Peregrine falcons, which can dive at 200 mph (90 m/s), grab prey birds from the air. The impact usually kills the prey. Suppose a 480 g falcon diving at 75 m/s strikes a 240 g pigeon, grabbing it in her talons. We can assume that the slow-flying pigeon is stationary. The collision between the birds lasts 15 ms.

Answers

Answer 1

Answer:

What is the average force of collision?

Explanation:

The velocity of the combined mass after impact is found by conservation of momentum

0.480(75) + 0.240(0) = (0.480 + 0.240)v

v = 50 m/s

An impulse results in a change of momentum

FΔt = mΔv

F = mΔv/Δt

for the pigeon

F = 0.240(50 - 0)/0.015

F = 800 N

for the falcon

F = 0.480(50 - 75)/0.015

F = -800 N

Answer 2

The final speed(v) of the Peregrine falcons and pigeon is 50 m/s

The objective of this question is to determine the final speed(v) of the Peregrine falcons and pigeon

From the information given:

the mass of the falcon m_f = 480 gthe speed of the falcon v_f = 75 m/sthe mass of the pigeon m_p = 240 gthe collision time = 15 ms = 0.015 s

According to the conservation of momentum, we can say that the totality of momentum before and after the collision is the same. As such;

[tex]\mathbf{m_fv_f = (m_f+m_p) v}[/tex]

[tex]\mathbf{480 g \times 75 m/s = (480 + 240) \ g \times v}[/tex]

[tex]\mathbf{v = \dfrac{480 g \times 75 m/s}{(480 + 240) \ g}}[/tex]

[tex]\mathbf{v = \dfrac{36000 \ m/s}{ 720}}[/tex]

v = 50 m/s

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Related Questions

73 ml of water is followed by 25 ml of juice. What is the percent strength of juice?

Answers

Total liquid: 73 + 25 = 98 ml

Percent juice = (25/98) x 100 = 25.5 %

How does friction help us in walking.​

Answers

Depending on the position and angle of our foot, this reaction contact force applied on our foot helps us to move forward as well as saves us from slipping and falling. This is how friction helps walking, in simple words.

give with an example a cause where the velocity of an object is zero but its acceleration is not zero .

Answers

Answer:

At the highest point when you toss a ball into the air.

Explanation:

At the higest point of a trajectory of a ball, the velocity is zero for a split second and there is no speed and direction. However, there still is acceleration of -10 m/s^2 because the force of gravity is still acting upon it at that point.

Hi there!

An example of this could be when a ball is thrown vertically into the air and reaches the TOP of its trajectory.

When an object is thrown with a vertical velocity, the acceleration due to gravity results in a decrease in its positive (upward) velocity until it reaches its highest point, where the instantaneous velocity = 0 m/s and the object begins to fall back down (negative velocity).

Additionally, throughout its entire trajectory, the ball experiences an acceleration due to gravity of g = 9.8 m/s², even at its highest point where there is a velocity = 0 m/s.

To get the dimmest bulbs with two batteries and two bulbs you would connect the batteries in ____ and the bulbs in ____.

Answers

Answer:

batteries in parallel connection and bulbs in serial connection

To get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.

What is Parallel and series circuits?

When two-terminal components and electrical networks that can be connected in series or parallel. This will result in two terminals in the electrical network, and may themselves participate in a series or parallel topology. When a two-terminal "object" is an electrical component or electrical network is a matter of perspective.

A circuit is said to be in series when the same current flows through all the components in the circuit where the current has only one path. A circuit is said to be parallel when there are multiple paths for the electric current to flow through it where the components which are part of the parallel circuit will have a constant voltage across all their ends.

Thus, to get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.

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a student lifts a toy car from a bench and places the toy car at the top of a slope describe an energy transfer that occurs when the student lifts the toy car from the bench and places the toy car at the top of the slope.

Answers

Answer:

Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope.

