Vanessa made a list comparing the differences between the outer coverings in plants and animals, but she made a mistake. Read the table and then choose the statement that explains why the list is incorrect. Structure Description of Function Spines Protects the plant from predators and helps increase water loss Skin Protects the animal from changes in weather and its environment The skin is a reproductive organ. The skin offers little protection against the weather. The spines attract predators. The spines help decrease water loss.

Answers

Answer 1

The skin is NOT a reproductive organ and the spines attract predators are the incorrect statements.

Why the list is incorrect?

The list is incorrect because skin is not a reproductive organ and the spines do not attract predators towards each other. Skin protects the body from the outer environment whereas the spines repel the predators from the plant body.

So we can conclude that the skin is NOT a reproductive organ and the spines attract predators are the incorrect statements.

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Related Questions

A single frictionless roller-coaster car of mass m = 750 kg tops the first hill with speed v= 15 m/s at height h = 40 m as shown

Answers

The speed of cars at A and B will be 25 m/sec.

What is speed?

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.

Given data;

Mass of roller coaster,m =750 kg

Height at location 1,h₁ = 40 m

Height at location 2,h₂ = 302 m

The velocity of car A,[tex]\rm v_A[/tex]

The velocity of the car B, [tex]\rm v_B[/tex]

The velocity of the car A and B is the same is found as;

[tex]\rm v_A = v_B = v = \sqrt{v_0^2 + 2g(h_1-h_2} )\\\\ v = \sqrt{15^2 + 2 \times 10 (40-30)} \\\\ v = \sqrt{225+400} \\\\ v = 25 \ m/sec[/tex]

Hence the speed of cars at A and B will be 25 m/sec

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The water is flowing through the horizontal constricted pipe. The pressure at one end is 4500Pa, speed is 3m/s and area of cross-section is A, Calculate the speed and pressure at another end where the area of cross-section is A/3.​

Answers

The speed and pressure at another end will be 9 m/sec and -31500 pascals.

What is gauge pressure?

The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.

Given data in problem is;

For horizontal pipe, Z₁=Z₂

Pressure at end 1, P₁ = 4500Pa

Speed at end 1, V₁ = 3 m/sec

Speed at end 2, V₂ = ? m/sec

Area of cross-section at end 1,A₁ =A

Area of cross-section at end 1,A₂ = A/3

From the continuity equation;

A₁V₁=A₂V₂

A× 3  = (A/3)×V₂

V₂ = 9 m/sec

From Bernoulli's equation;

[tex]\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2 \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} = \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} = \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 = - \ 31,500 \ pa[/tex]

Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.

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An extended body (not shown in the figure) has its center of mass at the origin of the reference frame. In the case below give the direction for the torque τ with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r.

Options are:
-Y
x
Z
-Z

Answers

+ X axis.

The equation for torque is:
[tex]\tau = r \times F[/tex]

Torque is the CROSS-PRODUCT of 'r' and 'F'.

We can use the right-hand rule to determine the direction of torque.

Looking at the diagram, with an open hand, point your fingers towards the top of the picture (Along the +Z axis in the direction of the 'r' vector.)

Then, curl your fingers in the direction of the force vector, (along the -Y axis).

Your thumb will stick out in the direction of the torque, or the +X axis.

You can also just eliminate any options that involve axes that have already been used for 'r' and 'F'. In this case, there are components in the Z and Y axis, so this leaves the 'X' axis to be considered.

A 60kg man jumps off a 30 foot building. Assuming he falls from rest and there is negligible air resistance, how long will it take him to reach the ground?​

Answers

Answer:

1.37 seconds before.... splat !

Explanation:

y = y0 + v0 t + 1/2 a t^2      y0 = 30       y = 0

                    v0 = 0 ( from rest)

                            a = - 32.2 ft/s^2

0 = 30 - 1/2 * 32.2 * t^2

t=1.366 seconds                    Note that   mass   is irrelevant.

Hello!
I have an equation where I can't really find To (Initial temperature). The equation is from the formula Newton's law of cooling. To is the initial temperature of the object and ln is the natural log. The -0.00124 is the constant k multiplied by the time.
I just need to find the initial temp.

