Please provide an explanation.

Thank you!!

Please Provide An Explanation. Thank You!!

Answers

Answer 1

Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(c) left will decrease, right will increase

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support.  (Remember that the magnitude of torque is force times the perpendicular distance.  Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer 2

Answer:

a.  Left support = Right support = 22 kNb.  Left support = 36 kN     Right support = 29 kNc.  Left support force will decrease     Right support force will increase.d.  Right support = 43 kN

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned  as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)  

                                  0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

c)

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support,  Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)  

                                  0.6

Right support = 43 kN


Related Questions

Determine the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration).

Answers

This question is incomplete, the complete question is;

Two blocks, M1M1 and M2,M2, are connected by a massless string that passes over a massless pulley as shown in the figure. M2,M2, which has a mass of 25.0 kg,25.0 kg, rests on a long ramp of angle =25.0∘.θ=25.0∘. Friction can be ignored in this problem.

Determine the mass 1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration).

Answer:

the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration) is 10.57 kg

Explanation:

Given that;

m₂ = 25 kg

θ = 25°

Now at equilibrium, T = m₁g ------------------lets say equ 1

and also T = m₂gsinθ

therefore

m₁g = m₂gsinθ

m₁ = m₂sinθ

so we substitute

m₁ = 25 × sin(25)

m₁ = 25 × 0.4226

m₁ = 10.565 ≈ 10.57 kg

therefore the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration) is 10.57 kg

What type of research based on approach that used to describe variables rather than to test a predicted relationship between variables?

Answers

Answer:

Correlational research can be used to see if two variables are related and to make predictions based on this relationship.

Find the gravitational potential energy of an 84 kg person standing atop Mt. Everest at an altitude of 8848 m. Use sea level as the location for y

Answers

Answer:

[tex]E=7.28\times 10^6\ J[/tex]

Explanation:

Given that,

Mass of a person, m = 84 kg

The person is standing at a top of Mt. Everest at an altitude of 8848 m

We need to find the gravitational potential energy of the person. We know that the gravitational potential energy is possessed due to the position of an object. It is given by :

E = mgh, g is the acceleration due to gravity

[tex]E=84\ kg\times 9.8\ m/s^2\times 8848\ m\\\\E=7283673.6\ J\\\\E=7.28\times 10^6\ J[/tex]

So, the gravitational potential energy of the person is [tex]7.28\times 10^6\ J[/tex]

In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a long, straight line of current. At a perpendicular distance of 21 m from the bolt, a magnetic field of 8.4 x 10-5 T is measured. How much charged has flowed during the lightning bolt?

Answers

Answer: 10.58 C has flowed during the lightning bolt

Explanation:

Given that;

Time of flow t = 1.2 × 10⁻³

perpendicular distance r = 21 m

Magnetic field B = 8.4 x 10⁻⁵ T

Now lets consider the expression for magnetic field;

B = u₀I / 2πr

the current flow is;

I = ( B × 2πr ) / u₀

so we substitute

I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷

=  0.01107792 / 0.000001256

= 8820 A

Hence the charge flows during lightning bolt  will be;

q = It

so we substitute

q = 8820 × 1.2 × 10⁻³

q = 10.58 C

therefore 10.58 C has flowed during the lightning bolt

A truck with a mass of 1330 kg and moving with a speed of 15.0 m/s rear-ends a 805 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.

Answers

Answer:

The Speed of the vehicles is 9.34m/s

Explanation:

For an elastic collision the two bodies move with similar velocities after collision

Given

M1=1330kg

V1=15m/s

M2=805kg

V2=0(the car is parked on neutral)

The formula is

M1V1+M2V2=(M1+M2)V

1330*15+805*0=(1330+805)V

19950+0=2135V

2135V=19950

divide both sides by 2135

V=19950/2135

V=9.34m/s

If car A is at 40km/h and car B is at 10km/h in the opposite direction, what is the velocity of the car A relative to the car B?. ​

Answers

Explanation:

The velocity of car A relative to car B is (10km/h+40km/h)=50km/h

Define reflection of sound?​

Answers

The reflection of sound is the movement of sound waves bouncing off of a surface and back into another direction, hope this helped :)

distance is constant and time increseas

Will Speed increase or decrease?

Answers

Answer:

The speed will be decrease

The speed will decrease

What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

Answers

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s

Answers

Answer:

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

Explanation:

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

[tex]v = r\cdot \omega[/tex] (Eq. 1)

Where:

[tex]r[/tex] - Radius of rotation of the particle, measured in meters.

[tex]\omega[/tex] - Angular velocity, measured in radians per second.

[tex]v[/tex] - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

[tex]\omega = \frac{\theta}{t}[/tex] (Eq. 2)

Where:

[tex]\theta[/tex] - Angular displacement, measured in radians.

