The density of ice is 0.93 g/cm3. what is the volume, in cm3, of a block of ice whose mass is 5.00 kg? remember to select an ans
1 answer:
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Answer:
times
Explanation:
First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.
Age of the universe:
(1 followed by 17 zeroes)
Lifetime of the top quark:
(we moved the decimal point 24 places to the right)
Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:
<h2>Answer:</h2>
(a) v(t) = [10.0t - 12.0t²] m/s and a(t) = [10.0 - 24.0t ] m/s² respectively
(b) -28.0m/s and -38.0m/s² respectively
(c) 0.83s
(d) 0.83s
(e) x(t) = 1.1573 m [where t = 0.83s]
<h2>Explanation:</h2>The position equation is given by;
x(t) = 5.0t² - 4.0t³ m --------------------(i)
(a) Since velocity is the time rate of change of position, the velocity, v(t), of the particle as a function of time is calculated by finding the derivative of equation (i) as follows;
v(t) = dx(t) / dt = =
[ 5.0t² - 4.0t³ ]
v(t) = 10.0t - 12.0t² --------------------------------(ii)
Therefore, the velocity as a function of time is v(t) = 10.0t - 12.0t² m/s
Also, since acceleration is the time rate of change of velocity, the acceleration, a(t), of the particle as a function of time is calculated by finding the derivative of equation (ii) as follows;
a(t) = dx(t) / dt = =
[ 10.0t - 12.0t² ]
a(t) = 10.0 - 24.0t --------------------------------(iii)
Therefore, the acceleration as a function of time is a(t) = 10.0 - 24.0t m/s²
(b) To calculate the velocity at time t = 2.0s, substitute the value of t = 2.0 into equation (ii) as follows;
=> v(t) = 10.0t - 12.0t²
=> v(2.0) = 10.0(2) - 12.0(2)²
=> v(2.0) = 20.0 - 48.0
=> v(2.0) = -28.0m/s
Also, to calculate the acceleration at time t = 2.0s, substitute the value of t = 2.0 into equation (iii) as follows;
=> a(t) = 10.0 - 24.0t
=> a(2.0) = 10.0 - 24.0(2)
=> a(2.0) = 10.0 - 48.0
=> a(2.0) = -38.0 m/s²
Therefore, the velocity and acceleration at t = 2.0s are respectively -28.0m/s and -38.0m/s²
(c) The time at which the position is maximum is the time at which there is no change in position or the change in position is zero. i.e dx / dt = 0. It also means the time at which the velocity is zero. (since velocity is dx / dt)
Therefore, substitute v = 0 into equation (ii) and solve for t as follows;
=> v(t) = 10.0t - 12.0t²
=> 0 = 10.0t - 12.0t²
=> 0 = ( 10.0 - 12.0t ) t
=> t = 0 or 10.0 - 12.0t = 0
=> t = 0 or 10.0 = 12.0t
=> t = 0 or t = 10.0 / 12.0
=> t = 0 or t = 0.83s
At t=0 or t = 0.83s, the position of the particle will be maximum.
To get the more correct answer, substitute t = 0 and t = 0.83 into equation (i) as follows;
<em>Substitute t = 0 into equation (i)</em>
x(t) = 5.0(0)² - 4.0(0)³ = 0
At t = 0; x = 0
<em>Substitute t = 0.83s into equation (i)</em>
x(t) = 5.0(0.83)² - 4.0(0.83)³
x(t) = 5.0(0.6889) - 4.0(0.5718)
x(t) = 3.4445 - 2.2872
x(t) = 1.1573 m
At t = 0.83; x = 1.1573 m
Therefore, since the value of x at t = 0.83s is 1.1573m is greater than the value of x at t = 0 which is 0m, then the time at which the position is at maximum is 0.83s
(d) The velocity will be zero when the position is maximum. That means that, it will take the same time calculated in (c) above for the velocity to be zero. i.e t = 0.83s
(e) The maximum position function is found when t = 0.83s as shown in (c) above;
Substitute t = 0.83s into equation (i)
x(t) = 5.0(0.83)² - 4.0(0.83)³
x(t) = 5.0(0.6889) - 4.0(0.5718)
x(t) = 3.4445 - 2.2872
x(t) = 1.1573 m [where t = 0.83s]