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dsp73
1 year ago
11

A flowerpot falls from a window sill 36.5 m

Physics
1 answer:
hram777 [196]1 year ago
7 0

When a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

From Newton's third equation of motion,

v^2 = u^2 + 2gh

where,

h is the height of the object or body from ground

u is the initial velocity of the body or object

v is the final velocity of the body or object

g is the acceleration due to gravity

Now, as we know that

Flowerpot is at rest. So, u = 0

g = 9.81m/s^2

h = 36.5m

By substituting all the values, we get

v^2 = 2 × 9.81 × 36.5

= 716.13

v = 26.7m/s

Thus, we concluded that when a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

learn more about Newton's equation of law of motion:

brainly.com/question/8898885

#SPJ9

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Answer: C. the elimination of special privileges for members of the first and second estates

Explanation:

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From the options given, the effect of the French Revolution was the the elimination of special privileges for members of the first and second estates. Therefore, the correct option is C.

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3 years ago
Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
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To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

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Calculate the displacement of an object at 2.0 seconds when thrown straight up with an
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Answer:

<h3>30m</h3>

Explanation:

Velocity is the change of rate of displacement with respect to time.

velocity = displacement/time

Given

initial velocity = 15 m/s.

time taken =2 secs

Required

Displacement of the object

From the formula;

Displacement = Velocity * time

Displacement = 15 * 2

Displacement = 30m

<em>Hence the displacement of the object is 30m</em>

8 0
3 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
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<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

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     Substituting

                      s = ut + 0.5 at²

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The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
3 years ago
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