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liraira [26]
3 years ago
8

A motorboat traveling on a straight course slows

Physics
1 answer:
never [62]3 years ago
3 0

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

  • 65 km/h → 18.0556 m/s
  • 35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

  • v² = v₀² + 2aΔx

Substitute the known values into the equation.

  • (9.72222)² = (18.0556)² + 2a(45)
  • 94.52156173 = 326.0046914 + 90a
  • -231.4831296 = 90a
  • a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

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REY [17]

Answer:

Explanation:

El impulso aplicado a la pelota produce una variación en su momento lineal.

J = m (V -Vo)

Conviene elegir positivo el sentido de la velocidad final.

J = 0,100 kg [40 - (- 20)] m/s = 6 kg m/s

Saludos Herminio

7 0
2 years ago
I need someone to help me work through this
Sergio039 [100]
The average rate of change of distance over the time interval 
3 ≤ t ≤ 6 represents the coin's average velocity over that interval.
3 0
2 years ago
Una carga de -10Mc está situada a 20cm delante de otra carga de 5 Mc. Calcular la fuerza electrostática en Newton ejercida por u
tino4ka555 [31]

Answer:

a)  force between them is attraction,   b)  F = 1.125 10⁻² N

Explanation:

In this case the electric force is given by Coulomb's law

          F =k \frac{q_1q_2}{r^2}

           

In the exercise they give us the values ​​of the loads

          q1 = - 10 mC = -10 10⁻³ C

          q2 = 5 mC = 5 10⁻³ C

           d = 20 cm = 0.20 m

let's calculate

          F = 9 10⁹ \frac{10 \ 10^{-3} \ 5 \ 10^{-3}}{0.20^2}

          F = 1.125 10⁻² N

To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction

7 0
3 years ago
Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 C
Len [333]

Answer: 71.93 *10^3 N/C

Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:

∫E*dr=Q inside/εo  Q inside is given by: λ*L then,

E*2*π*r*L=λ*L/εo

E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C

6 0
2 years ago
Need help on this thank you
Semmy [17]

Answer:

TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.

7 0
3 years ago
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