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Harrizon [31]
2 years ago
11

a certain car is capable of accelerating at a rate of 10.60m/s^2. how long does it take for this car to go from a speed of 55/mi

/h to a speed of 60 mi/hr
Physics
1 answer:
irina [24]2 years ago
7 0
Use the equation:
v = u + at
because it has all the variables given and need to find. First you must make sure units are the same. Acceleration is given in m/s^2. Check to make sure this is correct before using the following solution. Convert acceleration from meters per second squared to miles per hour squared.

look up conversion meters to miles
1 mile = 1609.34 meters
10.60 m/s^2 * 1mile/1609.34m = 0.0069 mi/s^2
there are 3600 seconds in an hour, then there are 3600s*3600s in 1 hr^2
0.0069 mi/s^2 * 3600s*3600s/hr^2 = 85362 mi/hr^2

Then using: v = u +at
60 = 55 + 85362t
60 - 55 = 85362t
5 = 85362t
5/85362 = t
5.9 x 10^-5 hr = t
very small value in hours, multiply by 3600 to convert to seconds.

5.9 x 10^-5 hr * 3600s/hr = 0.21 seconds
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The distance an object falls, from rest, in gravity is

                         D  =  (1/2) (G) (T²)

                        'T' is the number seconds it falls.

In this problem,

                         0.92 meter = (1/2) (9.8) (T²)

Divide each side by  4.9 :   0.92 / 4.9 = T²

Take the square root
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                                                  0.433 sec = T    

The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor.  BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.

The pencil is in the air for  0.433 second.
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from the edge of the table.
 
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Using the strap at an angle of 31.0° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15.0
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Answer:

<h2>154.73N</h2>

Explanation:

The question is incomplete. Here is the complete question.

Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.

Check the diagram related to the question in the attachment below for better understanding.

The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.

The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).

Ty = 15sin31°

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W = mass * acceleration due to gravity

W = 15.0*9.8

W = 147N

The normal force is therefore expressed as;

N = Ty + W

N = 7.73 + 147

N = 154.73N

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