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Liula [17]
3 years ago
11

A rubber block weighing 60. newtons is resting on a horizontal surface of dry asphalt. what is magnitude of the minimum force ne

eded to start the rubber block moving across the dry asphalt?
Physics
1 answer:
levacccp [35]3 years ago
8 0
First you must look for the coefficient of friction of the dry asphalt. For this case, searching in google, the value is 0.85. Then, you must find the normal force by making a free-body diagram. In the diagram, you sum up forces in vertical direction and find that the normal force is 60N. Then, by sum of forces in a horizontal direction, the friction force will be given by:F = (60) * (0.85) = 51N.51N is the magnitude of the minimum force needed to start the rubber block moving across the dry asphalt
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Explain what the core muscles do and why it's important to have a strong set of core muscles.
Alika [10]

Answer:

Since your spine runs by your core, having a strong core will help protect and stabilize your spine, and you're also are able to control movements such as walking or standing better with a strong core.

7 0
3 years ago
A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of
jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

7 0
3 years ago
What is the momentum of a 1550kg car that is traveling 38.0 m/s?
konstantin123 [22]

Answer:

p = 58,900 kg m/s

Explanation:

p = m × v

p = 1,550 × 38.0

p = 58,900 kg m/s

5 0
2 years ago
A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with
Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

            Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

            Em_{f} = K = ½ m v²

            Em₀ = Em_{f}

            ½ m v₀² + m g h = ½ m v²

            v² = v₀² + 2 g h

             

Let's calculate

             v = √(1.23² + 2 9.8 1.69)

             v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

            W = ΔK

            -fr x = 0- ½ m v²  

               

Newton's second law

              N- W = 0

     

The equation for the friction is

               fr = μ N

               fr = μ m g

We replace

             μ m g x = ½ m v²

             x = v² / 2μ g

Let's calculate

            x = 5.89² / (2 0.255 9.8)

            x = 6.94 m

6 0
3 years ago
Read 2 more answers
A wave with a period of 0.008 second has a frequency of
coldgirl [10]

The frequency of a wave is the reciprocal of its period.

A period of 0.008 sec means a frequency of

         1 / 0.008 sec  =  125 per sec .  (125 Hz)

8 0
3 years ago
Read 2 more answers
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