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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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What is the speed of a wave that has a frequency of 45 Hz and a wavelength of 0.1 meters?
Yuri [45]
<span>λν=c
(wavelength x frequency = speed)

speed = 45 x 0.1
= 4.5 m/s</span>
6 0
3 years ago
Read 2 more answers
You are traveling 70 mph (31.3 m/s) and slam on the brakes to avoid hitting another car. How far do you travel if it takes you 8
Arlecino [84]

You traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

<u>Given the following data:</u>

  • Initial velocity, U = 31.3 m/s  
  • Time, t = 8.5 seconds.

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find the distance traveled, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

S = ut + \frac{1}{2} at^2

Where:

  • S is the distance travelled.
  • u is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the parameters into the formula, we have;

S = 31.3(8.5) + \frac{x}{y} (9.8)(8.5^2)\\\\S = 266.05 + 4.9(72.25)\\\\S = 266.05 + 354.025

<em>Distance, S</em><em> = </em><em>620.075 meters.</em>

Therefore, you traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

Read more: brainly.com/question/8898885

6 0
2 years ago
20 characters or more
andreev551 [17]

Answer:

yes 20 characters or more

Explanation:

3 0
2 years ago
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0
3 years ago
Can I also get help on this??
Xelga [282]

Answer:

25

Explanation:

8 0
2 years ago
Read 2 more answers
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