G.P.E = m*g*h

K.E = (m*v^2)/2

where

m = mass of toy car (kg)

g = gravity (m/s^2)

h = heigh of your car from the bottom (m)

v = velocity of the toy car as it reaches the bottom (m/s)

Equate K.E to G.P.E

G.P.E = K.E

m*g*h = (m*v^2)/2

make v the subject of the formula

v = (2*g*h)^(1/2)

Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v

v = (2*9.81*2)^(1/2)

v = 6.264 m/s

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time?

Free-fall Acceleration is -10 m/s^2

Answers

Answer:

Explanation:

s = s₀ + v₀t + ½at²

s = 0 + 0(15) + ½(6)(15²)

s = 675 m

Not sure what the free fall acceleration is needed for, but if the object is dropped from a high enough point, it will travel in 15 seconds

s = ½10(15²) = 2250 m  if air resistance is ignored

Why is the sky blue and why do we get a sunset

Answers

Answer:

Small particles of dust and pollution in the air can contribute to (and sometimes even enhance) these colors, but the primary cause of a blue sky and orange/red sunsets or sunrises is scattering by the gas molecules that make up our atmosphere. Large particles of pollution or dust scatter light in a way that changes much less for different colors.

Explanation:

A gold doubloon 6.1 cm in diameter and 2.0 mm thick is dropped over the side of a pirate sheep.  When it comes to rest on the ocean floor at a depth of 770 m, how much has its volume changed?​

Answers

The definition of volume modulus and the variation of pressure with depth allows to find the result for the variation of the volume of the coin is:

ΔV = 2.15 10⁻⁸ m³

The pressure with the depth is given by the relation

         P = P₀ + ρ g h

Where P is the pressure, ρ is the density anf h depth.

The size of the bodies is determined by the distance of their atomic and molecular bonds, therefore the size of the bodies changes under external interations, in the case of hydrostatic pressure a constant called volumetric modulus is defined.

            [tex]B = - \frac{\Delta P}{\frac{\Delta V}{Vo} } \\\Delta V = - \frac{\Delta P }{B} \ V_o[/tex]

Where ΔP is the pressure change, V₀ and V are the volume change and the initial volume of the body, the negative sign is introduced so that the volumetric modulus is a positive quantity.

They indicate the diameter and thickness of the coin (d = 6.1 cm and e =0.20 cm) on the sea surface and the depth to which it is submerged

h = 770 m

Let's look for the volume of the coin.

          V₀ = π r² h = [tex]\pi \ \frac{d^2}{4} \ e[/tex]  

          V₀ = [tex]\pi \ \frac{0.061^2 }{4} \ 0.002[/tex]  

          V₀ = 5.84 10-6 m³

Let's find the pressure at the depth of y = 770 m,  the density of sea water is ρ = 1025 kg / m³, the pressure at the surface is the atmospheric pressure P₀ = 1 10⁵ Pa, the volumetric modulus of water is B = 0.21 10¹⁰ Pa.

          P = 1 10⁵ + 1025 9.8 770

          P = 1 10⁵ + 7,735 10⁶

          P = 7.84 10⁶ Pa

Let's calculate

          ΔV =[tex]- \frac{1 \ 10^5 - 7.84 \ 10^6 }{0.21 \ 10^{10}} \ 5.845 \ 10^{-6}[/tex]  

          ΔV = 2.15 10-8 m³

In conclusion using the definition of volume modulus and the variation of pressure with depth we can find the result for the variation of the volume of the coin is:

ΔV = 2.15 10-8 m³

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This question has two parts. First, answer part A. Then, answer part B. Part A: Which statement best summarizes the theme of the text? O A. You do not always recognize what is most valuable. O B. Keep your friends close, but your enemies closer. W O C. Fine possessions do not make a fine person. h 0 D. The best things come in small packages. Part B: Which evidence from the text best supports your answer in part A? O A. "The other animals of the forest bowed to him, and they often spoke of his antlers with admiration." B. "The Stag was so engrossed that he did not notice that a Lion had crept up alongside him." C. ". . . but the branches of some of the trees hung low, and vines curled around them." D. "The long legs that I hated would have saved me, but the antlers that I loved have led to my destruction!"​

