Answer:
The tension in the part of the cord attached to the textbook is 6.58 N.
Explanation:
Given that,
Mass of textbook = 2.09 kg
Diameter = 0.190 m
Mass of hanging book = 3.08 kg
Distance = 1.26 m
Time interval = 0.800 s
Suppose,we need to calculate the tension in the part of the cord attached to the textbook
We need to calculate the acceleration
Using equation of motion
Put the value into the formula
We need to calculate the tension
When the book is moving with acceleration
Using formula of tension
Hence, The tension in the part of the cord attached to the textbook is 6.58 N.
Answer:
if your going 15000 mph all you can do is accelerate or go fast unless there is a opposite
Explanation:
Answer:
a) The astronauts would see the real length of the meter stick, i.e. L₀
b) The length of the meter stick as measured by the stationary observer will be 
Explanation:
a) Let the proper length of the meter stick be L₀
The meter stick and the astronauts on the on the space ship are on the same moving frame, therefore, they will see the exact length of the meter stick, that is, L₀
b) A stationary observer watching the space ship and meter stick travel past them will see a contracted length of the meter stick
The original length = L₀
Let the speed of the space ship = v
The contracted length, L, is related to the original length in the frame of rest by
L = L₀/γ......................(1)
Where γ = 
....................(2)
Substituting equation (2) into (1)
K so I'm not completely sure about this...
but I believe the answer would be C. It will float upward
because the buoyancy, in the image, is stronger than the Gravitational pull.
Let me know if that is right or wrong plz
Hope it's right and helps tho!
Answer:
a) I = 2279.5 N s
, b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N
Explanation:
The impulse is equal to the variation in the amount of movement.
I =∫ F dt = Δp
I = m
- m v₀
Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero
² = V₀² - 2g y
² = 0 - 2 9.8 30.0
= √588
= 24.25 m/s
a) We calculate the impulse
I = 94 24.25 - 0
I = 2279.5 N s
b) Let's join the other expression of the impulse to calculate the average force
I = F t
F = I / t
F = 2279.5 / 6 10⁻³
F = 3.80 10⁵ N
just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n
c) I = m - m v₀
I = 94 8 - 94 (-24.25)
I = 3125.5 N s
d) F = I / t
F = 3125.5 / 6 10⁻³
F = 5.21 10⁵ N
