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Elis [28]
3 years ago
8

A car from the beginning was traveling at 27.8m/s when he stepped on the brakes and stopped. According to the DMV, this would le

ave a skid mark of 60m. What is his acceleration?
Physics
1 answer:
Yuliya22 [10]3 years ago
3 0

Answer:

-6.44 m/s²

Explanation:

Given:

Δx = 60 m

v₀ = 27.8 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (27.8 m/s)² + 2a (60 m)

a = -6.44 m/s²

You might be interested in
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold
lubasha [3.4K]

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

       fr - Wₓ = m    a                      (1)

Y axis  

       N- W_{y} = 0

       N = W_{y}

let's use trigonometry to find the components of the weight

        sin θ = Wₓ / W

        cos θ = W_{y} / W

        Wₓ = W sin θ

        W_{y} = W cos θ

the friction force has the formula

         fr = μ N

         fr = μ Wy

         fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

           F = fr

           F = (μ  mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

4 0
3 years ago
in which kind of radioactive decay would the number of protons in the resulting nucleus be more than in the initial nucleus?
9966 [12]

Answer:

A related type of beta decay

Explanation:

4 0
2 years ago
20 Accelerated motion is represented by a<br> line on a nonlinear distance time graph.
Varvara68 [4.7K]

Curved line

Explanation:

Acceleration of motion is represented by a curved line on a non-linear distance-time graph.

The acceleration of a non-linear motion is depicted using a parabola which is a curve. This implies that the velocity is constantly changing and the distance covered by the body is also changing with equal amount of time.

  • A plot of this will give a parabola. This can be further established using one of the equations of motion below:

    x = u + \frac{1}{2}at ²

This is a quadratic function where:

x is the distance

u is the initial velocity

t is the time

a is acceleration

A quadratic function gives a curved line which is a parabola.

Learn more:

Acceleration brainly.com/question/10932946

#learnwithBrainly

3 0
3 years ago
Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d
BartSMP [9]

La longitud <em>final</em> del puente de acero es 100.018 metros.

Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (L), en metros:

L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})] (1)

Donde:

  • L_{o} - Longitud inicial del puente, en metros.
  • \alpha - Coeficiente de dilatación, sin unidad.
  • T_{o} - Temperatura inicial, en grados Celsius.
  • T_{f} - Temperatura final, en grados Celsius.

Si tenemos que L_{o} = 100\,m, \alpha = 11.5\times 10^{-6}, T_{o} = 8\,^{\circ}C y T_{f} = 24\,^{\circ}C, entonces la longitud final del puente de acero es:

L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]

L = 100.018\,m

La longitud <em>final</em> del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

5 0
2 years ago
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi
natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0
3 years ago
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