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3 years ago

Which of the following is closer to the glomerulus?

Physics

2 answers:

VLD [36.1K]3 years ago

8 0

A - proximal convoluted tubule

Leviafan [203]3 years ago

7 0

Answer:

proximal convoluted tubule

Explanation:

this is one of the closest parts to the glomerulus/kidney as it serves a major role to the urinary system, it wraps around the glomerulus.

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A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum "is 2.51 s". The temp

Artyom0805 [142]

Answer:

0.0034 sec

Explanation:

L = initial length

T = initial time period = 2.51 s

Time period is given as

$T = 2\pi \sqrt{\frac{L}{g}}$

$2.51 = 2\pi \sqrt{\frac{L}{9.8}}$

L = 1.56392 m

L' = new length

ΔT = Rise in temperature = 142 °C

α = coefficient of linear expansion = 19 x 10⁻⁶ °C

New length due to rise of temperature is given as

L' = L + LαΔT

L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)

L' = 1.56814 m

T' = New time period

New time period is given as

$T' = 2\pi \sqrt{\frac{L'}{g}}$

$T' = 2\pi \sqrt{\frac{1.56814}{9.8}}$

T' = 2.5134 sec

Change in time period is given as

ΔT = T' - T

ΔT = 2.5134 - 2.51

ΔT = 0.0034 sec

5 0

3 years ago

A meteoroid is traveling east through the atmosphere at 18. 3 km/s while descending at a rate of 11.5 km/s. What is its speed, i

Annette [7]

Answer:

The speed of meteoroid is 21.61 km/s in south-east.

Explanation:

Given that,

A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.

We need to draw a diagram

Using Pythagorean theorem

$AC^2=AB^2+BC^2$

$AC^2=(18.3)^3+(11.5)^2$

$AC=\sqrt{(18.3)^2+(11.5)^2}$

$AC=21.61\ km/s$

Hence, The speed of meteoroid is 21.61 km/s in south-east.

6 0

3 years ago

Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi

Oksanka [162]

<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>

6 0

3 years ago

For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution

Ann [662]

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

$v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}$

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

$a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}$

3 0

2 years ago

Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone

anygoal [31]

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

7 0

3 years ago