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3 years ago

Which of the following relationships is symbiotic?

1 answer:

0 0

There are 3 types of symbiotic relationships which are mutualism, commensalism and parasitism

mutualism is when both of the organisms (partners) have benefits so answer for that one is B

commensalism is when one of the partners or species has benefits while the other partner doesn’t get benefits and isn’t harmed in the process. answer for that one is E

parasitism is when one organism (known as parasites) has benefits while the other one suffers (known as the host) answer so that one is C

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What is the difference between radial acceleration and tangential acceleration and how do you calculate both of these accelerati

sergey [27]

Answer:

Tangential acceleration is in the direction of velocity - along the circumference of a circle if the object is undergoing circular motion

a = (V2 - V1) / T

Radial acceleration is perpendicular to the direction of motion if the object is not moving in a straight line (perhaps along the circumference of a circle)

a = m V^2 / R = m ω^2 R where R is the radius vector of the velocity - note that the Radius vector is directed from the center of motion to the object and for circular motion would be constant in magnitude but not in direction

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2 years ago

Whats mechanical energy mean

serious [3.7K]

Mechanical energy is the sum of kinetic and potential energy in an object that is used to do work. In other words, it is energy in an object due to its motion or position, or both.

Hope this helped XD

5 0

3 years ago

Read 2 more answers

A) In the figure below, a cylinder is compressed by means of a wedge against an elastic constant spring = 12 /. If = 500 , deter

Radda [10]

Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline. Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left. Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

6 0

3 years ago

5. If you walk 70 meters north in 5 minutes and 20 meters south in 15 seconds. What was the average

Tpy6a [65]

Answer:

kjfvkuzfvnv

Explanation:

3 0

3 years ago

A particle leaves the origin with a speed of 2.1 times 106 m/s at 30 degrees to the positive x axis. It moves in a uniform elect

Verizon [17]

Answer:

-1449.69404 N/C

Explanation:

u = Velocity of particle = $2.1\times 10^6\ m/s$

$\theta$ = Angle = 30°

x = Distance = 1.5 cm

m = Mass of electron = $9.11\times 10^{-31}\ kg$

q = Charge of electron = $-1.6\times 10^{-19}\ C$

In the case of projectile motion

$x=utcosA\\\Rightarrow t=\dfrac{x}{ucosA}$

The force of on the particle will balance the Electric force

$ma=qE\\\Rightarrow a=\dfrac{qE}{m}$

Now

$y=utsin\theta-\dfrac{1}{2}at^2\\\Rightarrow y=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2$

If y = 0

$0=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow utsin\theta=\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow t=\dfrac{2musin\theta}{qE}$

$\dfrac{x}{ucosA}=\dfrac{2musin\theta}{qE}\\\Rightarrow E=\dfrac{2mu^2sin\theta cos\theta}{xq}\\\Rightarrow E=\dfrac{2\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\times sin30\times cos30}{1.5\times 10^{-2}\times (-1.6\times 10^{-19})}\\\Rightarrow E=-1449.69404\ N/C$

The electric field is -1449.69404 N/C

8 0

3 years ago