Which of the following organisms has an adaptation that will allow it to survive in tundra biome? *
A.)Plants with roots that are short and grows sideways with hairy stems and small leaves.
B.)Plants that have broad leaves to capture sunlight and long roots to penetrate the soil.
C.)Animals with thin fur that allows them to get rid of heat efficiently.
D.)Animals with long tongues for capturing prey and sticky pads for climbing trees.

Answer:

40 m/s^2

Explanation:

Mass= 0.40 kg

Force= 16 N

Therefore the acceleration can be calculated as follows

F = ma

16= 0.40 × a

16= 0.40 a

a= 16/0.40

a= 40 m/s^2

Hence the acceleration is 40 m/s^2

Answer:

transition metal, and inner transition metals groups are numbered 1-18 from left to right

Answer:

478.75 J

Explanation:

W=force* displacement

constant speed= (a=0) net F=0

Horizontal component of tension

Tcosx

125Ncos40= 95.76 N

W= (95.76 N)(5 m)

=478.75 J

The work done in moving the crate across the given distance is 478.75 J.

The given parameters;

The work done in moving the crate is the product of the horizontal component of the tension and the distance through which the crate is moved.

The work-done in moving the crate is calculated as;

W = Tcos(Ф) x d

W = 125cos(40) x 5

W = 478.75 J.

Thus, the work done in moving the crate across the given distance is 478.75 J.

Learn more here: https://brainly.com/question/19498865

Answer:

Making an observation

Explanation:

The use of senses as a tool to gather information is described as making an observation.

While making an observation, the senses must be at alert.

Therefore, the act of using senses or tools to gather information is called making an observation.

Answer:

When the mouse broke free from the hawk, its vertical velocity is zero since the mouse just fell from the grips of the hawk

We are given:

initial vertical velocity (v) = 0 m/s

initial horizontal velocity (u) = 10.1 m/s

height from the ground (h) = 15 m

acceleration due to gravity (a) = 10 m/s/s

final vertical velocity (vf) = v m/s

Solving for Final vertical velocity:

from the third equation of motion

(vf)² - (v)² = 2ah

replacing the variables

v² - 0² = 2(10)(15)

v² = 300

v = √300

v = 10√3     OR    17.3 m/s

Answer:

C

Explanation:

Static electricity is what makes your hair stand up when you rub a balloon against it or gives you a shock from your doorknob. In static electricity, electrons are moved around mechanically (i.e. by someone rubbing two things together).

HOPE THIS HELPED AND LETTER D MADE ME LAUGH

Answer:

Negative

Explanation:

duh

Answer:

:)

Explanation:

negative.

go add the snap carmel.bratz

Answer:

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} \\[/tex]

From the question

mass = 100 g

volume = 50 mL

We have

[tex] density = \frac{100}{50} = \frac{10}{5} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

Explanation: someone help me please you want free points right

Answer:

A south pole attracts a north pole :)

Explanation:

Look up "Everything You Need To Know About Math In One Big Fat Notebook pdf." It's the best thing I've ever been given, I have it with me in math class all the time and I've aced every test. I have it with me right now and it has everything I've ever been taught about math in it so it might help you.

Answer:

The final speed of the second package is twice as much as the final speed of the first package.

Explanation:

Free Fall Motion

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

[tex]v=gt[/tex]

And the distance traveled downwards is:

[tex]\displaystyle y=\frac{gt^2}{2}[/tex]

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]

Replacing into the first equation:

[tex]\displaystyle v=g\sqrt{\frac{2y}{g}}[/tex]

Rationalizing:

[tex]\displaystyle v=\sqrt{2gy}[/tex]

Let's call v1 the final speed of the package dropped from a height H. Thus:

[tex]\displaystyle v_1=\sqrt{2gH}[/tex]

Let v2 be the final speed of the package dropped from a height 4H. Thus:

[tex]\displaystyle v_2=\sqrt{2g(4H)}[/tex]

Taking out the square root of 4:

[tex]\displaystyle v_2=2\sqrt{2gH}[/tex]

Dividing v2/v1 we can compare the final speeds:

[tex]\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}[/tex]

Simplifying:

[tex]\displaystyle v_2/v_1=2[/tex]

The final speed of the second package is twice as much as the final speed of the first package.

