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andrey2020 [161]
4 years ago
14

Over a span of 6.0 seconds, a car changes it's speed from 89 km/h to 37 km/h. What is its average acceleration in meters per sec

ond squared?
a. 2.4
b. -8.7
c. -2.4
d. -6.2
Physics
1 answer:
scoundrel [369]4 years ago
3 0

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr

Acceleration = (-52 km/hr) / (6 sec)

Acceleration = (-26/3) km/(hr·sec)

Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²

(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)

= -26,000/(3 · 3600) m/s²

<em>Acceleration = -2.41 m/s²</em>

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One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:
Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

      a) The free body diagram is in the attachment

      b) The normal force is N = 737 N

      c) The mass of the table is  10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

          ∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

             N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

                N -  W_m -w_{table} = 0

                N =  W_m +w_{table}

                N = M_m g + w_{table}  

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

                N = 65 9.8 + 100

                N = 737 N

c) To calculate the mass of the table we use the relation

                W = m_{table} g

                m_{table} = \frac{w_{table}}{g}

                m_{tabble}= \frac{100}{9.8}  

               m_{table}e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

      a) The free body diagram is in the attachment

      b) The normal force is N = 737 N

      c) The mass of the table is  10.2 kg

Learn more here:  brainly.com/question/19860811

7 0
3 years ago
Initially, a particle is moving at 5.25 m/s at an angle of 35.5° above the horizontal. Three seconds later, its velocity is 6.0
ivolga24 [154]

Answer:

 a =( -0.32 i ^ - 2,697 j ^)  m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

  v₀ₓ = v₀ cos θ

  v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

   v_{oy} = v₀ sin θ

 v_{oy}= 5.25 sin35.5

v_{oy} = 3.05 m / s

Final

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vₓ = 3.31 m / s

v_{y} = v₀ sin θ

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v_{y} = -5.04 m / s

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4 years ago
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natali 33 [55]
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3 years ago
The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G
Charra [1.4K]

Answer:

Atomic Size and Mass:

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7 0
3 years ago
The diagram shows a ballistic pendulum. A 200 g bullet is fired into the suspended 4 kg block of wood and remains embedded insid
adoni [48]

Question

What was the initial momentum of the bullet before collision?

Answer:

10 Kg.m/s

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