[tex]t[/tex] - Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

[tex]v = \frac{r\cdot \theta}{t}[/tex] (Eq. 3)

From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:

[tex]s = r\cdot \theta[/tex]

[tex]v = \frac{s}{t}[/tex]

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

What is the maximum current flow possible through a 12 Ohm resistor from a 120V source?

Answers

Answer:

I=10.0A

Explanation:

V=RI(120)=(12)II=10.0A

Before digital filmmaking, what tool was used to control the speed of movement on the screen after filming?

Answers

Answer:

The appropriate response is "Optical printer ".

Explanation:

A photographic printer used mostly for optical aberrations, comprised simply of either a camera that captures the frame to expand, minimize, deform, respectively. through magnifying lenses. A projector that always, as distinct from some kind of touch printer, transferred the image to something like the printing supply.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answers

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

What does g stand for

Group of answer choices

gravity

The acceleration of gravity

The force of gravity

Answers

Answer:

the acceleration of gravity.

Answer:

g stand for the acceleration of gravity .

Explanation:

HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the
motion of the ball?

Answers

Answer:

Option C.

Explanation:

To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:

1. Determination of the velocity.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Final velocity (v) =.?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 100)

v² = 0 + 1960

v² = 1960

Take the square root of both side.

v = √(1960)

v = 44.27 m/s

2. Determination of the time taken.

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Time (t) =.?

h = ½gt²

100 = ½ × 9.8 × t²

100 = 4.9 × t²

Divide both side by 4.9

t² = 100 / 4.9

Take the square root of both side

t = √(100 / 4.9)

t = 4.52 s

From the above illustration,

Initial time (t1) = 0 s

Final time (t2) = 4.52 s

Initial velocity (u) = 0 m/s

Final velocity (v) = 44.27 m/s

Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.

what causes chest pain is it by eating peppery food? ​

Answers

Answer:

Yes that is almost always the problem!

Explanation:

Answer:

the answer is Acid reflux hope this helped!

what is the meaning of the word physics​

Answers

Answer:

the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

Explanation:

mark as brainliest

describe the energy conversion that occurs in a diesel engine

Answers

A diesel engine is a type of heat engine that uses the internal combustion process to convert the energy stored in the chemical bonds of the fuel into useful mechanical energy. ... First, the fuel reacts chemically (burns) and releases energy in the form of heat.

A recipe gives the instructions below
After browning the meat pour off fat from the pan to further reduce fat use a strainer.

what type lf separation methods are described in the recipe

A decantation and screening
B distillation and screening
C decantation and centrifugation
D distillation and filtration

Answers

Answer:

A. decantation and screening

Explanation:

Decantation is the one of the process of separating the mixture. In this process the precipitated liquid is separated from the solid. According to the given instruction for the recipe, the fat which is in liquid state is separated from meat. In the process of screening, more liquid is separated by placing the mixture on the screen. Here, the gravity plays an important role for the process of separation.  

Answer:

a

Explanation:

What type of force holds atoms together in a crystal?

Answers

Answer:

Covalent Bond

Explanation:

i took the test , mark me brainliest.

Answer: Electrical

Explanation: Atoms are tied together by electrical bonding forces.

A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration of the car.

Answers

Answer:

[tex]a=2.4\ m/s^2[/tex]

Explanation:

Given that,

The initial speed of a car, u = 0

Time, t = 18 s

Distance, d = 390 m

We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

or

[tex]d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2[/tex]

So, the acceleration of the car is [tex]2.4\ m/s^2[/tex].

what is the summary for Electrons and protons​

Answers

Explanation:

the link enjoy

A person drops a marble from the top of a skyscraper. After falling four floors the marble has gained a certain speed. How many more floors will the marble have to fall to triple this speed?

a. 8
b. 12
c. 32
d. 48

Answers

Answer:

B. 12

Explanation:

4 x 3 = 12

How much energy is used by a 1000 W microwave that operates for 5
minutes?

Answers

Answer:

300000 Joules

Explanation:

Recall that the unit Watts is Joules per second, derived from the quotient of energy per unite of time. Therefore, to calculate the energy used on the given time, we need to multiply the power used (1000 W) times the time used. Then we need to express the 5 minutes in seconds to get our answer in Joules:

5 minutes = 5 * 60 seconds = 300 seconds

Final energy calculation gives:

E = 1000 W * 300 sec= 300000 Joules

calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun

Answers

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

  = 149.668 * 10^8 m

next we will express the distance between the earth and the sun

[tex]r = \frac{a(1-E^2)}{1+Ecos\beta }[/tex]   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

[tex]\beta[/tex] = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

[tex]v^2 = \frac{4\pi^2 }{r_{c} }[/tex]   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

([tex]V^2 = (\frac{4\pi^{2} }{149.626*10^9})[/tex]

therefore : V = 1.624* 10^-5 m/s

Calculate the effective charges on the H and F atoms of the HF molecule in units of the electronic charge, e.

