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3 years ago

5

you would most likely find a weather Ballon an it's highest altitude in which layer above earth (A) Exosphere (B) Mesospher (C)

Stratosphere (D) Thermosphere

2 answers:

laila [671]3 years ago

7 0

U will find it in the Exosphere

Trava [24]3 years ago

6 0

The answer to this question is

C. stratosphere...

this picture should help you

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A rock thrown with speed 7.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 18.0 m bef

Iteru [2.4K]

Answer:

height from where rock was thrown is 27.916 m

Explanation:

speed = 7.50 m/s

θ = 30°

g= 9.8 m/s²

horizontal distance = 18 m

time require for vertical displacement

$time = \frac{distance}{velocity} \\t = \frac{18}{7.5\ cos30^0}$

t = 2.8 sec

now for calculation of height

s = ut + 0.5 a t²

-h = v sinθ× t + 0.5 ×(-9.8)× (2.8²)

-h = 7.5 sin30°× 2.8 - 0.5 ×(9.8)× (2.8²)

-h = -27.916 m

h= 27.916 m

height from where rock was thrown is 27.916 m

5 0

3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago

For a standing wave to form in a medium, two waves must

Slav-nsk [51]

For a standing wave to form, two waves must be traveling in opposite directions and cause destructive interference.

4 0

3 years ago

Why are objects that fall near earth’s surface rarely in free fall?

Sloan [31]

Answer:

Because of the presence of air resistance

Explanation:

When an object is in free fall, ideally there is only one force acting on it:

- The force of gravity, W = mg, that pushes the object downward (m= mass of the object, g = acceleration of gravity)

However, this is true only in absence of air (so, in a vacuum). When air is present, it exerts a frictional force on the object (called air resistance) with upward direction (opposite to the motion of free fall) and whose magnitude is proportional to the speed of the object.

Therefore, it turns out that as the object falls, its speed increases, and therefore the air resistance acting against it increases too; as a result, the at some point the air resistance becomes equal (in magnitude) to the force of gravity: when this happens, the net acceleration of the object becomes zero, and so the speed of the object does not increase anymore. This speed reached by the object is called terminal velocity.

3 0

3 years ago

In a car lift, compressed air exerts a force on a piston with a radius of 2.62 cm. This pressure is transmitted to a second pist

marin [14]

Answer:

847.45 N

Explanation:

F₁=force of exerted by smaller piston

F₂=force of exerted by larger piston=1.44×10⁴ N

A₁=Area of smaller piston= 2.62 cm =0.0265 m

A₂=Area of larger piston= 10.8 cm =0.108 m

Pressure exerted by both the pistons will be equal

$P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times 0.108^2}\pi\times 0.0262^2\\\Rightarrow F_1=847.45\ N$

Hence, force exerted to lift a 14400 N car is 847.45 N

8 0

4 years ago