Refraction occurs because of changes in the ____________

In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each ha

eduard

Average force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the net change in momentum of the hailstones in a time interval of $\Delta t$

In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:

$p_i =m u sin \theta$ is the initial momentum of one hailstone, with

m = 7 g = 0.007 kg is the mass

$u=4.5 m/s$ is the initial speed

$\theta=25^{\circ}$ is the angle with the window

The final momentum is

$p_f = mv sin \theta$

where

v = 4.5 m/s is the final speed (the collision is elastic so the speed is equal, while the direction changes)

$\theta=-25^{\circ}$ (after the rebound, the direction has changed)

So the change in momentum of 1 hailstone is

$\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s$

We are interested only in the magnitude, so

$\Delta p = 0.0266 kg m/s$

There are 595 hailstones hitting the window in 49 s, so the total change in momentum is

$\Delta p = 595\cdot 0.0266 = 15.8 kg m/s$

And therefore, the average force on the window is

$F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N$

Learn more about force:

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3 0

3 years ago

A jet plane lands with a speed of 100 m/s and can

kiruha [24]

Answer:

a) t = 20 [s]

b) Can't land

Explanation:

To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.

a)

$v_{f}=v_{i}-(a*t)$

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = desacceleration = 5 [m/s^2]

t = time [s]

Note: the negative sign of the equation means that the aircraft slows down as it stops.

0 = 100 - 5*t

5*t = 100

t = 20 [s]

b)

Now we can find the distance using the following kinematics equation.

$x -x_{o}=(v_{o}*t)+\frac{1}{2}*a*t^{2}$

x - xo = distance [m]

x -xo = (0*20) + (0.5*5*20^2)

x - xo = 1000 [m]

1000 [m] = 1 [km]

And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land

4 0

4 years ago

Communication with submerged submarines via radio waves is difﬁcult because seawater is conductive and absorbselectromagnetic wa

FrozenT [24]

Answer:

D. 4000 km

Explanation:

f = Frequency of wave that is being transmitted = 76 Hz

$\lambda$ = Wavelength of wave that is being transmitted

v = The velocity of electromagnetic waves through air = $3\times 10^8\ m/s$

Velocity of a wave is given by

$v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{3\times 10^8}{76}\\\Rightarrow \lambda=3947368.42105\ m=3947.36842105 km\approx 4000\ km$

Hence, the approximate wavelength of the waves is 4000 km

8 0

3 years ago

A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei

tekilochka [14]

Assume no air resistance, and g = 9.8 m/s².

Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0

3 years ago

The angle of reflection is the angle the ____ to the reflecting surface.

Arlecino [84]

Answer:

The angle of reflection is the angle the reflected rays make with a perpendicular line to the reflecting surface.

Explanation:

Reflection It is the change of direction suffered by a luminous ray when hitting the surface of an object. The angle of reflection is that which is formed by the reflected ray and the normal vector to the study surface

7 0

4 years ago

Refraction occurs because of changes in the _____________