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3 years ago

Florida law applies the same protections given to pedestrians to any person riding or leading an animal upon a roadway

Physics

1 answer:

dmitriy555 [2]3 years ago

5 0

Answer:

True

Explanation:

The Florida law states that a person driving an animal-drawn vehicle has the same protections that are applicable to a person driving a vehicle and the protections that are applicable to pedestrians also apply to a person riding or leading an animal upon a roadway or the shoulder of it. According to this, the answer is that the statement is true.

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A double insulated drill ____. A. has two ground wires to insure proper grounding B. generally has a plastic case with the elect

Len [333]

Answer:

Option B is correct.

A double insulated drill generally has a plastic case with the electrical connections and motor insulated within the tool

Explanation:

A double insulated drill's design in real life has plastic casing as the insulator & it wouldn't be a drill without its motor now, would it!?

8 0

3 years ago

Carlos was camping and getting cold as the sun went down. He wanted to light a fire for warmth and light. However, he discovered

storchak [24]

1. The chemical reaction produced by Carlo's fire is exergonic because energy is "going out". As the reaction proceeds, entropy increases as the energy stored in the dry wood and leaves are used up as fuel to create the fire which produces low quality light and warmth.

2. This reaction is a classic example of an exothermic reaction. Exothermic reactions are characterized with the presence of heat and light in the products. Combustion reactions are always exothermic in nature.

3. Catalyst are substances that are used to speed up reactions by lowering the activation requirement. Catalysts aren't consumed in the reaction and can still be chemically retrieved afterwards. In this situation, the leaves cannot be retrieved after the reaction ends. The leaves speed up the heating of the wood but it does not behave as a catalyst.

6 0

3 years ago

A juggler throws a ball straight up into the air. The ball remains in the air for a time (t) before it lands back in the juggler

natka813 [3]

Answer:

$9.8 m/s^2$ , downward

Explanation:

There is only one force acting on the ball during its motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity (downward)

According to Newton's second law,

$F=ma$

where F is the net force on the object and a is its acceleration. Rearranging for a,

$a=\frac{F}{m}$

As we said, the only force acting on the ball is gravity, so F = mg and the acceleration of the ball is:

$a=\frac{mg}{m}=g$

Therefore, the ball has a constant acceleration of $9.8 m/s^2$ downward for the entire motion.

5 0

3 years ago

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

3 years ago

Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de

vovangra [49]

Answer: 52.9 metros.

Explanation:

Podemos escribir la fuerza de fricción cinética como

F = μ*N

donde N es la fuerza normal entre el coche y el suelo, cuya magnitud es igual al peso en esta situación.

F = μ*m*g

donde m es la masa del coche y g es 9.8m/s^2

y sabemos que μ = 0.8

Por la segunda ley de Newton, sabemos que:

F = m*a

fuerza es igual a masa por aceleración.

a = F/m

entonces la aceleración causada por la fuerza de rozamiento es:

F = 0.8*m*g

a = F/m = (0.8*m*g)/m = 0.8*g.

Entonces ya encontramos la aceleración, hay que recordar que esta aceleración es en sentido opuesto a la sentido de movimiento, entonces podemos escribir la aceleración como:

a(t) = -0.8*g

Para la velocidad, podemos integrar sobre el tiempo para obtener.

v(t) = -0.8*g*t + v0

donde v0 es la velocidad inicial del auto = 28.7m/s

v(t) = -0.8*g*t + 28.8m/s

Ahora podemos encontrar el tiempo necesario para que la velocidad del coche sea cero, en ese momento, como deja de moverse, ya no tendremos rozamiento cinético, entonces no habrá aceleración y el coche se detendrá completamente.

v(t) = 0m/s = -0.8*9.8m/s^2*t + 28.8m/s

7.84m/s^2*t = 28.8m/s

t = (28.8m/s)/(7.84m/s^2) = 3.63 segundos.

Ahora vamos a la ecuación de movimiento, donde asumimos que la posición inicial del coche es 0m, así que no tendremos constante de integración.

p(t) = -(1/2)*(0.8*9.8m/s^2)*t^2 + 28.8m/s*t

Ahora podemos evaluar la posición en t = 3.63 segundos, y esto nos dara la distancia que el coche se movio mientras frenaba.

p(3.63s) = -(1/2)*(0.8*9.8m/s^2)*(3.63s)^2 + 28.8m/s*(3.63s) = 52.9 metros.

6 0

3 years ago