Answers

Answer:

[tex]T_0=80695.17162...[/tex]

Explanation:

Given equation:

[tex]\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124[/tex]

To solve the given equation:

[tex]\textsf{Apply log rules}: \quad e^{\ln (x)}=x[/tex]

[tex]\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}[/tex]

Multiply both sides by T₀:

[tex]\implies T_0-100=T_0e^{-0.00124}[/tex]

Add 100 to both sides:

[tex]\implies T_0=T_0e^{-0.00124}+100[/tex]

Subtract [tex]T_0e^{-0.00124}[/tex] from both sides:

[tex]\implies T_0-T_0e^{-0.00124}=100[/tex]

Factor out the common term T₀:

[tex]\implies T_0(1-e^{-0.00124})=100[/tex]

Divide both sides by [tex](1-e^{-0.00124})[/tex]

[tex]\implies T_0=\dfrac{100}{1-e^{-0.00124}}[/tex]

Carry out the calculation:

[tex]\implies T_0=\dfrac{100}{1-0.99876...}[/tex]

[tex]\implies T_0=\dfrac{100}{0.001239231...}[/tex]

[tex]\implies T_0=80695.17162...[/tex]

Please help me answer this

Answers

im j answer this so i can

The cannon on a battleship can fire a shell a maximum distance of 36.0 km.
(a) Calculate the initial velocity of the shell.
m/s

(b) What maximum height does it reach? (At its highest, the shell is above a substantial part of the atmosphere--but air resistance is not really negligible as assumed to make this problem easier.)
m

(c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface at the ship?
mDoes your answer imply that error introduced by the assumption of a flat earth in projectile motion is significant here? (Select all that apply.)
The error is insignificant compared to the distance of travel.
The error is significant compared to the distance of travel.
The error could be significant compared to the size of a target.
The error is insignificant compared to the size of a target.

Answers

(a)The initial velocity of the shell will be 594.27 m/sec

(b)The maximum height it reaches will be 9000 m.

c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.

d)The error could be significant compared to the size of a target. Option C is correct.

What is velocity?

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

m is the mass of the block = Kg.

u is the initial velocity of fall = m/sec

h is the distance of fall =  m

g is the acceleration of free fall = m/sec²

v is the hitting velocity of =?

a)

The range of the projectile is;

[tex]\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27 \ m/sec[/tex]

b)

The maximum height of the projectile is;

[tex]\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m[/tex]

c)

The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;

[tex]\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m[/tex]

d)

An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.

The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.

Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.

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A 21.0 kg uniform beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

If the angle between the beam and the cable is θ = 66.0° what is the tension in the cable?

What is the vertical component of the force exerted by the hi.nge on the beam?

Answers

The tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

Tension in the cable

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (21 x 9.8)/(2 x sin66)

T = 112.64 N

Vertical component of the force

T + F = W

F = W - T

F = (9.8 x 21) - 112.64

F = 93.16 N

Thus, the tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

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2 red and 2 blue overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running through it. 2 green balls sit on the white line, and a line leading a bracket around the balls is labeled A. A line leading to a bracket overlapping the white line is labeled B. Which is a characteristic of the part of the atom marked "B”? It is very dense. It has a large mass. It is negatively charged. It is where neutrons move.

Answers

A characteristic of the part of the atom marked "B” is that it has a large mass. Option B. This is further explained below.

What is atom?

Generally,  Atom, the lowest possible unit into which matter may be split without releasing electrically charged particles.

In conclusion, the protons and neutrons are represented by the red and white spheres, respectively. Since it indicates both protons and neutrons, the line preceding the bracket around the balls (designated A) stands in for the atom's nucleus.The mass of the "B" section of the atom is very high.

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Answer:

it's b

Explanation:

i did the ed quiz

Three containers are filled with water to the same level. Identical marbles are dropped into all the three containers. In which container will the water level be the highest?

a. P

b. Q

c. R

d. The level will be the same in all the three containers.​

Answers

Answer:

P

Explanation:

,............................................ I have. a doubt too on D

Answer: D
The answer is d


What is the length of the y-component of the vector plotted below?

Answers

The length of the y-component of the vector plotted below is 1 unit.

We know this because this vector can be descomposed in two vectors (blue) that form an angle of 90° beetween them. The measures of this vectors must be able to form a right-angle triangle with the descomposed vector (see image).

The mass of a star is 1.210×1031 kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter suddenly shrinks to 0.350 times its present size. Assume a uniform mass distribution before and after.

Answers

The new period will be 2.486 days.

What is the period?

The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.