Answers

Answer: for part A Its A the ANSWER for part B its D

Explanation:

Answer:

a and b

Explanation:

A race car traveling at 100 m/s enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tires and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct

Answers

Answer:

The race car will crash into the outer wall

Explanation:

max fr = μsN = 1.1 mg = 11 m

mv2/R = m(100)2/(400) = 25 m > fr

the c component of vector a is 5.3 units, and it’s y component is -2.3 units. the angle that vector a makes with the +x axis is closest to
110
160
23
340
250

Answers

Answer:

340

Explanation:

Sorry I don't know how to do this one yet, I just found the answer in a textbook.

The angle that vector a makes with the +x axis is closest to 23.

What is direction of a vector?

The direction of a vector is represented tangent of angle equal to the ratio of the y component and the x component of the vector quantity.

tangent of angle = y/x

angle = tan⁻¹ (-2.3/5.3)

angle = 23.46°

Thus, the angle that vector makes with +x is 23.

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Does it appear that true average HAZ depth is larger for the high current condition than for the nonhigh current condition

Answers

Answer:

The data suggest that the true mean HAZ depth is larger when the current setting is higher.

Read the sentence from the text. “They are as glossy as satin or sunlight reflecting off water!" What does the word glossy mean in the sentence? O A. pointed o B. shiny O C. small O D. strong​

Answers

Answer:

b Shiny

Explanation:

Trust me it's right

The answer is B. Shiny

what is the acceleration of the cart at t=8 seconds?
a) 0 m/s^2
b) 10 m/s^2
c) 20 m/s^3
d) -20m/s^2​

Answers

ANSWER:

What is the acceleration of the cart at t=8 seconds?

a) 0 m/s^2b) 10 m/s^2c) 20 m/s^3d) -20m/s^2

Hence the answer us letter a) 0 m/s^2.

That's all I know, Hope it help :)

Acceleration of a Car A car traveling along a straight road at accelerated to a speed of over a distance of ft. What was the acceleration of the car, assuming that it was constant

Answers

Answer:

how many feet?

Explanation:

tell types of instruments

Answers

Answer:

Instrument Use

Accelerometer Measures acceleration

Altimeter Measures altitude of an aircraft

Ammeter Measures electric current in ampere

Anemometer Measures wind speed

An object is travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the acceleration?

Free-fall Acceleration is -10 m/s^2

I also need the Formula

Answers

Answer:

Explanation:

s = s₀ + v₀t + ½at²

50 = 0 + 0(4) + ½a(4²)

50 = 8a

a = 50/8 = 6.25 m/s²

Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s . As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight.

Answers

Answer:

Explanation:

0.120 rev/s(2π rad/rev) = 0.24π rad/s

At the highest point of the arc, gravity must supply the required centripetal acceleration. As the normal force is 1/4 of his normal weight, then 3/4 of gravity acceleration must be used as centripetal acceleration

0.75g = ω²R

R = 0.75(9.81) / (0.24π)²

R = 12.942198...

R = 12.9 m

The radius of the circle is equal to 12.95m which is rotating with an angular velocity of 0.120 rev/s.

What is vertical circular motion?

A body spins in a vertical circle so that its motion at different points is different from the motion of the body is said to be vertical circular motion.

The velocity and tension vary in maximum magnitude from the lowest to the highest position because of the effect of the gravitational force of the earth.

Given, the angular velocity of the Ferris wheel, ω = 0.120 rev/s

ω = 0.120 rev/s × 2π rad/rev

ω= 0.7536 rad/s

If r is the radius of the circle and 'm' is the mass of the jack.

From newton's second law of motion, the net force will be equal to

mg - N = mrω²

mg - (mg/4) = mrω²

r = 3g/4ω²

r = 3×9.81 / (4× 0.7536)

r = 12.95 m

Therefore, the radius of the circle in which the jack travels is equal to 12.95m.

