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eduard
2 years ago
10

5. Annie drags her little red wagon with a mass of 5.00 kg, up a hill that has an angle of

Physics
1 answer:
maxonik [38]2 years ago
6 0

Answer: 25.08\ m/s^2

Explanation:

Given

mass of wagon m=5 kg

elevation

\theta =30.9^{\circ}\\F=150\ N

Sin component of weight will oppose the applied force therefore we can write

F-W\sin \theta=ma\\where\\W=weight(mg)\\a=acceleration

150-5\times 9.8\times \sin (30.09)^{\circ}=5\times a\\125.43=5a\\a=25.08\ m/s^2

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Which of the following is one way of preventing poisoning?
Iteru [2.4K]
I think its number 1

7 0
3 years ago
A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa
NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

6 0
3 years ago
Read 2 more answers
What is density?
il63 [147K]
<span>a. the amount of matter in a given volume </span>
8 0
3 years ago
Read 2 more answers
Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation
Artemon [7]

Answer:

L =   μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

        E_{L} = - L dI / dt

 

Let's use Faraday's law

       E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

        - d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

        n d Ф_B = L dI

we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

       L =   μ₀ n r / 2I

4 0
3 years ago
An action force is 40 N to the right. The reaction force must be: A. 20 N left B. 40 N left C. 20 N right D. 40 N right
Sergeeva-Olga [200]

Answer:

The answer to your question is letter B.

Explanation:

To answer this question, we must remember the third law of motion of Newton that states that For every action, there is an equal and opposite reaction.

Then, if the action force is 40 N to the right, the reaction force must be 40 N to the left.

8 0
3 years ago
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