Answer:
The sled needed a distance of 92.22 m and a time of 1.40 s to stop.
Explanation:
The relationship between velocities and time is described by this equation: 
, where 
is the final velocity, 
is the initial velocity, 
the acceleration, and 
is the time during such acceleration is applied.
Solving the equation for the time, and applying to the case: 
, where 
because the sled is totally stopped, 
is the velocity of the sled before braking and, 
is negative because the deceleration applied by the brakes.
In the other hand, the equation that describes the distance in term of velocities and acceleration:
, where 
is the distance traveled, is the initial velocity, the time of the process and, is the acceleration of the process.
Then for this case the relationship becomes: 
.
<u>Note that the acceleration is negative because is a braking process.</u>
A star is located 5.9 light years from Earth.
We know that : 1 light year = 9.46 trillion kilometers.
We will calculate the distance in trillion kilometers multiplying the number of light years by 9.46:
5.9 * 9.46 = 55.814
Answer: The distance is 55.814 trillion km.
It is about 100oC at a pressure of 1.1 atmosphere. Hope this helps.
Use the distance swan and the time elapsed in that interval.
Average velocity = distance / time
Average velocity = [4.0 m + 3.0m] / 3.2 s = 2.1875 m/s
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