Answers

Answer:

Explanation:

Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is as seen below

HF ⇄ H⁺ + F⁻

There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ coloumbs.

The required value of effective charge on HF molecule, due to H and F is  1.602 × 10⁻¹⁹ Coulombs.

The given problem is based on the concept of effective charges. The net positive charge carried out by the electrons of atomic species, after forming a polyelectronic atom  is known as Effective charge.

As per the given problem, the Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is given as,

HF ⇄ H⁺ + F⁻

There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ Coulombs.

Thus, we can conclude that the required value of effective charge on HF molecule, due to H and F is 1.602 × 10⁻¹⁹ Coulombs.

Learn more about the effective charge here:

https://brainly.com/question/25002720

A studious physics student is interrupted by a swarm of bees and chased off a cliff. Since she has her calculator in hand she quickly punches in numbers to figure out the initial velocity she needs to make it into the lake below. The cliff is 10 m high and the lake is 15 m away from the edge of the cliff. Find the time it takes her to drop. Find her initial velocity,​

Answers

Answer:

The time is 1.4 sec

The initial velocity is 10.7 m/s.

Explanation:

Given that,

Height = 10 m

Distance = 15 m

We need to calculate the time

Using equation of motion

[tex]s=ut-\dfrac{1}{2}gt^2[/tex]

Put the value into the formula

[tex]10=0+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]t^2=\dfrac{2\times10}{9.8}[/tex]

[tex]t=\sqrt{\dfrac{2\times10}{9.8}}[/tex]

[tex]t=1.4\ sec[/tex]

We need to calculate the initial velocity

Using formula of velocity

[tex]v=\dfrac{d}{t}[/tex]

Put the value into the formula

[tex]v=\dfrac{15}{1.4}[/tex]

[tex]v=10.7\ m/s[/tex]

Hence, The time is 1.4 sec

The initial velocity is 10.7 m/s.

Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.

Answers

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The expression is  [tex]W_c =  P_o V_o ln (R_v)[/tex]  

Explanation:

Generally smallest workdone done by  a gas is mathematically represented as

          [tex]dW  =  PdV[/tex]

Generally for an isothermal process

    [tex]PV  =  nRT = constant [/tex]

=>   [tex]P = \frac{nRT}{V}[/tex]

Generally the total workdone is mathematically represented as

   [tex]W_c =  \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]

=> [tex]W_c = nRT  \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]

=>  [tex]nRT [lnV]   | \left \ {V_f}} \atop {V_o}} \right.[/tex]

=>  [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]

=>  [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]

From the question [tex]\frac{V_f}{V_o }  =  R_v[/tex]

=> [tex]W_c =  P Vln (R_v)[/tex]

at initial  state

[tex]W_c =  P_o V_o ln (R_v)[/tex]  

PLEASE PROVIDE AN EXPLANATION!!
THANK YOU.

Answers

Answer:

(a) 3.43 m/s

(b) 3.43 m/s

(c) 95.8 kPa

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

where P is pressure (either absolute or gauge), ρ is density, v is velocity, g is acceleration due to gravity, and h is elevation.

(a) Let's choose point 1 at the surface of the fluid in the container, and point 2 at point Z at the exit of the tube.  I'll say 0 elevation is at point Z, and I'll use gauge pressure.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 Pa + ½ ρ (0 m/s)² + ρ (9.8 m/s²) (0.60 m) = 0 Pa + ½ ρ v² + 0

ρ (9.8 m/s²) (0.60 m) = ½ ρ v²

5.88 m²/s² = ½ v²

v = 3.43 m/s

(b) The tube's cross section is constant, so the fluid's speed is the same at all points in the tube. v = 3.43 m/s.

(c) Use Bernoulli equation again, choosing point 2 to be at Y.  I'll say 0 elevation is at the surface of the fluid, and again use gauge pressure.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + 0 + 0 = P + ½ (700 kg/m³) (3.43 m/s)² + (700 kg/m³) (9.8 m/s²) (0.20 m)

0 = P + 4116 Pa + 1372 Pa

P = -5488 Pa

The gauge pressure is -5488 Pa, so the absolute pressure is 101,300 Pa + -5488 Pa = 95812 Pa, or approximately 95.8 kPa.

A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping distance of the car?

Answers

v² - u² = 2 ax

where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆x

x = (27 m/s)² / (16 m/s²)

x ≈ 45.6 m

The stopping distance of car achieved during the braking is of 45.56 m.

Given data:

The initial speed of car is, u = 27 m/s.

The final speed of car is, v = 0 m/s. (Because car comes to stop finally)

The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].

In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.

Therefore,

[tex]v^{2}=u^{2}+2(-a)s[/tex]

Here, s is the stopping distance.

Solving as,

[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]

Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.

Learn more about the kinematic equation of motion here:

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