PHYSICS COLLEGE

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Answers

Answer 1

Answer:

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

W = $F_(total) .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) .d =(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) = ((1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i))/(d)$

$F_(total=) ((1)/(2) X 62 X6^(2) -(1)/(2) X 62 X2^(2) )/(25)$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_(sprinter) = F_(total) + F_(wind) = 39.7 + 30 = 69.68 N$

Answer 2

Answer:

Force exerted by sprinter = 69.68 N

Explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

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A certain lightning bolt moves 40 C of charge. How many fundamental units of charge |qe| is this?

Answers

Answer to A certain lightning bolt moves 40.0 C of charge. How many fundamental units of charge | qe | is this? . ... charge, N is the total number of electron or protons that constitute total charge Q.

A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pull the bucket of water upwards. the total mass of the bucket of water is f2= 3.9kg. -Calculate how much work Wg in J gravity does on the bucket filled with water as the farmer lifts it up the well.
-Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.

Answers

Final answer:

To calculate the work done by gravity on the bucket of water as it is lifted up the well, multiply the weight of the bucket by the lifting distance. The net work done on the bucket by the force applied by the farmer and gravity is the sum of the work done by both forces. The net work is represented by the equation Wnet = W1 + Wg.

Explanation:

To calculate how much work gravity does on the bucket filled with water as the farmer lifts it up the well, we need to multiply the force of gravity (weight) by the vertical distance the bucket is lifted. The equation for work is W = Fd, where W is the work done, F is the force, and d is the distance. In this case, the force of gravity is the weight of the bucket, which can be calculated by multiplying the mass by the acceleration due to gravity (9.8 m/s^2).

So, the work done by gravity (Wg) on the bucket is Wg = Fg * d = (m * g) * d = (3.9 kg * 9.8 m/s^2) * d = 38.22 d Joules.

To calculate the net work done on the bucket by the two forces, we can use the equation Wnet = W1 + Wg, where W1 is the work done by force F1 and Wg is the work done by gravity. Since force F1 and the displacement (lifting distance) are both vertical, the work done by F1 is given by W1 = F1 * d.

Therefore, the net work done on the bucket by forces F1 and gravity is Wnet = F1 * d + Fg * d = (57.5 N) * d + (3.9 kg * 9.8 m/s^2) * d = (57.5 N + 38.22 d) Joules.

Learn more about Work done by forces here:

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A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?

Answers

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?

Answers

Answer:

The inductance is $L = 40\mu H$

Explanation:

From the question we are told that

The number of turns is $N = 163 \ turns$

The diameter is $D = 6.13 \ mm = 6.13 *10^(-3) \ m$

The length is $l = 2.49 \ cm = 0.0249 \ m$

The radius is evaluated as $r = (d)/(2)$

substituting values

$r = (6.13 *10^(-3))/(2)$

$r = 3.065 *10^(-3) \ m$

The inductance of the Tarik's solenoid is mathematically represented as

$L = (\mu_o * N^2 * A )/(l )$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^(-7) \ N/A^2$

A is the area which is mathematically evaluated as

$A = \pi r^2$

substituting values

$A = 3.142 * [ 3.065*10^(-3)]^2$

$A = 2.952*10^(-5) \ m^2$

substituting values into formula for L

$L = ( 4\pi *10^(-7) * [163]^2 * 2.952*10^(-5) )/(0.0249 )$

$L = 40\mu H$

Two narrow slits separated by 1.5 mm are illuminated by 514 nm light. Find the distance between adjacent bright fringes on a screen 5.0 m from the slits. Express your answer in meters using two significant figures.

Answers

The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Given data :

separation between slits ( d ) = 1.5 x 10⁻³ m

wavelength of light ( λ ) = 514 * 10⁻⁹ m

Distance from narrow slit ( D ) = 5.0 m

Determine the distance between the adjacent bright fringes

we apply the formula below

w = D * λ / d ---- ( 1 )

where : w = distance between adjacent bright fringes

Back to equation ( 1 )

w = ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³

= 1.7 * 10⁻³ M

Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Learn more about bright fringes calculations : brainly.com/question/4449144

Answer:

$1.7* 10^(-3)$ m

Explanation:

d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m

λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m

D = Distance of the screen from the narrow slits = 5.0 m

w = Distance between the adjacent bright fringes on the screen

Distance between the adjacent bright fringes on the screen is given as

$w = (D\lambda )/(d)$

$w = ((5.0)(514* 10^(-9)) )/(1.5* 10^(-3))$

$w = 1.7* 10^(-3)$ m

Two blocks of masses 1 = 700 and 2 = 1100 are connected by a cord of negligible mass and hung over a diskshaped pulley, as shown in the figure. The pulley has a mass of = 1.50 and a radius of = 14 , and rotates about a lightweight axle through its center. The axle itself is hung from the ceiling by two like cords of negligible mass and is held horizontally. The system is released from rest. a) Draw a free-body diagram for each of the blocks and the pulley separately. b) Find the magnitude of the acceleration of the blocks. c) Find the magnitude of the angular acceleration of the pulley. d) Find the magnitude of tensions in the cords, 1, 2, and 3. (See the figure.)