Answer:

D. The liquid is a mixture because it has a gas dissolved in a liquid.

Answer:

A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) µ_s = tan θ

D) µ_s = 0.4663

Explanation:

A) The forces acting on the car will be;

Force due to friction; F_f

Force due to Gravity; F_g

Normal Force; F_n

Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.

Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n

Thus, sum of forces about the vertical j^ direction gives;

ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0

Since F_f = µ_s × F_n ;

F_n•cos θ − mg + (µ_s × F_n × sin θ) =0

F_n = mg/[cos θ + (µ_s•sin θ)]

Also, sum of forces about the centre i^ direction gives;

ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r

Plugging in formula for F_n gives;

ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r

Making v the subject gives;

v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B) What we got in a above is the minimum speed the car can have while going round the turn.

The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.

Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;

v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

Thus the range is;

√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;

ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0

Thus;

mg(sin θ - µ_s•cos θ) = 0

Making µ_s the subject gives;

µ_s = sin θ/cos θ

µ_s = tan θ

D) If θ = 25.0°;

Thus;

µ_s = tan 25

µ_s = 0.4663

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   [tex]COP  =  \frac{T_i}{T_o - T_i}[/tex]

Here [tex]T_i[/tex] is the inside temperature

while  [tex]T_o[/tex] is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   [tex] \frac{T_i}{T_o - T_i}[/tex] heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         [tex]Q \ \alpha \ (T_o - T_i)[/tex]

=>        [tex]Q= k (T_o - T_i)[/tex]

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   [tex] \frac{T_i}{T_o - T_i}[/tex]  amount of heat

   E  unit of electricity will remove  [tex]Q= k (T_o - T_i)[/tex]

So

      [tex]E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }[/tex]

=>   [tex]E = \frac{k}{T_i} (T_o - T_i)^2[/tex]

given that  [tex]\frac{k}{T_i}[/tex] is constant

    =>  [tex]E \  \alpha  \  (T_o - T_i)^2[/tex]

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  [tex]T_i   =  30 ^oC[/tex]

 and      [tex]T_o  =  40 ^oC[/tex]

Hence  

     [tex]E = K (T_o - T_i)^2[/tex]

Here K stand for a constant

So  

        [tex]E = K (40 -  30)^2[/tex]

=>      [tex]E = 100K [/tex]

Now if  the  [tex]T_i   =  20 ^oC[/tex]

Then

       [tex]E = K (40 -  20)^2[/tex]

=>      [tex]E = 400 \ K[/tex]

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       [tex]Q = k (T_o - T_i )^{\frac{1}{2} }[/tex]

So

       [tex]E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }[/tex]

=>   [tex]E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }[/tex]

Assuming [tex]\frac{k}{T_i}[/tex] is a constant

Then  

     [tex]E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }[/tex]

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

Answer:

 N = 136.77 N

Explanation:

This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.

Let's use trigonometry to decompose the force F1

           cos θ = F₁ₓ / F₁

           sin θ = F_{1y} / F₁

           F₁ₓ = F₁ cos θ

           F_{1y} = F₁ sin θ

now let's apply Newton's second law to each axis

X axis

        F₁ₓ - F4 = m a

Y axis

        N + F3 + F_{1y} -F₂ -W = 0

the acceleration can be calculated with kinematics

        v = v₀ + a t

since the object starts from rest, the initial velocity is zero v₀ = 0

        a = v / t

        a = 20/3

        a = 6.667 m / s²

we substitute in the equation

        F₁ₓ = F₄ + m a

        F₁ₓ = 42 + 15  6,667

        F₁ₓ = 142 N

        F₁ cos θ = 142

        cos θ = 142/206 = 0.6893

        θ = cos⁻¹ 0.6893

        θ = 46.42º

now let's work the y axis

        N = W + F₂ - F₃ - F_{1y}

        N  = 15 9.8 + 144 -5 - 206 sin 46.42

        N = 286 - 149.23

        N = 136.77 N

Answer:

The tension in the web is 0.017738 N

Explanation:

Net Force

The net force exerted on an object is the sum of the vectors of each individual force applied to an object.

