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pychu [463]
3 years ago
12

What is the velocity of a 1.3 kg puppy with a forward momentum of 6 kg m/s​

Physics
1 answer:
Alex Ar [27]3 years ago
5 0

Answer:

by using p = mv equation we can find v,

6 = 1.3 v

4.615 = v

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You are given two vectors vector A = 4.9 at 31o vector B = 6 at 156o Angles are measured counterclockwise from the x-axis. What
Ket [755]

Answer:

   C_{y} = 4.96  and     θ' = 104,5º

Explanation:

To add several vectors we can decompose each one of them, perform the sum on each axis, to find the components of the resultant and then find the module and direction.

Let's start by decomposing the two vectors.

Vector A

             sin θ = A_{y} / A

             cos θ = Aₓ / A

             A_{y} = A sin  θ

             Ax = A cos θ

             A_{y} = 4.9 sin 31 = 2.52

             Ax = 4.9 cos 31 = 4.20

Vector B

           B_{y} = B sin θ

           Bx = B cos θ

           B_{y} = 6 sin 156 = 2.44

           Bx = 6 cos 156 = -5.48

The components of the resulting vector are

X axis

         Cx = Ax + B x

         Cx = 4.20 -5.48

         Cx = -1.28

Axis y

         C_{y} = Ay + By

         C_{y} = 2.52 + 2.44  

         C_{y} = 4.96

Let's use the Pythagorean theorem to find modulo

         C = √ (Cₙ²x2 + Cy2)

         C = Ra (1.28 2 + 4.96 2)

         C = 5.12

We use trigonemetry to find the angle

         tan θ = C_{y} / Cₓ

          θ’ = tan⁻¹ (4.96 / (1.28))

           θ’ = 75.5

como el valor de Cy es positivo y Cx es negativo el angulo este en el segundo cuadrante, por lo cual el angulo medido respecto de eje x positivo es

       θ’ = 180 – tes

        θ‘= 180 – 75,5

        θ' = 104,5º

7 0
3 years ago
consider the free-body diagram. if you want the box to move, the force applied while dragging must be greater than the
NeTakaya

Answer:

Force of static friction between the two surfaces

Explanation:

When two surfaces come into contact, they exert a force that resist the sliding of the two surfaces. This force is called static friction.

This force is given by the relation

                                       F_{s}=\mu_{s}\eta

Where,

                             μ - coefficient of static friction

                             η - normal force acting on the body

When a force acts on a body placed on a rough surface, it doesn't do any work if the applied force was less than the force of static friction.

So, in order to move the body, the applied force should be greater than the force of static friction.

6 0
3 years ago
If I work out rotational energy to be 102.2J which equals kg.M/s^2, and I hadn't factored time into it, would that be Joules per
Marina CMI [18]

Answer:

0.057 joules is needed to create the total rotational energy each second.

Explanation:

The energy rate is the ratio of total energy to time, which coincides with the definition of power at constant rate:

\dot W = \frac{\Delta E}{\Delta t}

\dot W = \frac{102.2\,J}{\left(30\,min\right)\cdot \left(60\,\frac{s}{min} \right)}

\dot W = 0.057\,\frac{J}{s}

\dot W = 0.057\,W

0.057 joules is needed to create the total rotational energy each second.

7 0
3 years ago
What is the only "power" unit of measurement in light energy?
n200080 [17]
I believe it wattage or watts
8 0
3 years ago
Ben Rushin is waiting at a stop light. Turns green, ben accelerated from rest at a rate of 6.00 m/s squared for a time of 4.10 s
Lera25 [3.4K]
D= vt +.5at^2
since he started at rest, v (initial velocity) is 0
so d=.5at^2
d = .5 (6m/s^2) (4.1s)^2 
then put that into a calculator.

4 0
3 years ago
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