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3 years ago

What types of landforms are located in Florida? plaz i need help FAST

2 answers:

8 0

The coastal plains generally consist of flat land, fronted by barrier islands, sandy beaches, coral reefs and sandbars. The low rolling-hills of the uplands area stretch across the Florida Panhandle. Here, the state's highest point, Britton Hill, rises to 345 ft.

nordsb [41]3 years ago

4 0

There are beaches.

You didnt add anny choices

Hope this helps Buddy!

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A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i

NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = $\frac{1}{2}mu^2$

Final kinetic energy = $\frac{1}{2}mv^2$

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2$

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2$

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0

4 years ago

20 points?!!!!!! please help I need this asap

IgorLugansk [536]

$F - F_f = ma$

We are trying to find F.

F - 15 = 30(1.5)

F - 15 = 45

F = 60 N

6 0

3 years ago

According to the inverse square law of light, how will the apparent brightness of an object change if its distance to us triples

aleksley [76]

According to the inverse square law of light, apparent brightness will decrease by a factor of 9. Use the formula $B=L/(4*pD^2)$ , to check it.

4 0

3 years ago

What was the direction of the ball’s velocity

Tju [1.3M]

Part of the question is missing. Here it is:

A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table. The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity just before it hit the floor?

Answer:

$\theta=-75.7^{\circ}$

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

$s=ut+\frac{1}{2}at^2$

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s$

Now we can find the final vertical velocity of the ball, using:

$v_y=u+at$

And susbtituting t = 0.52 s, we find

$v_y = 0 +(9.8)(0.52)=5.1 m/s$

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

$v_x = \frac{0.7}{0.52}=1.3 m/s$

So now we can find the direction of the ball's velocity using:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}$

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

$\theta=-75.7^{\circ}$

8 0

3 years ago

The speed of light in air is 3.0 x 10^8 m/s. The speed of light in particular glass is 2.3 x 10^8 m/s. Use the information to de

olga_2 [115]

Answer:

33.61°

Explanation:

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

v is the velocity in the medium (2.3 × 10⁸ m/s)

c is the speed of light in air = 3.0 × 10⁸ m/s

So,

n = 3.0 × 10⁸ / 2.3 × 10⁸

n = 1.31

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 25.0° )

${\theta_r}$ is the angle of refraction ( ? )

${n_r}$ is the refractive index of the refraction medium (air, n=1)

${n_i}$ is the refractive index of the incidence medium (glass, n=1.31)

Hence,

$1.31\times {sin25.0^0}={1}\times{sin\theta_r}$

Angle of refraction = $sin^{-1}0.5536$ = 33.61°

5 0

3 years ago