You are a member of an alpine rescue team and must get a box of supplies, with mass 3 kg , up an incline of constant slope angle 30.0, so that it reaches a stranded skier who is a vertical distance 4 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 0.05. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s2 .What is the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier

Answers

Answer 1

Answer:

v = 9.04 m / s

Explanation:

For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.

          W = Em_f - Em₀         (1)

Starting point. Lower slope

        Em₀ = K = ½ m v²

highest point. Where is the skier at a height h

        Em_f = U = m g h

The work of rubbing

        W = -fr x

the negative sign is because the friction force opposes the movement.

Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular

let's use trigonometry to break down the weight

        cos θ = W_y / W

        sin θ = Wₓ / W

        W_y = W cos θ

        Wₓ = W sin θ

Y axis

        N - Wₓ = 0

        N = mg sin  θ

X axis

         fr = m a

the friction force has the expression

         fr = μ N

         fr = μ mg sin θ

we look for the job

         W = - μ mg sin θ  x

where x is the distance along the slope

       

we substitute in 1

         -μ mg sin θ x = mg h - ½ m v²

let's use trigonometry to find the distance x

        tan 30 = h / x

        x = h / tan 30

we substitute

          -   [tex]\mu \ mg \ sin \theta \ \frac{h}{tan 30} \ x[/tex] = m gh - ½ m v²

we use  

          tan 30 = sin30 / cos30

         

          v² = 2g h + 2 μ g h cos 30

          v = [tex]\sqrt{ 2gh \ (1+ cos 30}[/tex]

let's calculate

          v = [tex]\sqrt{ 2 \ 9.8 \ 4 \ (1 + 0.05 \ cos \ 30)}[/tex]

          v = 9.04 m / s


Related Questions

HELP ITS DUE IN 4 MINUTES

Answers

Answer:

the nephew

Explanation:

because we need to First find the trails that is on top of all the others in order to find who was the last to leave, since the person that was last to leave would most likely step on another person's trail that has been theirs before them.

by looking at this picture I noticed that the butler's footprint was the latest one but the dogs footprint steps over it therefore making it the last to leave, and the person who owns a dog is her nephew so therefore the nephew is the one that stole the butterfly trophy.

Which of the following represents a concave lens?
A. -di
B. +di
C. -f
D. +f

Answers

Answer:

The answer is option D. +f

Option D represents a concave lens. There are two types of lenses, one is a concave lens while the other is the convex lens.

What is the definition of a concave lens?

A concave lens deviates a direct beam from the source into a reduced form.  At minimum, one interior face of concave lenses is curved.

Because it is curved round inwards at the center and bulges outwards through the edges, causing the light to diverge, a concave lens is also known as a diverging lens.

It forms an upright , virtual picture both real and virtual pictures are formed from the concave lens.

The image formed from the positive side of the focus of the lens.+f shows the positive side of the focus of the lens.f is the focal length.

Hence, option D represents a concave lens

To learn more about the concave lens, refer to the link;

https://brainly.com/question/2919483

#SPJ2

PLZZZ HELP!!! 80PTS!!!
You are given a toy car and told it has a density of 0.54 g/mL and a volume of 38.2 mL. Calculate impulse, given that the toy car's acceleration is 10 m/s^2 and time is 10 s. WRITE THE NUMBER ONLY and DO NOT ROUND.

Answers

Answer:

Impulse = [tex]206.28 * 10 = 2062.8[/tex] g.m/s

Explanation:

Impulse is equal to the product of force and time

Here

Density of car =  [tex]0.54[/tex]  g/mL

Volume of car [tex]= 38.2[/tex] mL

Mass of the car is equal to the product of density and volume of car

Mass of the car [tex]= 0.54 * 38.2[/tex] [tex]= 20.628[/tex] grams

Acceleration of the car is [tex]10[/tex] m/s^2

Force is equal to product of mass and acceleration

F = [tex]20.628 * 10 = 206.28[/tex] g .m/s^2

Impulse = F * t

Impulse = [tex]206.28 * 10 = 2062.8[/tex] g.m/s

A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. When a rock is put into the glass, the volume level of the water changes to 450 mL and the scale reading changes to 9.22 N. What is the specific gravity of the rock

Answers

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N  weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

The specific gravity of the rock in the given water volume is 0.2.