Given data;

Mass of a star,m= 1.210×10³¹ kg

The time period for one rotation of the star, T = 20.30 days

D' = 0.350 D

R' = 0.350 R

From the law of conservation of angular momentum;

[tex]\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days[/tex]

Hence, the new period will be 2.486 days.

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power is measured in units of watts what are watts ?

Answers

Answer:

the watt is the unit of power or radiant flax

Select the correct answer. Which segment of this graph shows a decreasing velocity? Graph representing Time in seconds on the x-axis and position in miles on the y-axis. The line starts at point A (0, 20), and passes points B (20, 20), C (40, 40), D (60, 60), E (80, 60), and F (100, 80) A. A to B B. B to C C. C to D D. D to E E. E to F

Answers

The portion that shows a decreasing velocity is  D to E.

What is a position - time graph?

The position time graph is used to describe the motion of a body. We know that the slope of a velocity time graph is the velocity of the graph.

If we look at the graph as shown in the image attached to the answer, we can see that the portion that shows a decreasing velocity is  D to E.

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Answer:

D - E

Explanation:

hope this help

I need a super tragic past for my desired reality, the reality I'm going to is Naruto.

Thanks

Answers

To create a tragic past for your character, you would have to write a plot that shows a sad ending to a good story.

What is a Tragic Story?

This refers to a type of story that has negative emotions such as sadness, death, etc.

Hence, we can see that for example, you could write about two lovers and how despite all the struggles they faced, eventually died in mysterious circumstances.

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The mass of a star is 1.210×10[tex]^{31}[/tex] kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter suddenly shrinks to 0.350 times its present size. Assume a uniform mass distribution before and after.

Answers

Since the diameter of the star suddenly shrinks to 0.350 times its present size, the new period of the star is 4.2 days

To find the new period, we need to know Kepler's third law of planetary motion

What is Kepler's third law of planetary motion?

Kepler's third law of planetary motion states that the square of the period of orbit of a planet, T equals the cube of its radius, R.

So, T² ∝ R³

and T₂²/T₁² = R₂³/R₁³ where

T₁ = initial period of star = 20.30 days, T₂ = final period of star after shrinking, R₁ = initial radius of star and R₂ = final radius of star after shrinking

Given that the diameter suddenly shrinks to 0.350 times its present size and its radius is proportional to its diameter. So, its radius shrinks 0.350 times its initial size. Also, since there is a uniform mass distribution before and after, the mass remains the same.

So, R₂ = 0.350R₁

The new period of the star

Making T₂ subject of the formula, we have

T₂ = [√(R₂/R₁)³]T₁

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

T₂ = [√(R₂/R₁)³]T₁

T₂ = [√(0.350R₁/R₁)³]20.30 days

T₂ = [√(0.350)³]20.30 days

T₂ = [√0.042875]20.30 days

T₂ = [0.2071]20.30 days

T₂ = 4.2 days

So, the new period of the star is 4.2 days

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You add chocolate syrup to a glass of milk and stir it. The resulting chocolate milk would be classified as:

A. Homogeneous mixture
B. Element
C. Heterogeneous mixture
D. Compound

Answers

Answer is b I think

3. The following sound waves have what velocity? (convert cm into m in your calculations.)
a. 20.0 Hz, λ of 17.2 m.
b. 200.0 Hz, λ of 1.72 m.
c. 2,000.0 Hz, λ of 17.2 cm.
d. 20,000.0 Hz, λ of 1.72 cm.

Answers

The sound waves with the frequency 20 Hz, 200 Hz, 2,000 Hz, and 20,000 Hz have a velocity of 344 m/s.

What are sound waves?

A sound wave is a sample resulting from the motion of strength traveling thru a medium (inclusive of air, water, or every other liquid or stable matter) because it propagates from the supply of the sound. Sound waves are created and convey strain waves, for example, a cellphone.

The frequency can be given as:

Frequency = Velocity/ Wavelength

Velocity = Frequency * Wavelength

The velocity of the following sound waves is given as:

Frequency = 20 Hz

Wavelength = 17.2 m

Velocity = 20 * 17.2 m/s

Velocity = 344 m/s

Frequency = 200 Hz

Wavelength = 1.72 m

Velocity = 200 * 1.72 m

Velocity = 344 m/s

Frequency = 2,000 Hz

Wavelength = 17.2 cm = 0.172 m

Velocity = 2,000 * 0.172 m/s

Velocity = 344 m/s

Frequency = 20,000 Hz

Wavelength = 1.72 cm = 0.0172 m

Velocity = 20,000 * 0.0172 m/s

Velocity = 344 m/s

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Complete the sentence to describe how light behaves as it relates to its nature (waves or particles). light travels as and interacts with matter as .

