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Lelu [443]
3 years ago
10

19) Which two structures are not found in animal cells? A) vacuoles and ribosomes B) chloroplasts and ribosomes C) cell walls an

d chloroplasts D) endoplasmic reticulum and cell walls
Physics
2 answers:
LUCKY_DIMON [66]3 years ago
7 0

The answer to your question is: C) Cell walls and chloroplasts


(I hope this helps!)


Molodets [167]3 years ago
5 0

Cell walls and Chloroplasts

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What is the mechanical advantage of this system?<br> А<br> 2<br> B<br> 3<br> C<br> 4
faltersainse [42]
I think the answer will be B tell me if it’s right after
4 0
3 years ago
approximately what force,FM must the extensor muscle in the upper arm exert on the lower arm to hold a 7.6kg shot put? assume th
MAXImum [283]
Lets do the sum of the forces about the elbow joint.

Fm = Force of Muscle; Fe = Force Elbow; Fb = Force Ball

Sum Force about Joint = (-2.5)Fm + 12.5Fe + 30Fb = 0

(-2.5)Fm + 12.5(2.8) + 30(6.9) = 0

Fm = 96.8kg

Fm = 96.8 * 9.8 = 948.6N

Do you understand why the -2.5 is negative?
<span> Because I put the origin at the joint. So when you go left it is negative and when you go right it is positive. </span>
7 0
3 years ago
Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the p
Dominik [7]

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = \frac{2\times r^2(d-\rho)}{9\times n}

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = [tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}[/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= \frac{100}{387\times10^{-5}}

= 2.6 x 10⁴ s.

5 0
3 years ago
When does a falling object reach terminal velocity?
Ratling [72]

At some speed, the drag or force of resistance will equal the gravitational pull on the object. At this point the object ceases to accelerate and continues falling at a constant speed called the terminal velocity (also called settling velocity).

5 0
3 years ago
The components of a 15 meters per second velocity at an angle of 60 degrees above the horizontal are?
mafiozo [28]

Answer:

x-component of velocity: 7.5 m/s

y-component of velocity: 13 m/s

Explanation:

This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.

The procedure to find the x-velocity is very similar, but instead of using sine, we will use the cosine of theta. Recall that cosΘ=adjacent/hypotenuse. Once again plugging this scenario's numbers into that, you end up with cos60 = vₓ/15. Multiplying this out gives you vₓ = 15cos60. Once again, plug this into your calculator. 7.5 m/s should be your answer. This is the velocity in the x-direction.

By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.

8 0
2 years ago
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