The given parameters;

initial volume of the water, = 375 mlweight of the water, = 7.6 Nfinal volume of water = 450 mlchange in scale reading = 9.22 N

The specific gravity of the rock is calculated as follows;

[tex]S.G = \frac{weight \ in \ air}{Weight \ in \ water} \\\\S.G = \frac{450 - 375}{375} \\\\S.G = 0.2[/tex]

Thus, the specific gravity of the rock in the given water volume is 0.2.

Learn more here:https://brainly.com/question/19142897

Which of the following statements is NOT correct about sea breezes?
A) In a sea breeze, the sand is much warmer than the water, therefore creating a low pressure above the sand.
B) The breeze/wind moves from the sea to the land.
C) The breeze/wind moves from the land to the sea.
D) There is high pressure above the sea, and low pressure above the sand.

Answers

I think the correct answer is B.

The breeze/wind moves from the land to the sea. This statement is NOT correct about sea breezes.

What is  sea breeze?

Local wind patterns known as sea breezes flow from the sea to land during the day. When there is no strong large-scale wind system and it is very hot or very cold during the day or at night, sea breezes and land breezes alternate along the coasts of large lakes or oceans.

Because the sea breeze's surface flow ends over land, an area of low-level air convergence is created. Locally, this convergence frequently causes air to rise, which promotes the formation of clouds. Showers over land in the afternoon may result from such clouds.

Hence, the sea  breeze/wind moves from the land to the sea. This statement is NOT correct about sea breezes.

Learn more about sea breeze here:

https://brainly.com/question/13015619

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A 1500 kg car traveling at 30 m/s hits a stationary 1200 kg car. If they stick together on collision, what is the final velocity of the two cars?

Answers

Answer:

16.67 m/s

Explanation:

Let that velocity be v.

Using conservation of momentum:

Initial momentum = final momentum

momentum of car1 = momentum of cars

mass1 x velocity1 = (m1 + m2)v

1500*30 = (1500 + 1200)v

45000/2700 = v

16.67 m/s = v

HELP ITS DUE IN 4 MINUTES

Answers

Answer:

The nephew

Explanation:

The nephew has a seeing eye dog which means the nephews tracks are the dogs tracks, and the dog tracks are not covered by anything which means the dog was last and stole the butterfly.

Hope this helps!

Calculate the size of the magnetic field 20 m below a high voltage power line. The line carries 450 MW at a voltage of 300,000 V. Group of answer choices 0.237 T 0.0237 T 0.474 T 2.37 T 0.237 J

Answers

Answer:

[tex]1.5 \times 10^{-5} \mathrm{~T}[/tex].

Explanation:

Power carried by the line [tex]=P=450 \mathrm{MW}=450 \times 10^{6} \mathrm{~W}[/tex]

Voltage across the line Volts

Current flowing in the line =i

Size of magnetic field =B

Distance from the line

Formula Used:

Current flowing is given as

[tex]i=\frac{P}{\Delta V}[/tex]

Magnetic field by the current carrying wire is given as

[tex]B=\left(\frac{\mu}{4 \pi}\right)\left(\frac{2 i}{r}\right)[/tex]

Inserting the values

 [tex]B=\left(10^{-7}\right)\left(\frac{2(1500)}{(20)}\right) \\ B=1.5 \times 10^{-5} \mathrm{~T}[/tex]

Conclusion:

Thus, the magnetic field comes out to be [tex]1.5 \times 10^{-5} \mathrm{~T}[/tex].

can someone help me with science:

The cells within a multicellular organism are of a uniform shape.

True
or False

Answers

Answer:

trur

Explanation:

gvidttcyhjddgdhjfjdjsjs

Which of the following is an
example of thermal energy?
A. turning on a lamp so you can read
B. rubbing your hands together to get warm
C. kicking a ball

Answers

b. rubbing your hands together to get warm

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 520 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tensions in the rope to the left and to the right of the mountain climber.

