A baseball player is running to second base at 5.03 m/s. when he is 4.80 m from the plate he goes into a slide. the coefficient
2 answers:
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<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]
Answer:
1) Charge chord resistance is 75 Ω
2) Charge chord resistance is 6.33 Ω
Explanation:
1) To answer the question, we note that the the formula voltage is found as follows;
V = IR
Therefore,
2) Where the voltage, V = 19.5 V and the current, I = 3.33 A, we have;
Initial resistance R₁ = 19.5 V/(3.33 A) = 5.86 Ω
However, to reduce the current to 1.6 A, we have;
Therefore, where the resistance is found by the sum of the total resistance we have;
= R₁ + Charge chord resistance
∴ 12.1875 = 5.86 + Charge chord resistance
Hence, charge chord resistance = 6.33 Ω
A photoelectric cell is an electronic device which is used to convert light energy into electric energy.The operation of this device is based on photoelectric effect.
Light of suitable frequency i.e greater or equal to threshold frequency will fall on the cathode maintained at negative potential.The electron emission will take place and these electrons are drifted towards the anode which is at positive potential.
Here,only those radiations will be capable of emitting electrons irrespective of surface barrier of metals whose energy is greater than the work function.
We know that the radiation having long wavelength has least energy as energy and wavelength are inversely proportional to each other.
Here h is the Planck's constant,c is the velocity of light.
Here we have been given red light and blue light.
In the visible spectrum of radiation, the red light has longer wavelength than all other colors of light.Hence blue light has more energy as it's wavelength is less as compared to red light.
Hence, the blue light will activate the most and red the least.