Explanation:
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A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900
2 answers:
Answer:
Final velocity of the car, v = 17 m/s
Explanation:
It is given that,
Initial velocity of the car, u = 5 m/s
Acceleration of the car,
Time, t = 6 s
We need to find the final velocity of the car after 6 seconds. It can be calculated using first equation of motion as :
, v is the final velocity of the car
v = 17 m/s
So, the final velocity of the car is 17 m/s. Hence, this is the required solution.
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Answer:
The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;
(b) 4, 3, -3, 1/2.
Explanation:
To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;
4, 4, 4, 1/2
4, 3, -3, 1/2
4, 3, 0, 0
4, 5, 7, -1/2
4, 4, -5, 1/2
Let us label them as a to as follows
(a) 4, 4, 4, 1/2
(b) 4, 3, -3, 1/2
(c) 4, 3, 0, 0
(d) 4, 5, 7, -1/2
(e) 4, 4, -5, 1/2
Next we note the rules for the assignment and arrangement of quantum numbers are as follows
Number Symbol Possible values
Principal Quantum Number .......n........................1, 2, 3, ......n
Angular momentum quantum
number...............................................l.........................0, 1, 2, .......(n - 1)
Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l
Spin Quantum Number.................m.....................+1/2, -1/2
We are meant to analyze each of the arrangement for acceptability.
Therefore for (a),
we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.
Therefore (a) is not acceptable.
(b) Here we note that
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable
The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.
(c) Here we have
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable
The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 0 ∉ (+1/2, -1/2) → not acceptable
Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.
(d) Here we have;
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable
The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = -1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.
(e) Here we have;
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable
The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.
Answer:
The net force = 0
Explanation:
The given information includes;
The mass of the crate = 250 kg
The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)
In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.
The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate
The weight of the crate, ↓ = 250 kg × 9.81 m/s² = 2,452.5 N
The force the helicopter should provide to just lift the crate, ↑ = The weight of the crate = 2,452.5 N
The net force, =
↑ -
↓ = 2,452.5 N - 2,452.5 N = 0
The net force = 0.