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maxonik [38]
3 years ago
14

A rubbit gets down from a rump which its /\x=0.85m in 0.5s, The rubbit's mass is 2kg, what is the net Force?

Physics
1 answer:
Ira Lisetskai [31]3 years ago
6 0

Answer: 13.6 N

Explanation:

The equation of motion for the rabbit is:

\Delta x=V_{ox}t+\frac{1}{2}a_{x}t^{2} (1)

Where:

\Delta x=0.85 m is the distance traveled by the rabbit

V_{ox}=0 m/s is the rabbit's initial velocity, assuming it started from rest

t=0.5 s is the time

a_{x} is the acceleration

Isolating a_{x}:

a_{x}=\frac{2 \Delta x}{t^{2}} (2)

a_{x}=\frac{2 (0.85 m)}{(0.5)^{2}} (3)

a_{x}=6.8 m/s^{2} (4)

On the other hand, the force F_{x} is given by:

F_{x}=m.a_{x} (5)

Where m=2 kg is the mass of the rabbit

Substituting (4) in (5):

F_{x}=(2 kg)(6.8 m/s^{2}) (6)

Finally:

F_{x}=13.6 N

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Para el siguiente conjunto de medidas, calcule EL ERROR RELATIVO PORCENTUAL: 1.34 m, 1.35 m, 1.37 m y 1.36 m
ivanzaharov [21]

Answer:

Ver explicacion abajo

Explanation:

En este caso para poder calcular el error relativo porcentual, es necesario calcular primero el error absoluto, que se calcula de la siguiente forma:

Error absoluto = Resultado exacto - aproximación

Sin embargo, no tenemos el resultado exacto de las medidas, pero podriamos conocerlo tomando el promedio de estas medidas y este es el que tomaremos como el verdadero resultado de las medidas:

Promedio de medidas = 1.34 + 1.35 + 1.37 + 1.36 / 4

Promedio de medidas = 1.355 m

Ya que tenemos el promedio, podemos calcular el error absoluto de cada medida y luego el error relativo porcentual:

Ea1 = 1.355 - 1.34 = 0.015

Ea2 = 1.355 - 1.35 = 0.005

Ea3 = 1.37 - 1.355 = 0.015

Ea4 = 1.36 - 1.355 = 0.005

Ya que tenemos los 4 errores absolutos, es posible calcular el porcentual:

%error relativo = (Error absoluto / resultado exacto) * 100

Aplicando la expresión con cada uno de los valores tenemos:

%Er1 = (0.015/1.34) * 100 = 1.12%

%Er2 = (0.005/1.35) * 100 = 0.37%

%Er3 = (0.015/1.37) * 100 = 1.09%

%Er4 = (0.005/1.36) * 100 = 0.37%

Espero que te sirva.

4 0
3 years ago
The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C. How many grams of chlorophyll, C55H72MgN4O5, a nonvolatile,
mixer [17]

Answer:

29.4855 grams of chlorophyll

Explanation:

From Raoult's law

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 457.45 mmHg ÷ 463.57 mmHg = 0.987

Mass of solvent (diethyl ether) = 187.4 g

MW of diethyl ether (C2H5OC2H5) = 74 g/mol

Number of moles of solvent = mass/MW = 187.4/74 = 2.532 mol

Let the moles of solute (chlorophyll) be y

Total moles of solution = moles of solute + moles of solvent = (y + 2.532) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.987 = 2.532/(y + 2.532)

y + 2.532 = 2.532/0.987

y + 2.532 = 2.565

y = 2.565 - 2.532 = 0.033

Moles of solute (chlorophyll) = 0.033 mol

Mass of chlorophyll = moles of chlorophyll × MW = 0.033 × 893.5 = 29.4855 grams

8 0
3 years ago
a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame
natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}

where;

L = length of the bed

\mu = viscosity

U = superficial velocity

\epsilon = void fraction

dp = equivalent spherical diameter of bed material (m)

\rho = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6       --------------   equation (1)

24A + 576B = 24.1    ---------------  equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

       B = 0.017014

From equation (1)

12A + 144B  = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

A = \frac{9.6 -144(0.017014}{12}

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

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Find the order of magnitude of your age in seconds
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