Answer:
V = 0.9 m/s
Explanation:
The parameters given are:
Initial velocity U = 6.4 m/s
Time t = 0.64s
Height h = 2.05 m
To find the final velocity, let us use third equation of motion
V^2 = U^2 - 2gH
Since the ball is going upward, g will be negative
Substitute all the parameters into the formula
V^2 = 6.4^2 - 2 × 9.8 × 2.05
V^2 = 40.96 - 40.18
V^2 = 0.78
V = sqrt( 0.78)
V = 0.883 m/ s
V = 0.9 m/ s approximately
To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.
From the definition of logarithm we know that,

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

We can express this also as,

By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.
So this can be expressed as

Since the definition of the base logarithm 10 of 10 is equal to 1 then

The value of the given logarithm is equal to 6
We use the formula,

Here, h is the variable represents the height of the flare in feet when it returns to the sea so, h = 0 and u is the initial velocity of the flare, in feet per second and its value of 192 ft/sec.
Substituting these values in above equation, we get
.
Here, t= 0 neglect because it is the time when the flare is launched.
Thus, flare return to the sea in 12 s.
Answer:
75joules
Explanation:
Workdone = force x distance
workdone = 25 x 3
workdone = 75joules
True : <span>There are numerous third-class </span>levers<span> in the human </span>body<span>; one example can be illustrated in the elbow joint</span>