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harkovskaia [24]
1 year ago
8

What is the final velocity of the ball that is dropped from a height of 200m?

Physics
1 answer:
Hoochie [10]1 year ago
8 0

The final velocity of the ball that is dropped from a height of 200m is v = 44.73 m/s .

<h3>What is velocity with example?</h3>

The rate at which an object is travelling in one direction is referred to as its velocity. an automobile traveling north on a highway, or a rocket taking off. Its velocity vector's absolute value always is equal to the motion's speed because it is a scalar.

<h3>Briefing:</h3>

Given the initial velocity of the ball (u) =  0

Distance travelled by the ball (s) = 200m

Acceleration (a) = 10 m/s²

As we know:

v² = u² + 2as

Putting values:

v² = 0+2 × (10 m/s²) × (200 m)

v = 44.73 m/s.

To know more about Velocity visit:

brainly.com/question/18084516

#SPJ9

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C

Explanation:

it uses a renewable source

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2 years ago
List between six and eight student organizations either offered by your school or offered
melamori03 [73]

Examples of student-led organizations are:

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The benefits of getting involved in any of these are many. They include but are not limited to:

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Learn more about student organizations in the link below:

8 0
2 years ago
FORENSIC SCIENCE--
nikklg [1K]

Answer:

D. 90

Explanation:

because even if there is a shadow of doubt the person should not be persecuted as it could lead to false imprisonment

7 0
2 years ago
• How does wind shape Earth’s surface?
ehidna [41]

Lamina and turbulent flow

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mentioning about lamina and turbulent flow we could say that both form in different period of time

6 0
3 years ago
The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t
frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

4 0
3 years ago
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