A. 36 mins 52.5 secs<br> b. 24mins 22.5 secs<br> c. 13 mins 0 secs<br> d. 9 mins 45 secs

Do y’all know the answers to them all the choices are at the drop down box.!!!

Alborosie

the second option is your answer to what a liquid is

4 0

3 years ago

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What structure on the south african coast has a range of 63 km and releases flashes every 30 seconds?

I am Lyosha [343]

Answer:

Cape Point lighthouse

Explanation:

The lighthouse is located South of Cape Town, at a point where boats turn to go around the tip of Africa.

The new lighthouse has the most powerful light of all the lighthouses in South Africa, with a power of 10 megacandelas. That's why it's being visible for up to 63 km around.

It's a very important lighthouse helping to conduct the traffic around dangerous waters where 2 oceans meet.

7 0

3 years ago

How many types air<br>I need a formula of calculating the s i unti of force

yKpoI14uk [10]

The si unit of force is newton.

so, F is eqal to m*g

4 0

3 years ago

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A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

Imagine you have a collection of identical flat-bottomed coffee filters that can be nested (stacked inside of each other) so tha

skad [1K]

Answer:

the terminal velocity of 14 nested coffee filters is 3.2 m/s

Explanation:

Given the data in the question;

we know that;

The terminal velocity is proportional to the square root of weight.

v ∝ √W

v = k√W

the proportionality constant depends upon the surface area and the density of the medium (like air). The coffee filters can be stacked such that the resulting area is roughly unchanged. So, the constant of proportionality k is also unchanged

v/√W = constant

v₂/√W₂ = v₁/√W₁

v₂ = v₁√(W₂ / W₁ )

given that;

v₁ = 0.856 m/s,

W₂ = 14W₁; meaning 14 coffee filters have 14 times the weight of a single coffee filter

so we substitute

v₂ = 0.856 √(14W₁ / W₁ )

v₂ = 0.856 √( 14( W₁/W₁)

v₂ = 0.856 √( 14(1)

v₂ = 0.856 √( 14 )

v₂ = 0.856 × 3.741657

v₂ = 3.2 m/s

Therefore, the terminal velocity of 14 nested coffee filters is 3.2 m/s

6 0

3 years ago

A. 36 mins 52.5 secs b. 24mins 22.5 secs c. 13 mins 0 secs d. 9 mins 45 secs