Answers

Light behaves as it relates to its nature (waves or particles). light travels as and interacts with matter as wave particles.

What is light?

Electromagnetic radiation that is within the region of the electromagnetic spectrum that the human eye can see is referred to as light or visible light.

The main characteristics of light include intensity, polarization, frequency or wavelength spectrum, and propagation direction. One of the fundamental constants of nature is its speed in a vacuum, which is 299 792 458 meters per second (m/s).

Light can be categorized according to two theories: the first categorizes light as particles, and the second categorizes light as waves.

The main difference between the wave and particle natures of light is that the former asserts that light can act as an electromagnetic wave while the latter asserts that light is made up of tiny particles known as photons.

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Answer: 1.waves 2.particles

Explanation: just did it!<3

Please help me with this answer

Answers

The  position of the car at the given time is 18 m.

Position of the car at t = 6.0 s

The position of the car at the given time is calculated as follows;

x = vt

where;

v is the velocity of the car; from the graph v = 3 m/st is time of motion = 6.0 s

x = 3 m/s x 6.0 s

x = 18 m

Thus, the  position of the car at the given time is 18 m.

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how mesuremerment of gravity using simple pendulum

Answers

Answer:

See below

Explanation:

Set up your pendulum

  measure its length  and time the period ( you could time 100 of them and divide the time result by 100 to get the period, T)

    then use  

             T = 2 pi  sqrt (L/g)        T = period    L = length   g = gravity

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0 m. (It is not a very smart idea to play soccer on the roof of tall buildings.)
The ball is kicked with a speed of v0 = 15.10 m/s at an angle of θ = 74.1° with respect to the horizontal. The mass of a size 5 soccer ball is m = 450 g. When the ball lands on the other building, its speed is 19.89 m/s.
How much energy was lost to air friction? The ball is kicked without a spin.

Answers

The energy lost to the air due to friction is 37.7 J.

Energy lost to the air due to friction

The energy lost to the air due to friction is determined from the change in kinetic energy of the ball.

ΔK.E = K.Ef - K.Ei

ΔK.E = ¹/₂mv² - ¹/₂mu²

ΔK.E =  ¹/₂m(v² - u²)

ΔK.E = ¹/₂(0.45)(19.89² - 15.1²)

ΔK.E = 37.7 J

Thus, the energy lost to the air due to friction is 37.7 J.

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What is my speed for this?

Answers

D.

Reason: Speed = distance/time

An extended body (not shown in the figure) has its center of mass at the origin of the reference frame. In the case below give the direction for the torque τ with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r.

Options are:
X
-Y
Z
-Z

Answers

The direction of torque τ this method is mathematically given as

D=X  

Option A is correct.

What is the direction of torque?

Generally, the equation for torque is  mathematically given as

τ = r X F

Hence to decipher the torque direction with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r

We utilize our right hand.Place our right-hand fingers along the path of rPlace our right-hand palm on F Then slowly we sweep r into F.The path or direction of the thumb will provide the direction of the torque.

In conclusion, the direction of this method is

D=X   Option A.

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A roller coaster has a mass of 450 kg. It sits at the top of a hill with height 49
m. If it drops from this hill, how fast is it going when it reaches the bottom?

Answers

Answer:

Approximately [tex]31\; {\rm m\cdot s^{-1}}[/tex] assuming that there is no friction on the rollercoaster, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].

Explanation:

Make use of the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where:

[tex]v[/tex] is the final velocity of the moving object,[tex]u[/tex] is the initial velocity of the object,[tex]a[/tex] is the acceleration of the object, and[tex]x[/tex] is the displacement of the object.

If there is no friction on the rollercoaster, the acceleration of the rollercoaster would be equal in magnitude to the gravitational field strength, [tex]g[/tex]:

[tex]a = g = 9.81\; {\rm m \cdot s^{-2}}[/tex].

