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kodGreya [7K]
3 years ago
11

compare 51.5 hectograms to 51500 decigrams. How do each of these values compare to a gram, and which represents a larger mass?

Physics
1 answer:
Olin [163]3 years ago
7 0
These values when compared seem to be the same. They are equal. If we convert them to the same units, they results to the value which is:
 
51.5 Hectograms = 5100 Grams

51,500 Decigrams = 5100 Grams


Hope this answers the question. Have  a nice day.

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Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de
ahrayia [7]

Answer:

a) El caudal de salida del chorro es 1.213\times 10^{-3}\,\frac{m^{3}}{s}.

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

\Delta z = \frac{v_{out}^{2}}{2\cdot g}

Donde:

\Delta z - Diferencia de altura, medida en metros.

g - Constante gravitacional, medida en metros por segundo al cuadrado.

v_{out} - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

v_{out} = \sqrt{2\cdot g \cdot \Delta z}

Si g = 9.807\,\frac{m}{s^{2}} y \Delta z = 0.3\,m, entonces la rapidez de salida del chorro es:

v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}

v_{out} \approx 2.426\,\frac{m}{s}

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

\dot V_{out} = v_{out}\cdot A_{t}

Donde:

v_{out} - Rapidez de salida del chorro, medida en metros por segundo.

A_{t} - Área transversal del orificio, medido en metros cuadrados.

\dot V_{out} - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que v_{out} = 2.426\,\frac{m}{s} y A_{t} = 5\,cm^{2}, el caudal de salida del chorro es:

\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}}  \right)

\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}

El caudal de salida del chorro es 1.213\times 10^{-3}\,\frac{m^{3}}{s}.

5 0
3 years ago
An object with a mass of 9 kg weighs 14.4 N on the Moon. What is the
erma4kov [3.2K]

Answer:

1.6 m/s²

Explanation:

Weight equals mass times acceleration due to gravity.

F = mg

14.4 N = (9 kg) g

g = 1.6 m/s²

7 0
4 years ago
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To help prevent people from getting the Ebola virus, countries around the world paid to help doctors develop a new vaccine. Now
sweet-ann [11.9K]

your answer would be

B complexity

I agree that it is B but would suggest that you also consider the implications of each answer. They are all technically correct.

A is correct simply due to the fact that the longer time passes without an ebola cure, the more people can be potentially infected with it, resulting in less healthy individuals that may be able to volunteer to test potential ebola vaccines. Scientists need healthy volunteers from affected regions in order for rapid clinical trials to occur.

B is correct because of the nature of epidemics. A combination of technical, social, economic, and geographical obstacles slow the progress and dissemination of ebola vaccine information.

C is correct because finance impacts everything from finding resources used in studying ebola to distributing vaccines to affected areas. Economic imbalances or money shortages negatively impact research.

D is correct due to many social factors (more of a subjective answer). Some people think that vaccine distributors discriminate against certain ethnic groups. Others think the governments/other organizations are not effective enough at managing the situation. Still others feel that the search for a cure is not being approached in the right way. In all, there are multiple ways to oppose different aspects of ebola research.

Ultimately, B is correct because it seems to capture the meaning of A, C, and D together. 

5 0
3 years ago
Simple Circuit and Ohm's Law Check-for-Understanding
Vanyuwa [196]

Answer:

current, only

Explanation:

current:I

voltage:U

resistance:R

formula: I=U/R

Increasing the battery cause the increasing in the voltage. Resistance does not normally change. And the current would increase.

4 0
3 years ago
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Two cars, A and B , travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)
OLga [1]

A) Car A is initially ahead

B) The two cars are at the same point at the times: t = 0, t = 2.27 s and

t = 5.73 s

C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s

D) The two cars have same acceleration at t = 2.67 s

Explanation:

A)

The position of the two cars at time t is given by the following functions:

x_A(t) = \alpha t + \beta t^2

with

\alpha = 2.60 m/s\\\beta = 1.20 m/s^2

Substituting,

x_A(t)=2.60t+1.20 t^2

And

x_B(t)=\gamma t^2 - \delta t^3

with

\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3

Substituting,

x_B(t)=2.80t^2-0.20t^3

Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:

x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m

x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m

So, car A is initially ahead.

B)

The two cars are at the same point when their position is the same. Therefore, when

x_A(t)=x_B(t)

which means when

2.60t+1.20t^2 = 2.80t^2-0.20t^3

Re-arranging the equation, we find

0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0

One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation

0.20t^2-1.6t+2.60=0

which has two solutions:

t = 2.27 s

t = 5.73 s

So, these are the times at which the cars are at the  same point.

C)

The distance between the two cars A and B is not changing when the velocities of the two cars is the same.

The velocity of car A is given by the derivative of the position of  car A:

v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t

The velocity of car B is given by the derivative of the position of car B:

v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2

Therefore, the distance between the two cars is not changing when the two velocities are equal:

v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0

This is another second-order equation, which has two solutions:

t = 1.00 s

t = 4.33 s

D)

The acceleration of each car is given by the  derivative of the velocity of the car A.

The acceleration of car A is:

a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40

While the acceleration of car B is:

a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t

So, the two cars have same acceleration when

a_A(t)=a_B(t)

And solving the equation, we find:

2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s

So, the two cars have same acceleration at t = 2.67 s.

Learn more about accelerated motion:

brainly.com/question/9527152

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brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
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