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BlackZzzverrR [31]
3 years ago
14

A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the

person and the kayak is 74 kg, what is the magnitude of the net force acting on the kayak?
Physics
1 answer:
Iteru [2.4K]3 years ago
6 0

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

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3 years ago
Which of the following are correct statements about the way an atom is put
Ann [662]

Answer:

valenc e shell

Explanation:

4 0
3 years ago
Electric field lines moves away from positive to wards negative?
Archy [21]

The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.

6 0
2 years ago
A single slit is illuminated by light of wavelengths λa and λb, chosen so that the first diffraction minimum of the λa component
agasfer [191]

Answer:

λ_A = 700 nm ,   m_B = m_a 2

Explanation:

The expression that describes the diffraction phenomenon is

         a sin θ = m λ

where a is the width of the slit, lam the wavelength and m an integer that writes the order of diffraction

a) They tell us that now lal_ A m = 1

         a sin θ = λ_A

coincidentally_be m = 2

          a sin  θ = m λ_b

as the two match we can match

         λ _A = 2 λ _B

         λ_A = 2 350 nm

         λ_A = 700 nm

b)

For lam_B

       a sin  λ_A  = m_B  λ_B

For lam_A

        a sin θ_A = m_ λ_ A

to match they must have the same angle, so we can equal

           m_B  λ_B = m_A  λ_A

           m_B = m_A  λ_A / λ_B

           m_b = m_a 700/350

           m_B = m_a 2

8 0
3 years ago
If the current in the circuit is 10.5 A,
sergiy2304 [10]

Answer:

E) 147 V

Explanation:

Ra+Rb+Rc = 14.0

V = IR

V = (10.5)(14.0)

V = 147 V

7 0
3 years ago
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