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2 years ago

A track coach records how long it takes a sprinter to run 100 meters the track coach is recording what dimension of behavior

1 answer:

5 0

Behavior has at least six dimensions, which are: frequency, duration, latency, topography, locus, and force. Since the coach is recording how long it takes, the track coach is recording the duration behavior because duration is a synonym for time. Duration is your answer.

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A car accelerates uniformly in a straight line

julia-pushkina [17]

The car travels a distance d from rest with acceleration a after time t of

d = 1/2 a t²

It covers 69 m with 2.8 m/s² acceleration, so that

69 m = 1/2 (2.8 m/s²) t²

t² = 2 (69 m) / (2.8 m/s²)

t ≈ 7.02 s

where we take the positive square root because we're talking about time *after* the car begins accelerating.

8 0

3 years ago

The gravitational force between two objects is 2400 N. What will be the gravitational force between the objects if the mass of o

Sindrei [870]

Answer:

4800N

Explanation:

F=GmM/r^2

F=infinite m

F=2F

=2(2400N)

=4800N

7 0

3 years ago

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig

atroni [7]

Answer:

The plane is 2353.7 mi from the starting position.

Explanation:

Please, see the attached figure for a graphic representation of the problem.

We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.

vector a = (xa, ya)

vector b = (xb, yb)

where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.

Then, the vector c = a + b will be:

c = (xa + xb, ya + yb)

The module of a vector is calculated using the following expression for a vector “v”:

module of v = $\sqrt{x^{2} + y^{2} }$

Then, the module of c will be:

module of c = $\sqrt{(xa + xb)^{2} + (ya + yb)^{2}}$ = distance from starting point

Then, we have to find the components of vectors “a” and “b”

The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:

module of a = $\sqrt{xa^{2} + ya^{2} }$ = distance traveled during the first 1.5 hours.

The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = speed

t = time

Considering x0 as the starting point (x0 = 0)

x = 675 mi/h * 1.5 h = 1012.5 mi

Then:

module of a = $\sqrt{xa^{2} + ya^{2} }$ = 1012. 5 mi

Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.

Then:

(1012.5 mi)² = xa²

xa = 1012. 5mi

a = (1012.5 mi, 0)

In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:

module of b = $\sqrt{xb^{2} + yb^{2} }$ = x = x0 + v * t

Considering x0 as the point at which the plane turns (x0 = 0)

x = 675 mi / h * 2 h = 1350 mi

Using trigonometry, we can calculate xb and yb (see figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In this case:

opposite = yb

adjacent = xb

hypotenuse = module of “b”

Then:

sin 10° = yb / module of “b”

sin 10° * module of “b” = yb

In the same way:

cos 10° * module of “b” = xb

Since module of “b” = 1350 mi

xb = 1329.5 mi

yb = 234.4 mi

b = (1329.5 mi, 234.4 mi)

The vector c = a+b can now be calculated:

c = (xa + xb, ya + yb)

c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)

The module of c will be:

module of c = $\sqrt{(2342 mi)^{2} + (234.4 mi)^{2} }$ = 2353.7 mi

The plane is 2353.7 mi from the starting position.

4 0

2 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

2 years ago