Answer:
Please find the explanation below
Explanation:
A hypothesis in science is a testable explanation that is yet to be tested via experimentation. It is a predictive statement or suggested solution to an observation. A hypothesis aims at finding a possible explanation/answer to a question, which is subject to testing. One important aspect of formulating a hypothesis is that it tends to connect the independent variable with the dependent/measurable variable.
The statement "RED IS A BEAUTIFUL COLOR" cannot be considered a hypothesis because it does not aim to answer a question that can undergo experimental testing. This statement can not be measured via experimentation. The statement is not a possible answer to a question but rather a personal opinion about something.
-- As she lands on the air mattress, her momentum is (m v)
Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down
-- As she leaves it after the bounce,
Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up
-- The impulse (change in momentum) is
Change = (60 kg-m/s up) - (300 kg-m/s down)
Magnitude of the change = <em>360 km-m/s </em>
The direction of the change is <em>up /\ </em>.
Answer:
y₀ = 10.625 m
Explanation:
For this exercise we will use the kinematic relations, where the upward direction is positive.
y = y₀ + v₀ t - ½ g t²
in the exercise they indicate the initial velocity v₀ = 8 m / s.
when the rock reaches the ground its height is zero
0 = y₀ + v₀ t - ½ g t²
y₀i = -v₀ t + ½ g t²
let's calculate
y₀ = - 8 2.5 + ½ 9.8 2.5²
y₀ = 10.625 m
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e