Answers

Answer:

The answer is "892.90 N"

Explanation:

Following are the solution to these question:

Calculating the vertical force of the summation that is equal to zero:

[tex]\to TL \cos 65 + TR \cos 80 -520 = 0\\\\\to 0.4226\ TL + 0.1736\ TR = 520\\[/tex]

Calculating the sum of horizontal forces that is equal to zero:

[tex]\to TL\sin 65 - TR \sin 80 = 0 \\\\\to 0.9063TL - 0.9846TR = 0\\\\\to TL = (\frac{0.9846}{0.9063})\ TR \ \ = 1.0866\ TR\\\\\to 0.4226(1.0866) \ TR +0.1736\ TR =520 \ N\\\\\to 0.6328 \ TR = 520 \\\\\to TR = 821.74 \ N \\\\\to TL = 1.0866 \times 821.74 = 892.90\ N[/tex]

While being thrown, a net force of 132 N acts on a baseball (mass = 140g) for a period of 4.5 x 10^-2 sec. what is the magnitude of the change in momentum of the ball?

Answers

Answer:

5.94 N·s

Explanation:

F = 132 N

t = 0.045 s

Impulse = Ft = (132 N)(0.045 s) = 5.94 N·s

how does the leader and the led' reflect on the experience of a typical african country​

Answers

Answer:

leader give enthusiasm and strength power to country they reflect to. be a successful and hyperbolic

Explanation:

leader is the main king of the country they reflect them to be honourable and be confident that is the way to get key on a typical African

What happens after the president gives the budget to Congress?

Answers

Answer: The president submits a budget to Congress by the first Monday in February every year. The budget contains estimates of federal government income and spending for the upcoming fiscal year and also recommends funding levels for the federal government.

Explanation:

Answer: The president submits a budget to Congress by the first Monday in February every year. The budget contains estimates of federal government income and spending for the upcoming fiscal year and also recommends funding levels for the federal government.

Your welcome!! :)

Brainliest!!! Write: Forces are all around us. Imagine that your teacher has asked you to teach a lesson to your peers about forces. Explain, in detail, how you experience forces in your everyday life. Give 3 examples

Answers

If you drag a bag across floor, you are experiencing a friction force

if you throw a paper or feather up, it floats side ways slowly. It is called air Resistance.

if you push or pull a Door, it is Normal force.

I hope this helped!

have an amazing Day!!

Calculate the average maximum height for all three trials when the mass of the bottle is 0.125 kg, 0.250 kg,
0.375 kg, and 0.500 kg.

Record your calculations in Table A of your Student Guide.

When the mass of the bottle is 0.125 kg, the average maximum height of the beanbag is
m.

When the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is
m.

When the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is
m.

When the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is
m.

Answers

Answer:

10

Explanation:

i dont know this is right answer

0.35

0.91

1.26

1.57

Explanation:

I did it on ed2020 and got it correct

An ice cube in a glass of water is pushed to the bottom of the glass and held there with a straw. Consequently, the buoyant force on the ice cube is now a. the same as when the cube was floating at the top. b. exactly balanced by the weight of the ice cube. c. exactly balanced by the force exerted by the straw. d. greater than when the cube was floating at the top.

Answers

Answer:d

Explanation:

Given

Initially, the ice cube is floating over the water

When the cube is pressed to touch the bottom, it is submerged fully

Therefore more buoyant force is acting on it

At first, a part of the volume is submerged in the water, so the buoyant force is less, but as the entire cube is immersed in the water, the buoyant force increases.

A vessel having a capacity of 0.05 m³ contains a mixture of saturated water an saturated steam at a temperature 245°C the mass of the liquid present is 10 kg. find the following: i- The pressure. ii- The mass. iii- The specific volume. iv- The specific enthalpy. v- The specific internal energy.

Answers

b because because i’m was not a big head of a baby i i would’ve got caught up in here and he did it again i

I need help! thank you

Answers

Answer:

it is coice A and choice D

A 3.0-kg and a 1.0-kg box rest side-by-side on a smooth, level floor. A horizontal force of 32 N is applied to the 1.0-kg box pushing it against the 3.0-kg box, and, as a result, both boxes slide along the floor. How hard do the two boxes push against each other

Answers

Considering both boxes as one body, it would have a total mass of 4.0 kg. By Newton's second law, the 32 N force applies an acceleration a such that

F = 32 N = (4.0 kg) a   →   a = 8.0 m/s²

and both boxes share this acceleration. (There is no friction, so the given force is the only one involved in the direction of the boxes' motion.)

