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Mamont248 [21]
3 years ago
10

A man throws a ball straight up to his friend on a balcony who catches it at its highest point. The ball was thrown with an init

ial velocity of 17.9 m/s2. What height does the ball reach?
Physics
1 answer:
Ede4ka [16]3 years ago
3 0

Answer:

The maximum height reached by the ball is 16.35 m.

Explanation:

Given;

initial velocity of the ball, u = 17.9 m/s

the final velocity of the ball at the maximum height, v = 0

The maximum height reached by the ball is given by;

v² = u² + 2gh

During upward motion, gravity is negative

v² = u² + 2(-g)h

v² = u² -  2gh

0 = u² -  2gh

2gh = u²

h = u² / 2g

h = (17.9)² / (2 x 9.8)

h = 16.35 m

Ttherefore, the maximum height reached by the ball is 16.35 m.

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Answer:

The balance between incoming energy from the sun and outgoing energy from Earth ultimately drives our climate. This energy balance is governed by the first law of thermodynamics, also known as the law of conservation of energy.

4 0
2 years ago
A speaker vibrates at a frequency of 200 hz. What is it's period ?
Katen [24]
200 Hz = 200 cycles per sec 

<span>1 cycle, the period = 1/200 = 0.005 seconds, or 5 milli seconds.</span>
8 0
3 years ago
An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of
Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

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6 0
3 years ago
The force between two charges, q, and 92, is F. If the distance between the
Sholpan [36]

Answer:

4F

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

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4 0
2 years ago
If we start with 1.000 g of cobalt-60, 0.675 g will remain after 3.00 yr. this means that the of is _____
Tju [1.3M]
<span>Cobalt-60 is undergoing a radioactivity decay.

The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span>           n </span>⇒ remaining mass of cobalt after 3 years
          T ⇒ decaying period
           t ⇒ half-life of cobalt.

So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
 3/t= 0.567

t = 3÷0.567
  = 5.290626524

the half-life of Cobalt-60 is 5.29 years. 

<span>           
</span><span>
</span>
8 0
3 years ago
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