A block, initially at rest, has a mass m and sits on aplane inclined at angle (theta). It slides a distance d before hitting a spring and compresses the spring by a maximum distance of xf. If the coefficient of kinetic friction between the plane and block is uk, then what is the force constant of the spring?

Answers

Answer 1

Answer:

Answer: k = ma + uk×mgcosθ/ xf

Explanation: The body is placed on a frictionless inclined ramp.

The weight of the object has 2 components, horizontal component (mgsinθ) and vertical component (mgcosθ).

The horizontal component of weight is responsible for making tje object slide down the plane even with no applied force.

So from newton's second law of motion

mgsinθ - uk×R = ma

Where uk = coefficient of kinetic friction.

R = normal reaction = mgcosθ

mgsinθ - uk×mgcosθ = ma

mgsinθ = ma + uk×mgcosθ

mgsinθ is the applied force in this case. This applied force compresses a spring.

According to hooke's law,

F =ke

Where F = ma + uk×mgcosθ, e =xf

F = applied force , e = extension and k = spring constant.

k = F/e

k = ma + uk×mgcosθ/ xf

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7. An engineer is using a wire that has a resistance of 1.5 . This resistance is too high for the application he is designing. The wire must be exactly 2.5 cm long. What two things could he do to reduce the wire's resistance

Answers

Answer and Explanation:

We know that resistance $R=(\rho l)/(A)$ from the given equation of resistance it is clear that resistance depends on resistivity length and area of the material but we can not change the length because it is given that the length must be 2.5 cm long.

So we can do two two things to reduce the resistance

increase the cross sectional area
decrease the resistivity of the material

If you have two objects of the same mass colliding at the same speed but opposite directions, what situation will happen in an inelastic collision? a. The objects will collide and bounce back with the same initial velocity.
b. The objects will collide and bounce back with a larger velocity.
c. The objects will collide and stay stationary.
d. The objects will collide and move forward in one direction.

Answers

c. The objects will collide and stay stationary.

If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?

Answers

Answer:

$heigth=2.86m$

Explanation:

Given data

time=0.19 s

distance=1.6 m

To find

height

Solution

First we need to find average velocity

$V_(avg)=(distance)/(time)\nV_(avg)=(1.6m)/(0.19s)\nV_(avg)=8.42m/s$

Also we know that average velocity

$V_(avg)=(V_(i)+V_(f))/2\n$

Where

Vi is top of window speed

Vf is bottom of window speed

Also we now that

$V_(f)=V_(i)+gt\nV_(f)=V_(i)+(9.8)(0.19)\nV_(f)=V_(i)+1.862$

Substitute value of Vf in average velocity

$V_(avg)=(V_(i)+V_(f))/2\nwhere\nV_(f)=V_(i)+1.862\nand\nV_(avg)=8.42m/s\nSo\n8.42m/s=(V_(i)+V_(i)+1.862)/2\n2V_(i)+1.862=16.84\nV_(i)=(16.84-1.862)/2\nV_(i)=7.489m/s\n$

Vi is speed of balloon at top of the window

Now we need to find time

$V_(i)=gt\nt=V_(i)/g\nt=7.489/9.8\nt=0.764s$

So the distance can be found as

$distance=(1/2)gt^(2)\n distance=(1/2)(9.8)(0.764)^(2)\n distance=2.86m$

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

$Speed = (Distance)/(Time)$

$Speed = (5)/(0.504)$

= 9.92 m/s

Now

$v^2 - u^2 = 2* g* h$

$9.92^2 = 2* 9.98 * h$

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.

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

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

A train travels due south at 20 m/s. It reverses its direction and travels due north at 20 m/s. What is the change in velocity of the train? 50 m/s, due south e50 m/s, due north 120 m/s, due south zero ms 40 m/s, due north

Answers

Answer:

40 m/s due north

Explanation:

Consider that the south direction a negative Y axis and north direction as + Y axis

v1 = 20 m/s South = 20 (-j) m/s

v2 = 20 m/s North = 20 j m/s

Change in velocity = v2 - v1 = 20 j - 20 (-j) = 40 j m/s

So, change in velocity is 40 m/s due north.

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