Answer:
The correct option is A)
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Explanation:
Given that Dante is leading a parade across the main street in front of city hall.
Let, Initial location of parade is 0i+0j
One block of city is one units on the XY- graph
Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south
New location of parade is 4i-3j
Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.
Final location of parade is (4i-3j)+(-1i+9j)=3i+6j
Displacement is given by
Displacement = (Final destination)-(Initial destination)
Displacement = (3i+6j)-(0i+0j)=3i+6j
Thus,
Magnitude of displacement =
= 6.71 m
Direction of displacement =
=
= 63.43 NE
Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east
Answer:
The magnitude of force must you apply to hold the platform in this position = 888.89 N
Explanation:
Given that :
Workdone (W) = 80.0 J
length x = 0.180 m
The equation for this work done by the spring is expressed as:
Making the spring constant the subject of the formula; we have:
Substituting our given values, we have:
The magnitude of the force that must be apply to the hold platform in this position is given by the formula :
F = 888.89 N
Answer:
B) Because the Space Station is constantly in free-fall around the Earth.
Explanation:
Anything that is falling experiences an upward force on them. For example when a person is going down in a lift they will experience something that is pushing them upwards. This happens due to the fact that the total acceleration the body is feeling is less than the acceleration due to graviity.
The force on a body which is falling is
Where,
m = Mass of object
g = acceleration due to gravity
a = acceleration the object is experiencing.
a = g. So, the force becomes zero and the object experiences weightlessness.
Hence, the astronauts in the space station experience weightlessness due to fact that the Space Station is constantly in free-fall around the Earth.
Answer:
Explanation:
Given;
original length of steel, L₁ = 1.00 m
original length of invar, L₁ = 1.00 m
coefficients of volume expansion for steel, = 3.6 × 10⁻⁵ /°C
coefficients of volume expansion for invar, = 2.7 × 10⁻⁶ /°C
temperature rise in both meter stick, θ = 20.5°C
Difference in length, can be calculated as:
L₂ = L₁ (1 + αθ)
L₂ = L₁ + L₁αθ
L₂ - L₁ = L₁αθ
ΔL = L₁αθ
Where;
ΔL is difference in length
α is linear expansivity =
Difference in length, for steel at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m
Difference in length, for invar at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m