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3 years ago

An olympic hurdler accelerates at a rate of 15.5 m/s2 . what is the rate in miles/min2 ?

Physics

2 answers:

Alborosie3 years ago

6 0

Answer: 34.68 mile/min^2

Explanation:

1 mile = 1609 meter

1 min = 60 s then squared will be 3600s

Mile/min^2 = 2.24

Give 15.5m/s^2

We multiply

15.5 by 2.24

= 34.68 mile/min^2

Wittaler [7]3 years ago

5 0

You have to use conversion units for this. For every 1 mile, there is an equivalent amount of 1,609 meters. For every 1 minute, there is an equivalent amount of 60 seconds. Using the dimensional analysis approach, the solution is as follows:

Acceleration = 15.5 m/s²*(1 mile/1,609 m)*(60 s/1 min)² = 34.68 miles/min²

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a car company wants to ensure its newest model can stop in 50 m when traveling at 30 m/s (which is about 108 km/h). if we assume

Brut [27]

A decrease in velocity is referred to as deceleration. If car is moving at 30 m/s and stop in 50 m .The value of deceleration is 11.56 ms−2.

<h3>How to calculate deceleration ?</h3>

While acceleration is motion in which an object's speed varies every second, deceleration is motion that causes an object to slow down.

We are aware that acceleration refers to an object's rate of increase in speed, and deceleration refers to an object's rate of decrease in speed. For instance, when we apply the brakes while driving, we benefit from the vehicle's ability to decelerate and slow down.

The Deceleration Formula is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing, if starting velocity, final velocity, and time taken are given.

velocity of car = 30 m/s

car need to stop in 50m

Deceleration a = v^2 – u^2 / 2s

= 0^2 - 50^2 / 2*30

= 11.56

Deceleration of the care = 11.56 ms−2

To learn more about deceleration refer :

brainly.com/question/75351

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8 0

1 year ago

What is tarzan's speed vf just before he reaches jane? express your answer in meters per second to two significant figures?

e-lub [12.9K]

Before swinging, T has only potential energy, (no speed)
Ui = mgh
Where h is the vertical displacement of T
From the laws of geometry,
cos45 = (L-h)/L
cos45 = 1-h/L
h/L = 1-cos45
h = L(1-cos45)

Therefore
Ui = mgL(1-cos45)

Proceeding the same way,
Twill raise to aheight of h' due to swing
h' = L(1-cos30)
The PE of T after swing is
Uf = mgh'
Uf = mgL(1-cos30)

Along with the PE , T has some kinetic energy results due to the moment.
Tf = 0.5*mv^2

According to the law of conservation of energy,
Ui = Uf+Tf
mgL(1-cos45) = mgL(1-cos30) + 0.5*mv^2
gL(co30-cos45) = 0.5*v^2
9.8*20*(co30-cos45) = 0.5*V^2
v = 7.89 m/s

<span>The speed f T after swing is 7.89 m/s</span>

3 0

3 years ago

Two positive charges are labeled q Subscript 1 baseline and q Subscript 2 baseline are 0.1 m apart. Two positively charged parti

Snowcat [4.5K]

Answer:

Explanation:

Let that point be at a distance x from q1

Then Kq1/x^2= Kq2/ (s-x)^2

Taking square roots and simplifying, x =s /[1+(q2/q1)^0.5]

Assuming an identical distance, the rigidity of Q on 2Q is equivalent in value to the rigidity of 2Q on Q. for that reason, had the area R been stored an identical, the two forces could be equivalent. inspite of the shown fact that, via fact the area is being decreased, we could constantly consult with the equation we use to calculate those forces: F = ok(Q1xQ2)/(R^2) because R is squared and is being halved, the final result's that's it being divided by potential of a million/4. for that reason, the rigidity would be expanded by potential of four, and be 4F.

4 0

4 years ago

A sled plus passenger with total mass m = 53.1 kg is pulled a distance d = 25.3 m across a horizontal, snow-packed surface for w

Alex Ar [27]

Answer:

Explanation:

Force of friction

F = μ mg

μ is coefficient of friction , m is mass and g is acceleration due to gravity .

If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional force

The vertical component of applied force will reduce the normal force or reaction force from the ground

Reaction force R = mg - f sin28.3

frictional force = μ R where μ is coefficient of friction

frictional force = μ x (mg - f sin28.3 )

This force should be equal to horizontal component of f

μ x (mg - f sin28.3 ) = f cos 28.3

μ x mg = f μsin28.3 + f cos 28.3

f = μ x mg / (μsin28.3 + cos 28.3 )

a )

work done by pulling force = force x displacement

f cos28.3 x d

μ x mg d cos28.3 / (μsin28.3 + cos 28.3 )

b ) Putting the given values

= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )

= 1796.76 / (.073 + .88 )

= 1885.37 J

c )

Work done by frictional force

= frictional force x displacement

= - μ x (mg - f sin28.3 ) x d

= - μ x mgd + f μsin28.3 x d

= - μ x mgd + μsin28.3 x d x μ x mg / (μsin28.3 + cos 28.3 )

d )

Putting the values in the equation above

- .155 x 53.1 x 9.8 x 25.3 +

.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)

= -2040.67 + 149.92 / .95347

= -2040.67 + 157.23

= -1883.44 J .

6 0

3 years ago

Which of the following correctly compares the uses of bar graphs versus pie charts?

slavikrds [6]

Bar graphs show data involved in distinct categories that do not overlap; whereas, pie charts show data as parts out of a whole (such as out of 100%).

6 0

3 years ago