Now consider just the smaller box. It is feeling the effect of the 32 N push in one direction and, as it comes into contact with the larger box, a normal force that points in the opposite direction. Let n be the magnitude of this normal force; this is what you want to find. By Newton's second law,

F = 32 N - n = (1.0 kg) (8.0 m/s²)

n = 32 N - 8.0 N

n = 24 N

Just to make sure that this is consistent: by Newton's third law, the larger box feels the same force but pointing in the opposite direction. On the smaller box, n opposes the pushing force, so points backward. So from the larger box's perspective, n acts on it in the forward direction. This is the only force acting on the larger box, so Newton's second law gives

F = 24 N = (3.0 kg) (8.0 m/s²)

I need help with science o.o:

Question 1:

An egg yolk is unicellular.

True
False
___________________________

Question 2:

Your body has more than a billion cells.

True
False
_____________________


Question 3:


Almost all cells are not microscopic.

True
False

please dont steal my points :>

Answers

Answer:

TrueTrueFalse

Explanation:

this is the answer

Answer:

1.true

2true

3.false

Explanation:

espero que te ayude

An ant crawls in a straight line at a constant speed of 0.24 m/s for a distance of 3.0 m, beginning in the corner of a square classroom. It then turns exactly 90 degrees to the right, and proceeds an additional 4.0 m, reaching the far corner of the same wall from which it began. If the second leg of the journey was crawled in half the amount of time as the first, what was the ant's average speed for the whole trip?

Answers

Answer:

vavg = 0.37 m/s

Explanation:

The average speed is just the relationship between the total distance traveled, and the total time required for that travel , as follows:

       [tex]v_{avg} = \frac{\Delta x}{\Delta t} (1)[/tex]

We know that for the first leg of the journey, the ant crawls at a constant speed of 0.24 m/s, moving 3.0 m.We can find the time required for this part, just applying the definition of average velocity, and solving for the time t (which we will call t₁), as follows:

       [tex]t_{1} =\frac{x_{1}}{v_{1} } = \frac{3.0m}{0.24m/s} = 12.5 s (2)[/tex]

From the givens, we know that the time for the second part is exactly the half of the value found in (2), so we can write the total time Δt as follows:

       [tex]\Delta t = t_{1} + \frac{t_{1} }{2} = 12.5 s + 6.25 s = 18.75 s (3)[/tex]

We also know that in the second leg of the journey, the ant traveled 4.0 m, which adds to the 3.0 m of the first part, making a total distance of 7.0 m.Per definition of average speed, we can write the following expression as in (1) replacing Δx and Δt by their values, as follows:

       [tex]v_{avg} = \frac{\Delta x}{\Delta t} = \frac{7.0m}{18.75m} = 0.37 m/s (4)[/tex]

Consult Multiple-Concept Example 5 to review the concepts on which this problem depends. A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 6 m from the bulb, determine (a) the average intensity of the light, (b) the rms value of the electric field, and (c) the peak value of the electric field.

Answers

Answer:

a. 0.332 W/m² b. 11.2 V/m c. 15.8 V/m

Explanation:

(a) the average intensity of the light,

Intensity, I = P/A where P = average power = 150.0 W and A = area through which the power emits = 4πr² where r = distance from bulb = 6 m.

So, I = P/A = P/4πr²

Substituting the values of the variables into the equation, we have

I =  P/4πr²

I =  150.0 W/4π(6 m)²

I =  150.0 W/4π(36 m²)

I =  150.0 W/452.39 m²

I =  0.332 W/m²

(b) the rms value of the electric field,

Since Intensity, I = E²/cμ₀ where E = rms value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = permeability of free space = 4π × 10⁻⁷ H/m.

Making E subject of the formula, we have

E² = Icμ₀

E = √(Icμ₀)

Since I = 0.332 W/m², substituting the other terms into the equation, we have

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(12.5 × 10)

E = √125 V/m

E = 11.18 V/m

E ≅ 11.2 V/m

(c) the peak value of the electric field.

