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2 years ago

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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

2 answers:

LUCKY_DIMON [66]2 years ago

7 0

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

(I hope this helps!)

Molodets [167]2 years ago

5 0

Cell walls and Chloroplasts

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Water flows from a large drainage pipe at a rate of 950 gal/min. What is this volume rate of flow in (a) m3/s , (b) liters/min,

Paladinen [302]

Answer:

$0.05997\ m^3/s$

$3596.1395\ L/min$

$2.11647\ ft^3/s$

Explanation:

$1\ gal/min=\dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=0.05997\ m^3/s$

Volume rate of flow is $0.05997\ m^3/s$

$1\ gal/min=3.78541\ L/min\\\Rightarrow 950\ gal/min=950\times 3.78541=3596.1395\ L/min$

Volume rate of flow is $3596.1395\ L/min$

$1\ gal/min=\dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=2.11647\ ft^3/s$

Volume rate of flow is $2.11647\ ft^3/s$

6 0

2 years ago

A wave has a frequency of 6 Hz and a speed of 30 m/s. What is the wavelength of the wave?

Helen [10]

Answer:

D. 5m

Explanation:

fλ = c, where f is frequency, λ is wavelength and c is speed.

6λ=30

λ=30/6=5

5 0

2 years ago

Una tortuga recorre una pista de 1,25 km que tiempo emplearia si lleara una velocidad de 0,2 km/h

Marina86 [1]

Responder:

<h2>5 horas </h2>

Explicación:

La velocidad se define como el cambio de distancia de un cuerpo con respecto al tiempo. Matemáticamente, Velocidad = Distancia / Tiempo

Dada la velocidad del coche = 0,25 km / hy la distancia = 1,25 km

El tiempo se expresa a partir de la fórmula como Tiempo = Distancia / Velocidad

La sustitución de los valores dados en la fórmula dará;

Tiempo = 1,25 km / 0,25 km / h

Tiempo = 5 horas

Por lo tanto, la tortuga tardará 5 horas en viajar a una velocidad de 0,2 km / h.

7 0

2 years ago

Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L

snow_lady [41]

Answer:

$Q=81.56\ W/m^2$

Explanation:

Given that

$T_1= 210 K$

$T_2= 150 K$

Emissivity of surfaces(∈) = 1

We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

$Q=\sigma (T_1^4-T_2^4)\ W/m^2$

So now by putting the values

$Q=\sigma (T_1^4-T_2^4)\ W/m^2$

$Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2$

$Q=81.56\ W/m^2$

So rate of heat transfer per unit area

$Q=81.56\ W/m^2$

8 0

3 years ago

A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with

ivanzaharov [21]

Answer:

11.72 mm

Explanation:

The gravitational potential energy equals the potential energy of the spring hence

$PE_{gravitational}=PE_{spring}$

$mgh=0.5kx^{2}$ where m is the mass of object, g is the acceleration due to gravity, h is the height, k is the spring constant and x is the extension of the spring

$mgdsin\theta=0.5kx^{2}$ where \theta is the angle of inclination and d is the sliding distance

Making x the subject then

$x=\sqrt {\frac {2mgdsin\theta}{k}}$

Substituting the given values then

$x=\sqrt{\frac {2\times 11\times 9.81\times 3\times sin 38}{2.9\times 10^{4}}}= 0.117240716\approx 11.72 mm$

8 0

2 years ago