answer to group of organisms that share similar characteristics and can reproduce among themselves producing fertile offspring

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Answer 1

Answer: It is species. that the answer buddy

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3. A block of mass m1=1.5 kg on an inclined plane of an angle of 12° is connected by a cord over a mass-less, frictionless pulley to a second block of a mass m2= 1.2 kg hanging vertically. If the hanging block falls 0.92min 1.23 s, what is the coefficient of friction between m1 and the inclined plane?

Please show work

Answers

Answer:

$\mu=0.377$

Explanation:

we need to start by drawing the free body diagram for each of the masses in the system. Please see attached image for reference.

We have identified in green the forces on the blocks due to acceleration of gravity ( $w_1$ and $w_2$ ) which equal the product of the block's mass times "g".

On the second block ( $m_2$ ), there are just two forces acting: the block's weight ( $m_2\,*\,g$ ) and the tension (T) of the string. We know that this block is being accelerated since it has fallen 0.92 m in 1.23 seconds. We can find its acceleration with this information, and then use it to find the value of the string's tension (T). We would need both these values to set the systems of equations for block 1 in order to find the requested coefficient of friction.

To find the acceleration of block 2 (which by the way is the same acceleration that block 1 has since the string doesn't stretch) we use kinematics of an accelerated object, making use of the info on distance it fell (0.92 m) in the given time (1.23 s):

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2$ and assume there was no initial velocity imparted to the block:

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2\n-0.92\,m=0\,-(1)/(2) a\,(1.23)^2\na=(0.92\,*\,2)/(1.23^2) \na=1.216 \,(m)/(s^2)$

Now we use Newton's second law in block 2, stating that the net force in the block equals the block's mass times its acceleration:

$F_(net)=m_2\,a\nw_2-T=m_2\,a\nm_2\,g-T=m_2\,a\nm_2\,g-m_2\,a=T\nm_2\,(g-a)=T\n1.2\,(9.8-1.216)\,N=T\nT=10.3008\,N$

We can round this tension (T) value to 10.3 N to make our calculations easier.

Now, with the info obtained with block 2 (a - 1.216 $(m)/(s^2)$ , and T = 10.3 N), we can set Newton's second law equations for block 1.

To make our study easier, we study forces in a coordinate system with the x-axis parallel to the inclined plane, and the y-axis perpendicular to it. This way, the motion in the y axis is driven by the y-component of mass' 1 weight (weight1 times cos(12) -represented with a thin grey trace in the image) and the normal force (n picture in blue in the image) exerted by the plane on the block. We know there is no acceleration or movement of the block in this direction (the normal and the x-component of the weight cancel each other out), so we can determine the value of the normal force (n):

$n-m_1\,g\,cos(12^o)=0\nn=m_1\,g\,cos(12^o)\nn=1.5\,*\,9.8\,cos(12^o)\nn=14.38\,N$

Now we can set the more complex Newton's second law for the net force acting on the x-axis for this block. Pointing towards the pulley (direction of the resultant acceleration $a$ ), we have the string's tension (T). Pointing in the opposite direction we have two forces: the force of friction (f ) with the plane, and the x-axis component of the block's weight (weight1 times sin(12)):

$F_(net)=m_1\,a\nT-f-w_1\,sin(12)=m_1\,a\nT-w_1\,sin(12)-m_1\,a=f\nf=[10.3-1.5\,*\,9.8\,sin(12)-1.5\,*1.216]\,N\nf=5.42\,N$

And now, we recall that the force of friction equals the product of the coefficient of friction (our unknown $\mu$ ) times the magnitude of the normal force (14.38 N):

$f=\mu\,n\n5.42\,N=\mu\,*\,14.38\,N\n\mu=(5.42)/(14.38)\n\mu=0.377$

with no units.

How do contact forces such as friction differ from gravitational and magnetic forces?

Answers

"Contact" means "touching". A contact force can't act on an object unless
they're touching. Friction is a contact force.

Gravitational forces pull masses together even if they're not touching.
Magnetic forces pull a piece of iron toward a magnet even if they're not touching.

Gravitational and magnetic forces are "non-contact" forces.

a wave in the ocean has an amplitude of 3m. Winds pick up suddenly, increasing the wave's amplitude to 6m. How does this change in amplitude affect the wave's frequency and the amount of energy it transports

Answers

Answer

Hi,

An increase in amplitude from 3m to 6 m increases the energy it transports. The frequency of the wave is not affected

Explanation

Amplitude is the height of a wave where as frequency is the number of waves that pass by each second. A wave with bigger amplitude has more energy than a wave with smaller amplitude. A point where more waves pass contains more energy that is transferred every second. The change in the amplitude of a wave does not change its frequency. However, frequency is inversely related to the wavelength of a wave.

Best Wishes!

a 63 kg object needs to be lifted 7 meters in a matter of 5 seconds. approximately how much horsepower is required to achieve this task

Answers

Power is defined as the rate of doing work or the work per unit of time. The first step to solve this problem is by calculating the work which can be determined by the equation:

W = Fd

where:

F = force exerted = ma
d = distance traveled
m = mass of object
a = acceleration

Acceleration is equivalent to the gravitational constant (9.81 m/s^2) if the force exerted has a vertical direction such as lifting.

W = Fd = mad = 63(9.81)(7) = 4326.21 Joules

Now that we have work, we can calculate power.

P = W/t = 4325.21 J / 5 seconds = 865.242 J/s or watts

Convert watts to horsepower (1 hp = 745.7 watts)

P = 865.242 watts (1hp/745.7 watts) = 1.16 hp

work = mgh

power = mgh/t = 63•9.8•7/5 = 864 watts

1 HP = 746 watts

so that is 834/746 = 1.2 HP

Help me what times would it e and where do i put a x to show where London will be in 12 hours time

Answers

Look at the picture I attached. It's a part of your sheet,
with some added information.

-- The line between the day-side and night-side of the Earth is the line
where day changes to night, or night changes to day. Even if we don't know
which way the Earth is turning in the picture, London is halfway between
sunrise and sunset, so it's about Midnight there.

-- Do you remember that the Earth spins all the way around in 24 hours,
and half-way around in 12 hours ?
So in 12 hours from now, London will be halfway around, on the day side,
and it will be about Noon there,

When you see where I placed London at Noon on my drawing, you may
wonder why it's so 'high up' on that side.

Remember that the Earth is not spinning arounf the sunrise/sunset line.
It's spinning around the 'axis' ... the line through the north and south poles.

This is a picture of the Earth during the dead of Winter in the northern hemisphere.
The north pole is tipped away from the sun. You can see that London spends
much more than half of each spin in the dark, and much less than half of it in
the light. This corresponds to the short days and long nights of Winter.

Which of the following is not a way that oil production could potentially affect aquatic viability?

Answers

Since you have not presented any choices that would tell us which is NOT a way that oil production could potentially affect aquatic viability, I'll just go ahead and answer what are ways that oil could affect aquatic viability.

-Offshore drilling contaminates and disturbs habitats of aquatic creatures
-Transport of oil across seas or oceans might lead to oil spill
-Waste products from energy factories using oil still would disturb habitat

answer: c- interruption of water flow

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