An object experiences a 8.9 N force and accelerates 6 m/sec2. What is the object's

A speeding car collides with an unlucky bug flying across the road. Which explains why the impact doesn’t equally damage the car

velikii [3]

The bug was a lot smaller than the car, that's for sure. The car is bigger and sturdier, while the bug is smaller and frail. The bug is so frail, that rather that putting a dent in the car, it splatters all over the car. The bug is very damaged (obviously), while the car just needs a good wash.

8 0

4 years ago

Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle

marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

$n_1sin\theta_1 = n_2sin\theta_2$

Where,

$\theta =$ Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

$n_1 sin\theta_1 = n_2sin\theta_2$

$(1.54) sin\theta_1 = (1.33)sin(90)$

$sin\theta_1 = \frac{1.33}{1.54}$

$\theta = sin^{-1}(\frac{1.33}{1.54})$

$\theta = 59.72\°$

Therefore the $\alpha_{max}$ would be equal to

$\alpha = 90\°-\theta$

$\alpha = 90-59.72$

$\alpha = 30.27\°$

Therefore the largest value of the angle α is 30.27°

3 0

3 years ago

It has made us think we have to live perfect lives?

Citrus2011 [14]

Answer: Is this a question? I dont understand.

Explanation:

3 0

3 years ago

He instruction booklet for your pressure cooker indicates that its highest setting is 11.1 psi . you know that standard atmosphe

blagie [28]

Given that,

Atmospheric Pressure = 14.7 psi

Cooking Pressure = 14.7 +11.1 = 25.8 psi

Take, Atmospheric Temperature = 25 °C

Cooking Temperature = ??

Since, we know that Gas equation is given by:

PV = nRT

or

P ∝ T

P1 / T1 = P2 / T2

14.7/ 25 = 25.8/ T2

T2 = 25*25.8/14.7

T2 = 43.87 °C

The cooking pressure will be 43.87 °C.

5 0

3 years ago

A football player runs from his own goal line to the opposing team's goal line, returning to his forty-yard line, all in 22.4 s.

Bumek [7]

I assume L=120 yards as the length of the football field.

1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
$S=120y+(120-40)y=120y+80y=200 y$
And the time taken is t=22.4 s, so the average speed of the player is
$v= \frac{S}{t}= \frac{200 y}{22.4 s}=8.93 y/s$

2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
$S=120 y-80 y=40 y$
and so, the average velocity is
$v= \frac{S}{t}= \frac{40 y}{22.4 s}= 1.79 y/s$

6 0

3 years ago