IDNLearn.com is your go-to resource for finding expert answers and community support. Our community provides accurate and timely answers to help you understand and solve any issue.
Sagot :
The kinematic relations allow finding the results for the different questions are:
1. The acceleraations is a = 375 m/s²
2. The time is t = 62.5 s
3. The velocity at the ground is v = -14.7 m / s
4. The time to the bottoncliff is 1.7 s
5. The height isy₀ = 122.5 m
6.
A) The velocity is v = 25.2 m / s
B) The height for t=1 s is y = 18.1 m
7.
A) Time in the air is t = 2 s
B) The final veloicty is v = - 19.6 m / s
Kinematics analyzes the movement of bodies, finding relationships between the position, velocity and acceleration of bodies.
Let's look for the answers to a series of questions:
1. They indicate the initial and final velocities of the body and the time to reach it is 20 s, ask how much the acceleration is worth
Let's use the relationship
v = v₀ + a t
a = [tex]\frac{v-v_o}{t}[/tex]
a = [tex]\frac{17500-10000}{20}[/tex]
a = 375 m/s²
2. What is the time it takes to travel a distance of 125 m with an acceleration of 4 m / s²
x = v₀ t + ½ a t²
As it leaves the direction its initial velocity is zero
x = ½ to t²
t = [tex]\sqrt{\frac{2x}{a} }[/tex]
t = [tex]\sqrt{\frac{2 \ 125}{4} }[/tex]
t = 62.5 s
3. A pencil falls from a height and reaches the floor in a time of 1.5 s what is the speed when reaching the floor
v = v₀ - g t
As it is released its initial velocity is zero
v = -g t
v = - 9.8 1.5
v = -14.7 m / s
The negative sign indicates that the velocity is directed downwards.
4. From the edge of a cliff 40 m high, an apple is thrown downward with an initial velocity of 15 m / s. How long does it take to get to the bottom?
y = y₀ + v₀ t - ½ g t²
In this case the initial velocity is negative because it is directed downwards and when it reaches the floor its height is zero, let us substitute
Suppose the acceleration is g = 10 m / s²
0 = 40 - 15 t - ½ 10 t²
0 = 40 - 15 t - 5 t²
let's solve the quadratic equation
t² + 3 t - 8 = 0
t = -3 + Ts 9 + 4 8/2
t = -3 + 6. 4/2
t₁ = - 4.9 s
t₂ = 1.7 s
The time must be a positive quantity, so the correct answer is 1.7 s
5. A stone is dropped from the ceiling and it takes a time 5 s to reach the floor, which is the height of the ceiling
y = y₀ +v₀ t - ½ g t²
as it is released from the ceiling its initial velocity is zero and the height upon reaching the floor is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
y₀ = ½ 9.8 5²
y₀ = 122.5 m
6. A man at a height of 30 m throws a ball downward at 7 m / s
A) The speed when reaching the ground
B) The height of the ball when a time of 1 s has fallen.
A) Let's use the kinematics relation
v² = v₀² - 2 g (y -y₀)
v = 7² - 2 9.8 (0 -30)
v = 25.2 m / s
B) y = y₀ + v₀ t - ½ g t²
y = 30 7 1 - ½ 9.8 1²
y = 18.1 m
7. A watermelon is dropped from rest from the top of a 20m cliff.
A) time in the air
B) ground speed
A) Let's use the relation
y = y₀ + v₀ t - ½ g t²
It is released its initial velocity is zero and when it reaches the ground its height is zero (y = 0)
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]
t = [tex]\sqrt{\frac{2 \ 20}{9.8} }[/tex]
t = 2 s
B) the speed when reaching the ground
v = v₀ - g t
v = 0 - 9.8 2
v = - 19.6 m / s
The negative sign indicates that the speed is down
In conclusion using the kinematic relations we can find the results for the different questions are:
1. The acceleraations is a = 375 m/s²
2. The time is t = 62.5 s
3. The velocity at the ground is v = -14.7 m / s
4. The time to the bottoncliff is 1.7 s
5. The height isy₀ = 122.5 m
6.
A) The velocity is v = 25.2 m / s
B) The height for t=1 s is y = 18.1 m
7)
A) Time in the air is t = 2 s
B) Te final veloicty is v = - 19.6 m / s
Learn more here: brainly.com/question/15068914
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Discover insightful answers at IDNLearn.com. We appreciate your visit and look forward to assisting you again.