The peak value of electric field, E' is gotten from E = E'/√2 where E = rms value of electric field.

So, E' = √2E

= √2 × 11.2 V/m

= 15.81 V/m

≅ 15.8 V/m

An amateur blacksmith wants to cool off a 42kg glowing piece of iron, specific heat 470and decides to toss it into a 5.0 kg iron bucket with 10.0 kg of room temperature(23 C) water in it. To his surprise the water completely evaporates away (meaning once it vaporizes it is no longer part of the system) and after some time he goes to pick up the bucket but finds that the bucket is at 150 C.

Required:
What was the initial temperature of the glowing piece of iron?

Answers

Answer:

The right approach is "1479°C".

Explanation:

The given values are:

Mass of iron piece,

[tex]m_p=42 \ kg[/tex]

Mass of iron bucket,

[tex]m_I=5 \ kg[/tex]

Mass of water,

[tex]m_w=10 \ kg[/tex]

Iron's specific heat,

[tex]C_I=470 \ J/Kg^{\circ}C[/tex]

Water's specific heat,

[tex]C_w=4186 \ J/Kg^{\circ}C[/tex]

Initial temperature,

[tex]t_I=23^{\circ}C[/tex]

Final equilibrium temperature,

[tex]T=150^{\circ}C[/tex]

Latent heat,

[tex]L_v=2260\times 10^3 \ J/Kg[/tex]

As we know,

The heat lost by the glowing piece of iron will be equal to the heat gain by the iron bucket as well as water, then

⇒ [tex]m_IC_I \Delta T=m_wC_w(100-23)+m_wL_v+m_bC_I(150-23)[/tex]

On substituting the given values, we get

⇒ [tex]42\times 420\times \Delta T=10\times 4186(100-23)+10(2260\times 10^3)+5\times 420(150-23)[/tex]

⇒ [tex]17640 \Delta T=3.22\times 10^6+2.26\times 10^7+2.667\times 10^5[/tex]

⇒          [tex]\Delta T=\frac{2.60867\times 10^7}{17640}[/tex]

⇒          [tex]\Delta T=1479^{\circ}C[/tex]

The chemical energy in your food is
transformed into what type of energy in
your moving muscles?
A. Light
B. Mechanical
C. Sound
D. Electromagnetic

Answers

Answer:

Electromanetic

Explanation:ESPERO TE AYUDE

Which statement best compares gamma rays to the other components of the electromagnetic spectrum?

Am I correct ???

Answers

Answer:

no

Explanation:

gamma rays do have the most frequency, but they also have the highest amount of energy because they happen the most often.

The plates of a vacuum-gap parallel plate capacitor have a 100.0 mm2 area, a vacuum gap of 5.00 mm and are connected to a 1.5-volt battery. After the capacitor is charged, the battery is disconnected from the capacitor. After the battery is disconnected, the plates are pulled apart until the vacuum gap is 7.50 mm. a. What are the initial and final energies stored in the capacitor

Answers

Answer:

E₀ = 2.0*10⁻¹¹ J = 0.2 pJ

Ef = 3.0*10⁻¹¹ J = 0.3 pJ

Explanation:

The energy stored between the plates of a parallel plate capacitor can be expressed in terms of the capacitance C and the potential difference between plates V as follows:

      [tex]E = \frac{1}{2} * C * V^{2} (1)[/tex]

When the capacitor is fully charged, the potential difference between plates must be equal to the voltage of the battery, 1.5 V.In a parallel plate capacitor, the value of the capacitance is independent of the applied voltage, and depends only on geometric constants and the dielectric constant of the medium between plates, as follows:

       [tex]C = \frac{\epsilon_{o}*A}{d} (2)[/tex]

We can find the initial value of C replacing in (2) by the givens below:A = 100.0 mm2d= 5.00 mmε₀ = 8.85*10⁻¹² F/m[tex]C_{o} = \frac{\epsilon_{o}*A}{d} = \frac{(8.85*(10)^{-12} F/m)*(10^{-4} m2)}{5.0*(10)^{-3}m} = 1.77*10^{-13} F (3)[/tex]With this value of C₀, and the value of the initial potential difference between plates (1.5 V), we can find the initial charge on the capacitor, starting from the definition of capacitance:[tex]C =\frac{Q}{V} (4)[/tex]Solving for Q in (4):[tex]Q = C_{o}* V = 1.77*10^{-13} F * 1.5 V = 2.65*10^{-13} C (5)[/tex]Finally, we can find the initial energy stored in the capacitor, replacing (3) and V in (1):

