The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang
2 answers:
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Answer:
5.843 m
Explanation:
suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.
lets consider that horizontal motion
distance = speed * time
time = 40/ 37 = 1.081 s
arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.
applying motion equation
(assume g = 10 m/s²)
Arrow misses the target by 5.843m ig the arrow us split horizontally
Answer:
v = 54.2 m / s
Explanation:
Let's use energy conservation for this problem.
Starting point Higher
Em₀ = U = m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 150)
v = 54.2 m / s
<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]