Answer:
Speed of the boat, v = 4.31 m/s
Explanation:
Given that,
Height of the bridge, h = 32 m
The model boat is 11 m from the point of impact when the key was released, d = 11 m
Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :
Here, u = 0 and a = g
t = 2.55 seconds
Let v is the speed of the boat. It can be calculated as :
v = 4.31 m/s
So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.
Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:
B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º
C) The maximum force is: F = 24 N
D) The time to stop the block is: t = 25 s
Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.
B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.
x-axis
-10 + 12 sin 60 + Fₓ = 0
Fₓ = 10- 12 sin 60 = -0.39 N
y-axis
12 cos 60 - 6 + F_y = 0
F_y = 6 - 12 cos 60 = 0 N
We can give the result of the force in two ways:
Let's use Pythagoras' theorem to find the modulus.
F = 0.39N
We use trigonometry for the angle.
tan θ= 0º
The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.
θ = 180 + θ'
θ = 180 + 0
θ = 180º
C) if the three forces can be moved and the maximum force occurs when they are all linear.
10+ 6 + 6 + F = 0
F = -24 N
D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.
a =
a = 4 m / s²
The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s
If we use kinematic relations.
v = v₀ - a t
Final velocity when stopped is zero
t =
t = 100/4
t = 25 s
In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:
B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º
C) The maximum force is: F = 24 N
D) The time to stop the block is: t = 25 s
Learn more about Newton's second law here: brainly.com/question/25545050
Answer:
He could jump 2.6 meters high.
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:
With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:
This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.