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Your question is incomplete, most probably the complete question was,

Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s. As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels?

After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s

Answers

The energy in the system is given by the initial potential energy at the point 1.

The linear velocity at point 3, is approximately 33.59 m/s.

Reasons:

The parameters are;

Height at point 1, h₁ = 83 m

Radius of the ring = 8 cm

Mass of the ring, M = 8 kg

Height at point 2, h₂ = 32 m

At point 2, we have;

Change in potential energy = Kinetic energy

Which gives;

(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²

Which gives;

v ≈ 22.37 m/s

At point 3, the rotational kinetic energy remains constant while the

translational kinetic energy increases as follows;

K.E. at point 3 = Initial kinetic energy + Change in potential energy

Which gives;

K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8

[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]

v₃ ≈ √(1128.15) ≈ 33.59

The linear velocity at point 3, v₃ ≈ 33.59 m/s

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The probable question parameters as obtained from a similar question online are;

Height at point 1, h₁ = 83 m

Radius of the ring = 8 cm

Mass of the ring, M = 8 kg

Height at point 2, h₂ = 32 m

1.25 is the closest to 1.04 or not I want to answer please. I think it's true, but I want to prove it scientifically, please.

Answers

Answer/Explanation:

It False, because if You Round Both of them..

1.25= 1.30

1.04= 1.00

it's like, 1 dollar and 4 cents; compared to 1 dollar and 25 cents. Obviously 25 cents is a lot more than 4 cents.

True or False: The basketball should be dribbled below the waist.

Answers

True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.

Saturn's mass is 5.68 x 1024 kg and its radius is 6.03 x 107 m. A. Calculate the gravitational field strength at Saturn's surface. (2 marks) B. Calculate the force of gravity at Saturn's surface on an object with a mass of 50 kg.

Answers

Hi there!

A.

We can calculate the gravitational field strength using the following equation:

[tex]g = \frac{Gm_p}{r^2}[/tex]

G = Gravitational Constant

mp = mass of planet (kg)

r = radius (m)

Plug in the given values:

[tex]g = \frac{(6.67*10^{-11})*(5.68*10^{24})}{(6.03*10^7)^2} = \boxed{0.104 N/kg}[/tex]

B.

The force can be calculated using:

[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]

Plug in the values:

[tex]F_g = \frac{(6.67*10^{-11})(5.68*10^{24})(50)}{(6.04*10^7)^2} = \boxed{5.209N}[/tex]

Answer:

[tex]\boxed {\boxed {\sf g=0.104 \ N/kg \ and \ F_g= 5.2 \ N }}[/tex]

Explanation:

A. Gravitational Field Strength

The gravitational field strength can be calculated using the following formula:

[tex]g= \frac{Gm}{r^2}[/tex]

G, or the universal gravitational constant, is 6.67 × 10⁻¹¹ N*m²/kg². The mass of Saturn is 5.68 × 10²⁴ kilograms. The radius of Saturn is 6.03×10⁷ meters.

Substitute these values into the formula.

[tex]g= \frac{ (6.67 \times 10^{-11} \ N*m^2/kg^2) (5.68 \times 10^{24} \ kg)}{(6.03 \times 10^{7} \ m )^2}[/tex]

Multiply the numerator and square the denominator.

[tex]g= \frac{ 3.78856 \times 10^{14} \ N *m^2/kg }{3.63609 \times 10^{15} \ m^2}[/tex]

Divide.

[tex]g= 0.1041932405 \ N/kg[/tex]

The original measurements of mass and radius have 3 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 1 in the ten-thousandth place tells us to leave the 4 in the thousandth place.

[tex]\boxed {g \approx 0.104 \ N/kg}[/tex]

B. Force of Gravity

The force of gravity is calculated using the following formula:

[tex]F_g= mg[/tex]

The mass of the object is 50 kilograms. We just calculated the gravitational field strength, which is 0.104 Newtons per kilogram. Substitute these values into the formula.

