × [tex](3.05^2 - 5.65^2)[/tex]

=  1.93401 J energy is dissipated as the puck slides over the rough patch.

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As we've discussed, one model of material deformation links stress and strain by a material constant called a modulus. For a material under lengthwise stress (tension or compression), this is Young’s modulus (Y) and the relation may be stated as:


The data below was collected for five different materials using tools which allowed for the direct measurement of the stress and strain. By applying the model above, rank the Young’s modulus for each of these five materials from largest to smallest.

Answers

Young’s modulus for each of these five materials from largest to smallest are mentioned below.

What is Young’s modulus?

Many substances lack linearity and elasticity after very minor deformation. Only materials that are linearly elastic are subject to the constant Young's modulus. In this case, the ratio of stress to strain, which corresponds to the material's stress, determines the Young's modulus.

What is data?

Facts such as numbers, words, measurements, observations, or even simple descriptions of objects are grouped together as data. Both qualitative and quantitative data are possible. Qualitative data is information that is descriptive (describes something), whereas discrete data can only take particular values (like whole numbers).

Therefore, Young’s modulus for each of these five materials from largest to smallest are mentioned above.

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When weight training you should aim to be able to lift the weight ____ to _____ times without strain or injury but with enough resistance to give the muscles a good workout. 1 to 3 30 to 40 8 to 12 Show​

Answers

When weight training, you should aim to be able to lift the weight 8 to 12 times without strain or injury but with enough resistance to give the muscles a good workout.

Will sound move faster or slower in water? Why

Answers

Answer:

In water, the particles are much closer together, and they can quickly transmit vibration energy from one particle to the next. This means that the sound wave travels over four times faster than it would in air, but it takes a lot of energy to start the vibration.

Explanation:

Answer:

slower as it is more dense in water

Explanation:

Other Questions
Solve for x in log3 81 = x. Find a formula for the exponential function passing through the points (-2,768) and (2,3) Sketch a graph of a function that satisfies the given limit statement:[tex]\lim_{x \to \ 2^+} h(x)=5[/tex] How many solutions does the system of linear equations represented in the graph have? Coordinate plane with one line that passes through the points negative 2 comma negative 3 and 0 comma negative 2 and another line that passes through the points 0 comma 3 and 1 comma 1. One solution at (1, 2) One solution at (2, 1) No solution Infinitely many solutions A. Set up accounts for the following: cash, accounts receivable, office equipment, accounts payable, notes payable, common stock, advertising revenue, want ad revenue, printing expense, advertising expense, utilities expense, salaries expense, rent expense, and delivery expense. Prepare journal entries in a general journal and record the foregoing transactions in the accounts Chapter 1 Questions-Of mice and menChapter OneComprehension CheckDirections: To give you a comprehensive understanding of all aspects of the novel, answerthe following questions using complete sentences on a separate sheet of paper. Be sure touse your Note-Taking chart to keep important notes for each chapter and to help youanswer the Comprehension Check questions.1. Describe the two men. How are they different physically? How are they differentmentally?2. Why doesnt George want Lennie to drink the water?3. Why does Lennie have a dead mouse in his pocket?4. To where are George and Lennie headed?5. Where are George and Lennie coming from?6. What is your impression of the relationship between the two men? Why do you thinkGeorge takes care of Lennie?7. What does George tell Lennie to do when they get to where they are going?8. When Lennie goes out to get wood for a fire, what does he bring back?9. Who used to give Lennie mice? Why did she stop?10. How is George resentful of Lennie? What does George wish he could do?11. What dream do George and Lennie share?12. Why do George and Lennie feel they are different from other men?13. Where does George tell Lennie to go if he gets in trouble? Why do you think Georgemakes it a point to be sure Lennie understands this?14. Despite being slow, Lennie knows how to make George feel guilty. Describe twosituations in which Lennie manipulates George and makes him feel bad. What doesthis reveal about their relationship?15. You may have noticed that Steinbeck uses a lot of slangeven some curse wordsand foul language. Why do you think Steinbeck does this? What effect does the useof this language have on the way the story is told? How might the story be different ifSteinbeck avoided this technique? Reading a sentence is an example of.A. decodingB. signalingC. visualizingD.communicating name one technological or economic revolution that resulted from the expansion of the cotton industryand how did the consequences of expanding the cotton industry impact slaves and indigenous americansi believe the answer 2 the 1st question is somewhere in the photo but i cant find it this is like gibberish to me why might volunteering be a good option post secondary Critical Thinking 2. Put yourself in the position of an entrepreneur who is developing a new product to introduce into the market (a new backpack). Briefly describe the product and each component of its value proposition. (3 Marks) (Minimum 250 words) what did social scientist think about social movements The AMA of the wedge is 4. If How do the force exerted on the wedge as time as the load. For each of the following scenarios, indicate whether in your opinion it is a violation of the HKICPA Code of Ethics. Specifically, for each scenario, choose one of the following options: (1)not likely to be a violation; (2) likely to be aviolation; or (3) clearly a violation. For each- scenario, clearly explain your reasoning. .....................heop EFG has coordinates E(2, 4), F(8, 0), and G(5, 3). What are the coordinates of the vertices of the image after a reflection in the y-axis? HELP! If two objects have 24kg m/s of momentum before they collide. How much momentum do the two objects have afterwards? what are the domain and range of the function f(x)=2^x+1 (Inserta los smbolos +, -,,0 +27 ___________3_________ 5 _______2 = 19 Type 2 paragraphs disscusing the importance of "Sharpening the saw" How much heat does it take to heat 100 g ice of ice from -7C to 0C