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Otrada [13]
3 years ago
10

A car starts from rest and accelerates uniformly over a time of 10.21 seconds for a distance of 210 m. Determine the acceleratio

n of the car.
Physics
1 answer:
uysha [10]3 years ago
6 0

Answer:

4.03 m/s/s

Explanation:

d=1/2 a t^2

210 = 1/2 a (10.21)^2

210 = 52.1 a

a = 4.03 m/s/s

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A laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) abo
Bess [88]

Answer:

a

    n =  1.119 *10^{18} \ photons

b

  P  =  1.6 \ W

Explanation:

From the question we are told that

    The wavelength is  \lambda  =  2780 nm =  2780 *10^{-9} \ m

     The  energy  is  E =  80 mJ  =  80 *10^{-3} \ J

This energy is mathematically represented as

     E   = \frac{n  *  h *  c }{\lambda }

Where  c is the speed of light with a value  c =  3.0 *10^{8} \ m/s

             h is the Planck's  constant with the value  h  =  6.626 *10^{-34} \ J \cdot s

             n is the number of pulses

So

      n =  \frac{E * \lambda }{h * c }

substituting values

       n =  \frac{80 *10^{-3} *  2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }

       n =  1.119 *10^{18} \ photons

Given that the pulses where emitted 20 times in one second then the period of the pulse is

       T  =  \frac{1}{20}

      T = 0.05 \ s

Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as

       P  =  \frac{E}{T}

substituting values

       P  =  \frac{ 80 *10^{-3}}{0.05}

        P  =  1.6 \ W

6 0
3 years ago
se lanza un cuerpo desde el origen con velocidad horizontal de 40 m/s, y con un ángulo de 60º. calcular la máxima altura y el al
EastWind [94]

Answer:

1. h = 244.8 m    

2. x = 564.8 m  

Explanation:

1. La altura máxima se puede calcular usando la siguiente ecuación:

v_{f}^{2} = v_{0}^{2} - 2gh     (1)                        

Where:

v_{f_{y}}: es la velocidad final = 0 (en la altura máxima)  

v_{0_{y}}: es la velocidad inicial horizontal en "y"

g: es la gravedad = 9.81 m/s²          

h: es la altura máxima =?

La velocidad incial en "y" se puede calcular de la siguiente manera:

tan(\theta) = \frac{v_{0_{y}}}{v_{0_{x}}}

v_{0_{y}} = tan(60)*40 m/s = 69.3 m/s                    

Resolviendo la ecuación (1) para "h" tenemos:

h = \frac{v_{0_{y}}^{2}}{2g} = \frac{(69.3 m/s)^{2}}{2*9.81 m/s^{2}} = 244.8 m          

2. Para calcular el alcance horizontal podemos usar la ecuación:

x = v_{x}*t

Primero debemos encontrar el tiempo cuando la altura es máxima (v_{f_{y}} = 0).

v_{f_{y}} = v_{0_{y}} - gt    

t = \frac{v_{0_{y}}}{g} = \frac{69.3 m/s}{9.81 m/s^{2}} = 7.06 s      

Ahora, como el tiempo de subida es el mismo que el tiempo de bajada, el tiempo máximo es:

t_{m} = 2*7.06 s = 14.12 s          

Finalmente, el alcance horizontal es:

x = 40 m/s*14.12 s = 564.8 m                                                            

Espero que te sea de utilidad!

7 0
3 years ago
Two passenger cars, car A and car B, of the same weight are put to a 30 mph head-on crash test. In the test, cars are pulled for
maksim [4K]

Answer:

the statements the correct one is A

Explanation:

Let's analyze this exercise, vehicles have the same mass and speed, so we can use the momentum impulse ratio

          I = ∫ F dt = Δp

the Δp is the same for both cars since they have the same mass and the same speeds, so the momentum is the same in both vehicles

When they indicate that vehicle A was reduced more than vehicle B, this implies that the force acted for a longer time, to have the largest reduction in size, therefore the impact force was less in car A than in car B

Resisting the statements the correct one is A

7 0
3 years ago
In a World Cup soccer match, Juan is running due north toward the goal with a speed of 7.60 m/s relative to the ground. A teamma
spayn [35]

Answer:

Part a)

v_{bj} = 11.03 m/s

Part b)

\theta = 4.57 degree East of South

Explanation:

Part a)

Velocity of Juan is given as

v_1 = 7.60 m/s \hat j

velocity of the ball is given as

v_2 = 12.9(cos31.4 \hat i + sin31.4\hat j)

now we have

v_2 = 11\hat i + 6.72\hat j

Part a)

We need to find velocity of ball with respect to Juan

so it is given as

v_{bJ} = \vec v_b - \vec v_j

v_{bj} = 11\hat i + 6.72 \hat j - 7.6\hat j

v_{bj} = 11\hat i - 0.88\hat j

magnitude of the speed is given as

v_{bj} = \sqrt{11^2 + 0.88^2}

v_{bj} = 11.03 m/s

Part b)

direction of velocity of the ball

tan\theta = \frac{v_y}{v_x}

tan\theta = \frac{-0.88}{11}

\theta = 4.57 degree East of South

3 0
3 years ago
Ashley decides to enter her pet turtle in a race. She knows her turtle can travel at a rate of 2 meters per hour. The race track
Genrish500 [490]
It would probably be B
7 0
3 years ago
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