[tex]F_g= (50 \ kg)(0.104 \ N/kg)[/tex]

Multiply. The units of kilograms cancel.

[tex]\boxed {F_g=5.20 \ N}[/tex]

If the voltage across a 5-F capacitor is 2*e^-3
V find the current and the power

Answers

Answer correct in this picture………….

6) An object is released from rest at the top of a ramp inclined at 30. degrees up from the horizontal. Due to friction, the ramp is only 20. % efficient. What is the object's speed after it slides down ALONG the ramp for 2.0 m? *

Answers

Answer:

Explanation:

I've been doing these types of problems for many years and I don't think I've ever seen an "efficiency" rating on a ramp.

I'm going to ASSUME that 20% efficient means that 80% of the Potential energy that gets converted becomes system internal heat energy.

Potential energy at the start of a 2.0 m slide

PE = mgh = mg2sin30 = mg2(½) = mg J

0.8mg J gets converted to heat and 0.2mg converts to kinetic energy

0.2mg = ½mv²

v² = 0.4g

v = √(0.4(9.8)) = 1.979898... ≈ 2.0 m/s

what would happen if gravity were to stop everywhere?

Answers

Answer:

everything will float up and go up to space and die

Explanation:

gravity keeps us down and once it stops everything will float up. And if it were to stop everywhere everything and everyone will die and everything will be destroyed.

A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round.
The figure shows a top view of a merry-go-round of radius capital R rotating counterclockwise. A sandbag is located on the merry-go-round at a distance lowercase r from the center.

Rank the following different combinations of m and r on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round.

Answers

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 20 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 0.5·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]

The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]

Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of  m·r² increases, the value of [tex]\omega _f[/tex] decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

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A baseball player notices the ball when it is 3.4 m above the
ground, traveling at 4.4 m/s. He wants to make the catch when
the ball is 1.5 m above the ground, how long does it take to reach
his glove?

Answers

Find the distance the ball travels:

3.4 meters - 1.5 meters = 1.9 meters

Now divide the distance the ball travels by the speed:

1.9 meters / 4.4 m/s = 0.43 seconds

Answer:

Explanation:

s = s₀ + v₀t + ½at²

There are an infinite number of solutions to this question as posed because we are not told the direction of the initial velocity.

Assuming ground is level and origin and UP the positive direction

The shortest amount of time possible is when the initial velocity is straight down

1.5 = 3.4 - 4.4t + ½(-9.8)t²

0 = -4.9t² - 4.4t + 1.9

t = (4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))

positive answer is

t = 0.32 s

The longest amount of time possible is when the initial velocity is straight up.

1.5 = 3.4 + 4.4t + ½(-9.8)t²

0 = -4.9t² + 4.4t + 1.9

t = (-4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))

positive answer

t = 1.22 s

If the initial velocity is horizontal, meaning no vertical velocity

1.5 = 3.4 + 0t + ½(-9.8)t²

-4.9t² = -1.9

t² = 0.38775...

t = 0.62 s

Any angle between UP and Down will have a different initial vertical velocity and result in a different time to catch height.

It appears from the comments on the other answer, that I have shown you how to arrive at three of the four possible solutions.  The initial direction is very important.

if the Periodic time of an oscillating object Triples then its frequency will?​

Answers

Answer:

it would decrease

Explanation:

f=1/T

Anita Knapp needs to get hay to cows in a frozen field using an airplane flying
80.0 m/s, at a height of 300,m. If at the last minute, how far from the cow would
she have to release the hay in order to hit the cow?*
756 m
626m
700m
575 m
Other:

Answers

Answer:

626m

Explanation:

Which region of electromagnetic spectrum will provide photons of the least energy

Answers

Answer:

Explanation:

Radio waves

Radio waves have photons with the lowest energies. Microwaves have a little more energy than radio waves. Infrared has still more, followed by visible, ultraviolet, X-rays and gamma rays.

Other Questions
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