       [tex]E_{o} = \frac{1}{2} * C_{o} * V_{o} ^{2} = \frac{1}{2} * 1.77*10^{-13}F*(1.5V)^{2} = 0.2 pJ (6)[/tex]

If we pull apart the plates until the vacuum gap is 7.50 mm, we will change the expression of C in (2), decreasing its value due to the expanded gap.Replacing in (2) the new value of the gap (7.50 mm), we can find the new value of C, as follows:[tex]C = \frac{\epsilon_{o}*A}{d} = \frac{(8.85*10^{-12}F/m)*10^{-4} m2}{7.5*10^{-3}m} = 1.18*10^{-13} F (7)[/tex] In order to find the final energy stored in the capacitor, we need also the value of the final potential difference between plates.Once disconnected from the battery, the charge on any of the plates must remain the same, due to the principle of conservation of the charge.So, since we have the value of Q from (5) and the new value of C from (7), we can find the new potential difference between plates as follows:[tex]V_{f} = \frac{Q}{C_{f}} = \frac{2.7*10^{-13}C}{1.18*10^{-13}F} = 2.25 V (8)[/tex]With the values of Vf and Cf, we can find the value of the final energy stored in the capacitor, replacing these values in (1):

       [tex]E_{f} = \frac{1}{2} * C_{f} * V_{f} ^{2} = \frac{1}{2} * 1.18*10^{-13}F*(2.25V)^{2} = 0.3 pJ (9)[/tex]

A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction acting on the box is a constant 20 N. How much work is done by the normal force pushing up on the box from the ground?
A. 10 J
B. 0 J
C. 100 J
D. 50 J

Answers

Answer: 0 j

Explanation:

It is given that,

Mass of the box, m = 10 kg

Force with which the box is pulled, F = 50 N

It is moved a distance of 4 m

Force of friction acting on the box, f = 20 N

We need to find the initial kinetic energy the box have. It is clear that the box is at rest initially. As there is no motion in the box at that time. The formula for the kinetic energy of the box is given by :

As v = 0

So, the initial kinetic energy of the box is 0. Hence, this is the required solution.  

The diagram below shows the beaks of five species of birds that developed over time from
one parent species. The five species of birds can be found living in the different areas. Which
of the following BEST explains why the beak shape of each species of bird developed
differently? * 5
(1 Point)
A. Birds in different areas were able to make different birdcalls and songs.
O
B. Birds in different areas were exposed to different food sources.
C. Birds in different areas had different materials to make nests.
o o
D. Birds in different areas had different predators to defend or hide from.

Answers

Answer:

Birds in different areas were exposed to different food sources.

Explanation:Im almost 100 percent this is it.

Answer:

the answer is B

Explanation:

i did test on preformance

Katelyn (55 kg) is practicing a drop jump in the biomechanics lab. She steps off a plyometrics box, lands on the force plate, and immediately jumps back up into the air. Right before she hits the force plate, her vertical velocity is 3.0 m/s downwards. After leaving the ground again, her vertical velocity is 3.5 m/s upwards. Katelyn was in contact with the ground for 0.4 seconds. (a) What was the impulse exerted on Katelyn when she was on the force plate

Answers

Answer:

J = 357.5 kg*m/s

Explanation:

The impulse exerted on Katelyn when she was on the force plate, is equal to the change in her momentum, according to Newton's 2nd Law.Assuming as the positive direction the upward direction (coincident with the positive y-axis) we can express the initial momentum as follows:

       [tex]p_{o} = m*v_{o} = 55 kg * (-3.0 m/s) (1)[/tex]

By the same token, the final momentum is as follows:

       [tex]p_{f} = m*v_{f} = 55 kg * (3.5 m/s) (2)[/tex]

As we have already said, the impulse J is just equal to the change in momentum, i.e., the difference between (2) and (1):

      [tex]J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)[/tex]

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