The initial velocity of this rollercoaster is [tex]0\; {\rm m\cdot s^{-1}}[/tex] (that is, [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) since the rollercoaster was initially stationary. The displacement of this rollercoaster would be [tex]x = 49\; {\rm m}[/tex] (same as the height of the hill.)

Rearrange the equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find an expression for the final velocity [tex]v[/tex] of this rollercoaster:

[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 9.81\; {\rm m\cdot s^{-2} \times 49\; {\rm m} + (0\; \rm m\cdot s^{-1}})^{2}}\\ &\approx 31\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

More than 700 people in the United States have been infected with monkeypox virus American vector: the epidemic is spreading like a snowball.

Answers

Answer: I don't understand if it is a question

Explanation:

More than 700 people in the United States have been infected with monkeypox virus American vector: the epidemic is spreading like a snowball.

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 96.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 27.0 kg.
(a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (Select all that apply.)
the wagon
the child in the wagon
the children outside the wagon


(b) Draw a free body diagram, including the weight and all other forces acting on the system. (Do this on paper. Your instructor may ask you to turn in this diagram.)
(c) Calculate the acceleration.
m/s2
(d) What would the acceleration be if friction is 21.0 N?

Answers

(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

Net force on the third child

Apply Newton's second law of motion;

∑F = ma

where;

∑F is net forcem is mass of the third childa is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

the wagon the children outside the wagon

Free body diagram

           →                 →              Ф                         ←

         1st child      friction       wagon                2nd child

Acceleration of the  child and wagon system

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

When the frictional force is 21 N

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

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Problem 3: A block of mass m= 15.5 kg rests on an inclined plane with a coefficient of
static friction of us = 0.11 between the block and the plane. The inclined plane is L = 6.8 m long and
it has a height of h = 3.05 m at its tallest point.
A. What angle in degrees does the plane make with respect to the horizontal? B. What is the magnitude of the normal force in newtons that acts on the block? C. What is the component of the force of gravity along the plane, fgx in newtons? D. Write an expression in terms of degrees, the mass, the coefficient of the static friction, and the gravitational constant, for the magnitude of the force due to static friction, just before the block begins to slide.

Answers

(a) The angle the plane makes with respect to the horizontal is 27⁰.

(b) The magnitude of the normal force acting on the block is 135.3 N.

(c) The component of the force of gravity along the plane, fgx is 68.96 N.

(d) The expression for the net force on the block is mgsinθ - μmgcosθ = 0.

Angle of the plane with respect to the horizontal

sin θ = h/L

where;

θ is the angle of inclinationh is height of the inclineL is length of the plane

sin θ = 3.05/6.8

sin θ = 0.4485

θ = arc tan(0.4485)

θ = 26.7⁰

θ ≈ 27⁰

Normal force on the block

Fₙ = mgcosθ

Fₙ = (15.5)(9.8)cos(27)

Fₙ = 135.3 N

Parallel component of the force on the plane

Fx = mgsinθ

Fx = (15.5)(9.8) sin(27)

Fx = 68.96 N

Net force on the block

∑F = 0

Fx - Ff = 0

mgsinθ - μmgcosθ = 0

Thus, we can conclude the following;

the angle the plane makes with respect to the horizontal is 27⁰, the magnitude of the normal force acting on the block is 135.3 N,the component of the force of gravity along the plane, fgx is 68.96 N, and the expression for the net force on the block is mgsinθ - μmgcosθ = 0.

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You cover the following displacements every day going to school: d1=50 m, E and d2=95 m, N. You do this for 12 minutes. a) What is your speed in m/s? b) What is your velocity in m/s?

Answers

The speed will be 0.2 m/s and the velocity will be 0 m/s.

Speed = Total Distance / Total time

We have given total distance as ( 50 + 95 ) metres and total time as 12 minutes or we can say 720 seconds.

Speed = 145/ 720 m/s

Speed = 0.2 m/s

Velocity = Total Displacement / Total time

As the initial and final is the home, hence the net displacement is 0 in that case.

In this case also the total time we have given is 12 minutes or we can say 720 seconds.

Velocity = 0 / 720 m/s

Velocity = 0 m/s

So to conclude with we can say that the speed is 0.2 m/s and the velocity is 0 m/s.

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The farther a star is away from Earth, the more it is

Answers

Answer:

black hole     -   one of the most dense objects in the universe

red giant       -    a very large star

white dwarf   -   an old, very dense hot star that is cooling

nebula           -   a mass of gas and dust

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