You can only score 1 point when shooting a free throw. Group of answer choices True False

Why aren’t the Appalachian Mountains still as tall as the Himalayas?

stealth61 [152]

Answer:

mountains are limited in their theoretical height by several processes. First is isostasy: the bigger a mountain gets, the more it weighs down its tectonic plate, so it sinks lower. ... Bottom line: mountains can get taller than Mount Everest in earth gravity, like the Appalachians probably did—but not much taller.

3 0

2 years ago

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An inventor claims to have invented a heat engine that receives 750kJ of heat from a source at 400K and produces 250kJ of net wo

IRISSAK [1]

Answer:

the claim is not valid or reasonable.

Explanation:

In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:

η(max) = 1 - T₁/T₂

where,

η(max) = maximum efficiency = ?

T₁ = Sink Temperature = 300 K

T₂ = Source Temperature = 400 K

Therefore,

η(max) = 1 - 300 K/400 K

η(max) = 0.25 = 25%

Now, we calculate the actual frequency of the engine:

η = W/Q

where,

W = Net Work = 250 KJ

Q = Heat Received = 750 KJ

Therefore,

η = 250 KJ/750 KJ

η = 0.333 = 33.3 %

η > η(max)

The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.

<u>Therefore, the claim is not valid or reasonable.</u>

3 0

3 years ago

A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same

Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

- 175 m/s · sin 36.7° / - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t

- 175 m/s · sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0

3 years ago

The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive

BartSMP [9]

Answer:

Part A

The intensity is $I = 618 W/m^2$

Part B

The intensity is $I_1= 81.884 W/m^2$

Explanation:

From the question we are told that

The intensity of the light detected by first eye is $I = 700 W/m^2$

Now at initial state according the question the light ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light )would detect when the head is rotated by 20° its previous orientation

This is mathematically evaluated as

$I = I_i cos^2 ( 20^o)$

$I = 700\ cos^2 (20)$

$I = 618 W/m^2$

Now the second question is to obtain the intensity the first eye (the first eye in this case is the one that is not focused on the light )would detect when the head is rotated by 20° its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

So

$I_1 = 700 \ cos^2 (70)$

$I_1= 81.884 W/m^2$

5 0

3 years ago

Consider two children sitting on a merry-go-round, with one closer to the outer edge and one closer to the center. show answer N

dolphi86 [110]

Answer:

They both have the same angular speed.

Explanation:

The mathematical formula for angular speed is:

$w=\frac{2\pi}{T}$

where $w$ is angular speed, $2\pi$ is a constant, and $T$ is the period (the time it takes the marry-go-round to complete a lap).

What we can see from the formula is that, since the $2\pi$ does not change its value, the angular speed depends only on the period T.

In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.

Thus, since the period for both is the same, the angular speed given by

$w=\frac{2\pi}{T}$ will also be the same

4 0

3 years ago