If the net force equals 0, then the object is at rest or moving at a constant speed.

The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.

A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:

T=W=m.g

The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:

[tex]T = 0.00181\ Kg\cdot 9.8\ m/s^2[/tex]

[tex]T=0.017738\ N[/tex]

The tension in the web is 0.017738 N

Answer:

1- the rate of flow of charge through a conductor is called current .whereas ,current density is the current per unit area of conductor

Answer:

It would take 5 seconds

Explanation:

I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.

V = d / t

34 = 170 / t

34 x t = 170

34t = 170

t = 170 / 34

t = 5

Answer:

Vy = 80.5 [m/s]

Explanation:

In order to solve this problem we must use the Pythagorean theorem.

V = 90 [m/s]

The components are Vx and Vy:

Therefore:

[tex]v=\sqrt{v_{x}^{2} + v_{y}^{2} }[/tex]

where:

Vy = 2*Vx ; because one is twice of the other.

[tex]90 = \sqrt{v_{x}^{2} +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s][/tex]

and the bigger vector is:

Vy = 40.25*2

Vy = 80.5 [m/s]

Answer:

Acceleration = -12/17 m/s ^2

Explanation:

Name the variables

V = final velocity( velocity after the train stops) u= intial velocity(velocity when the train was moving at constant speed, before brakes are applied)

A = acceleration( in this case, it is deceleration)

T = time ( time taken to fully stop)

Second Step: What equation should we use?

The 3 Big Motion in One Dimension Formulas are:

V = u+at

v^2 = u^2 + 2as

S = ut+ (1/2) at^2

For this question, we will be using the first one.

Third Step: Solve

V = 0 (the final velocity is 0, because the train stopped)

U = 60 (the train was going at 60 mps)

T = 85 seconds( 1 min and 25 sec)

0 = 60+ 85a

Answer:

We are given:

mass of the truck (m) = 3000 kg

velocity of the truck (v) = 25 m/s

Calculating the Momentum:

We know that the momentum of an object is equal to its mass times it's velocity

P = mv           (where P is the momentum of the truck)

Momentum of the truck:

replacing with the values of the truck

P = mv

P = 3000 * 25

P = 75000 kg m/s

Therefore, the momentum of the truck is 75000 kg m/s

Answer:Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.

Explanation:

Answer: its C

Explanation:

Answer:

5m

Explanation:

Answer:

5

Explanation:

because that's the average of how far it when the most

Explanation:

When the skater is dropped onto the ramp from above, the potential energy decreases and the kinetic energy increases.

Every time the skater bounces from the impact, thermal energy is gained, and both potential and kinetic energy are lost.

Answer:

The time for the change in the angular velocity to occur is 14.08 secs

Explanation:

From the question,

the angular acceleration is - 4.46 rad/s²

Angular acceleration is given by the formula below

[tex]\alpha =\frac{\omega -\omega _{o} }{t - t_{o} }[/tex]

Where [tex]\alpha[/tex] is the angular acceleration

[tex]\omega[/tex] is the final angular velocity

[tex]\omega _{o}[/tex] is the initial angular velocity

[tex]t[/tex] is the final time

[tex]t_{o}[/tex] is the initial time

From the question

[tex]\alpha[/tex] = - 4.46 rad/s²

[tex]\omega _{o}[/tex] = 0 rad/s (starting from rest)

[tex]\omega[/tex] = -31.4 rad/s

[tex]t_{o}[/tex] = 0 s

Now, we will determine t

From [tex]\alpha =\frac{\omega -\omega _{o} }{t - t_{o} }[/tex], then

[tex]-4.46 = \frac{-31.4 - 0}{t - 0}[/tex]

[tex]-4.46 = \frac{-31.4}{t}[/tex]

[tex]t = \frac{-31.4}{-4.46}[/tex]

t = 7.04 secs

This is the time spent in one direction,

Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t

Hence,

The time is 2×7.04 secs = 14.08 secs

This is the time for the change in the angular velocity to occur.

The question is missing parts. The complete question is here.

