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Ann [662]
2 years ago
4

In a crash test car is driven into solid wall at a speed of 35 mph. Figure below shows the car’s acceleration as it crashes into

the wall. What is the maximum force experienced by the 1650kg car?

Physics
1 answer:
Pavlova-9 [17]2 years ago
8 0

Net force on the car is given by

F = ma

here if acceleration is maximum then force must be maximum

so from graph the maximum value of acceleration is a = 300 m/s^2

F = 1650 * 300

F = 495000 N

<em>so the maximum force is 495000 N</em>

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La luz pasa del medio A al medio B formando un ángulo de 35° con la frontera horizontal entre ambos. Si el ángulo de refracción
zaharov [31]

Answer:

Índice de refracción entre los dos medios = 1,43

Refractive index between the two media = 1.43

Explanation:

El índice de refracción entre dos medios se explica mejor entendiendo primero la refracción.

Cuando las olas se mueven de un medio a otro, a menudo experimentan un cambio de dirección con respecto al medio en el que viajan.

Por lo tanto, el índice de refracción se expresa como el seno del ángulo de incidencia dividido por el seno del ángulo de refracción.

El seno del ángulo de incidencia y la refracción utilizados en esta fórmula de índice de refracción se miden respectivamente con respecto a la vertical.

En esta pregunta Ángulo de incidencia = 35° a la horizontal = (90° - 35°) a la vertical = 55° a la vertical.

Ángulo de refracción = 35°

Índice de refracción entre los dos medios

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 a 2 d.p.

¡¡¡Espero que esto ayude!!!

English Translation

The light passes from medium A to medium B at an angle of 35 ° with the horizontal border between the two. If the angle of refraction is also 35 °, what is the relative refractive index between the two media?

Solution

The refractive index between two media is best explained by first understanding refraction.

When waves move from one medium to another, they often experience a change in direction with respect to the medium in which they are travelling.

Hence, refractive index is expressed as the sine of angle of incidence dibided by the sine of angle of refraction.

The sine of angle of incidence and refraction used in this refractive index formula are both respectively measured with respect to the vertical.

In this question,

Angle of incidence = 35° to the horizontal = (90° - 35°) to the vertical = 55° to the vertical.

Angle of refraction = 35°

Refractive index between the two media

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 to 2 d.p.

Hope this Helps!!!

3 0
2 years ago
A squirrel in a tree drops an acorn. how long does it take the acorn to fall 20 feet?
mart [117]

We use the equation of motion,

S= ut+\frac{1}{2}at^{2}

Here, S is the height, u is initial velocity and a is acceleration.

Given, S = 20 \ ft S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft}  = 6.096 \ m

As  acorn falls from tree, therefore we take the value of a = 9.8 \ m/s^2 and initial velocity u = 0.

Substituting these values in equation of motion,

6.096 \ m = 0 \times t +\frac{1}{2} \times 9.8 m/s^2 (t)^2 \\\\\ t = 1.12 \ s

Thus, the time taken by the acorn to fall 20  feet ( 6.096 m ) is 1.12 s.

5 0
2 years ago
A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi
Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

h = ut + \frac{1}{2}\times a \times t^2

h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2

h = 313.6 m

Hence, the depth of the well is 313.6 m

4 0
2 years ago
A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels.
shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

engine power is 217 hp

70 % power reached to wheel

total mass ( car + driver) is 1530 kg

from the data given

2/3 rd of weight is over the wheel

w = 2/3rd mg

maximum force

F = \mu W

we know that F = ma

ma =  \mu (2/3 mg)

a_{max} = 2/3(1.00) (9.8) = 6.53 m/s^2

the new power is p  = 70\% P_[max} = 0.7 P_{max}

P =f_{max} v

0.7P_{max} = ma_{max} v

solving for speed v

v =0.7 \times \frac{P_{max}}{ma_{max}}

v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}

v = 11.5689 m/s

7 0
2 years ago
An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass
Otrada [13]
The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table. 
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. 
I = 7/5*mR^2 M = 7/5*m 
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2 
7 0
2 years ago
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