A sprinter starts from rest and accelerates to her maximum speed of 10.5 m/s in a distance of 11.0 m. (a) What was her acceleration, if you assume it to be constant? 0 Incorrect: Your answer is incorrect. Units are required for this answer. seenKey 5.01 m/s^2 (b) If this maximum speed is maintained for another 96.3 m, how long does it take her to run 107.3 m?

Answers

Answer 1

Answer:

Answer:

a)a=5.01m/s^2

b)t=11.26s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)

{Vf^{2}-Vo^2}/{2.a} =X(2)

X=Xo+ VoT+0.5at^{2} (3)

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

to solve the question a, we can use the ecuation number 2

Vo=0

Vf=10.5 m/s

x=11m

{Vf^{2}-Vo^2}/{2.a} =X

{Vf^{2}-Vo^2}/{2.x} =a

{10.5^{2}-0^2}/{2x11} =a

a=5.01m/s^2

to find the time we can use the ecuation number 1

Vf=Vo+a.t

t=(Vf-Vo)/a

t=(10.5-0)/5.01=2.09s

part b

in this case the spees is constant, so the movement is defined by the following ecuation

X=VT

t=x/v

t=96.3/10.5=9.17s

to find the total time we sum the times when the speed is constant and when the acceleration is constan

t=9.17+2.09

t=11.26s

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Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s , the induced emf in the second coil has a magnitude of 1.60×10−3 V . Part A What is the mutual inductance of the pair of coils? MM = nothing H Request Answer Part B If the second coil has 22 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ? ΦΦ = nothing Wb

Answers

To solve this problem we need to use the emf equation, that is,

$E=m(dI)/(dT)$

Where E is the induced emf

I the current in the first coil

M the mutual inductance

Solving for a)

$M=(E)/((dI)/(dT))\nM=(1.6*10^(-3))/(0.245)=6.53*10^(-3)H$

Solving for b) we need the FLux through each turn, that is

$\Phi=(MI)/(N)$

Where N is the number of turns in the second coil

$\Phi=(6.53*10^(-3)*1.25)/(22)=3.71*10^(-4)Wb$

An element emits light at two nearly equal wavelengths, 577 nm and 579 nm If the light is normally incident on a diffraction grating with 2000 lines/cm., what is the distance between the 3rd order fringes of the two wavelengths on a screen 1 m from the grating?

Answers

Answer:

Explanation:

d = width of slit = 1 / 2000 cm =5 x 10⁻⁶ m

Distance of screen D = 1 m.

wave length λ₁ and λ₂ are 577 x 10⁻⁹ and 579 x 10⁻⁹ m.respectively.

distance of third order bright fringe = 3.5 λ D/d

for 577 nm , this distance = 3.5 x 577 x 10⁻⁹ x 1 /5 x 10⁻⁶

= .403 m = 40.3 cm

For 579 nm , this distance = 3.5 x 579 x 1 / 5 x 10⁻⁶

= 40.5 cm

Distance between these two = 0.2 cm.

A nearsighted person has a far point of 40cm. What power spectacle lens is needed if the lens is 2cm from the eye

Answers

Answer:

The value is $p = - 2.63 \ Diopters$

Explanation:

From the question we are told that

The value of the far point is $a = 40 \ cm = 0.4 \ m$

The distance of the lens to the eye is $b = 2 \ cm = 0.02$

Generally

$1 Diopter = > 1 m^(-1)$

Generally the power spectacle lens needed is mathematically represented as

$p = (1)/(d_o ) + (1)/(d_i)$

Here $d_o$ is the object distance which for a near sighted person is $d_o = \infty$

And $d_i$ is the image distance which is evaluated as

$d_i = b - a$

=> $d_i = 0.02 - 0.4$

=> $d_i = -0.38 \ m$

$p = (1)/(\infty ) + (1)/(-0.38)$

=> $p = 0 - 2.63$

=> $p = - 2.63 \ Diopters$

A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?

Answers

Answer:

19.3 m/s

Explanation:

Take down to be positive. Given:

Δy = 19 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (19 m)

v = 19.3 m/s

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Answers

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is $\rm 0.430 \; kg\;m^2$ .

Given :

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope.
The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.
The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.
The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.

a) First, determine the acceleration of the B block.

$\rm s = ut + (1)/(2)at^2$

$\rm 1.8 = (1)/(2)* a* (2)^2$

$\rm a = 0.9\; m/sec^2$

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

$\rm \sum F=ma$

$\rm mg-T_b=ma$

$\rm T_b = m(g-a)$

$\rm T_b = 7* (9.8-0.9)$

$\rm T_b = 62.3\;N$

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

$\rm \sum F=ma$

$\rm T_a=ma$

$\rm T_a = 2.1* 0.9$

$\rm T_a = 1.89\;N$

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

$\rm \sum \tau = I\alpha$

$\rm T_br-T_ar = I\alpha$

$\rm I = ((T_b-T_a)r^2)/(a)$

Now, substitute the values of the known terms in the above expression.

$\rm I = ((62.3-1.89)(0.080)^2)/(0.90)$

$\rm I = 0.430 \; kg\;m^2$

For more information, refer to the link given below:

brainly.com/question/2287912

Answer:

(a) 62.3 N

(b) 1.89 N

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B. There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A. There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall building. Assume air resistance is negligible, and g = 9.8 m/s2. What is the magnitude of the horizontal displacement of the rock?

Answers

Answer:

The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Explanation:

Given that,

Initial speed = 11.5 m/s

Angle = 50.0

Height = 30.0 m

We need to calculate the horizontal displacement of the rock

Using formula of horizontal component

$v_(x)=u\cos\theta$

Put the value into the formula

$v_(x)=11.5*\cos50$

$v_(x)=7.39\ m/s$

Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Final answer:

The question is about determining the horizontal displacement of a projectile based on the given initial speed and projection angle and the height of the launch. This can be calculated using the equations of motion, specifically those pertaining to projectile motion.

Explanation:

In this problem, we're dealing with projectile motion. The stone being thrown is the projectile in this case. The horizontal displacement, also known as range, of a projectile can be defined using the formula: range = (initial speed * time of flight) * cosθ, where θ is the angle of projection. The initial speed is given as 11.5 m/s and the angle as 50 degrees. Now, we need to calculate the time of flight. This can be found by the formula: time of flight = (2 * initial speed * sinθ) / g. Considering g, the acceleration due to gravity, as 9.8 m/s², we can find the time of flight and thus calculate the range. Always remember that while the vertical motion of a projectile is affected by gravity, the horizontal motion remains constant.

Learn more about Projectile Motion here:

brainly.com/question/29545516

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