The human circulatory system is closed, that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuoous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart's four chambers comes briefly to rest before it is ejected by cotraction of the heart muscle. If the aorta (diameter [tex]d_{a}[/tex]) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches?

(a) [tex]\sqrt{d_{a}}[/tex]

(b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]

(c) [tex]2d_{a}[/tex]

(d) [tex]\frac{d_{a}}{2}[/tex]

Answer: (b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]

Explanation: The cross-sectional area of a vessel is a circle. So area is:

[tex]A=\pi.r^{2}[/tex]

Radius is half of a diameter, i.e.:

[tex]r=\frac{d}{2}[/tex]

Suppose [tex]d_{b}[/tex] is diameter of the branches, radius of one the branches is:

[tex]r_{b}=\frac{d_{b}}{2}[/tex]

Both branches are equal sized, which means, both have the same radius. So, combined area of both branches is:

[tex]A_{b}=\pi\frac{d_{b}^{2}}{4} +\pi\frac{d_{b}^{2}}{4}[/tex]

[tex]A_{b}=2\frac{d_{b}^{2}}{4}[/tex]

Area of aorta is

[tex]A_{a}=\pi.(\frac{d_{a}}{2})^{2}[/tex]

[tex]A_{a}=\pi.\frac{d_{a}^{2}}{4}[/tex]

Area of aorta is equal the combined area of the branches, then:

[tex]2\pi\frac{d_{b}^{2}}{4}= \pi\frac{d_{a}^{2}}{4}[/tex]

Rearraging:

[tex]d_{b}^{2}=\frac{d_{a}^{2}}{2}[/tex]

[tex]d_{b}=\frac{d_{a}}{\sqrt{2} }[/tex]

The diameter of one of the branches is [tex]\frac{d_{a}}{\sqrt{2}}[/tex].

Answer:

A soil

Explanation:

Sand is a granular material composed of finely divided rock and mineral particles. It is defined by size, being finer than gravel and coarser than silt. Sand can also refer to a textural class of soil or soil type; i.e., a soil containing more than 85 percent sand-sized particles by mass.

Answer:

mixture

sand is a mixture

Answer:

The thermal conductivity  [tex]k  = 1.4094   W/ m\cdot K[/tex]

Explanation:

From the question we are told that

  The  depth of the thermocouple from the surface is  x =  10 mm  = 0.01 m

   The  temperature is  [tex]T_f  =  100 ^o C[/tex]

   The  initial temperature is   [tex]T_i  =  30 ^o C[/tex]

    The  temperature of the thermocouple after t =  2 minutes( 2 * 60 =  120 \ seconds)  is   [tex]T_t  =  65 ^o C[/tex]

    The  density of the material  is  [tex]\rho =  2200 kg/m^3[/tex]

     The specific heat of the solid [tex]c_s  =  700 J/kg \cdot K[/tex]

Generally  the  equation for  semi -infinite medium  is mathematically as  

    [tex]\frac{T_s - T }{T_i - T} =  erf [\frac{x}{2 \sqrt{\alpha  * t} } ][/tex]

     [tex]\frac{65 - 100 }{30 - 100} =  erf [\frac{x}{2 \sqrt{\alpha  * t} } ][/tex]

        [tex]0.5 =  erf [\frac{0.01}{2 \sqrt{\alpha  * 120} } ][/tex]

Here [tex]\alpha[/tex] is a constant with unit [tex]m^2 /s[/tex]

  [tex]\frac{0.01}{ 2 (\sqrt{\alpha  *  120 } )}[/tex]   this is from the Gaussian function table  

    [tex]0.0 1 =  0.954 * (\sqrt{\alpha * 120  } )[/tex]

=>   [tex]\sqrt{\alpha  * 120  } =  \frac{0.01 }{0.954 }[/tex]

=>   [tex]\alpha =  9.1525 *10^{-7} \  m^2 /s[/tex]

Generally the thermal  conductivity is mathematically represented as

     [tex]k  =  \alpha  *  \rho * c_s[/tex]

      [tex]k  = 9.1525 *10^{-7}   *  2200 * 700[/tex]

     [tex]k  = 1.4094   W/ m\cdot K[/tex]