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2 years ago

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9) What will happen to the period of a pendulum if we change the rope of a pendulum with another which is four times with the in

itial? B. halved A. double C. tripled D increase four times .

1 answer:

Arturiano [62]2 years ago

5 0

Double

Explanation:

Since the period T of a pendulum is given by

$T = 2\pi \sqrt{\dfrac{l}{g}}$

By increasing the length of the pendulum by 4, the period becomes

$T' = 2\pi \sqrt{\dfrac{4l}{g}} = 2\left(2\pi \sqrt{\dfrac{l}{g}}\right) = 2T$

You can see that the period doubles when we increase the length by a factor of 4.

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A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

3 0

3 years ago

Read 2 more answers

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

PtichkaEL [24]

Answer:

533.33 nm

Explanation:

Since dsinθ = mλ for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.

Since the fringes coincide,

m'λ = m"λ'

λ' = m'λ/m"

= 10 × 640 nm/12

= 6400 nm/12

= 533.33 nm

8 0

2 years ago

What is the acceleration of a softball if it hits the ground with a force 0.50 kg and hits the catchers glove with a force of 25

Rom4ik [11]

Acceleration of the ball is $50 m/s^2$

Explanation:

The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass

a is the acceleration

For the ball in this problem, we have

m = 0.50 kg (mass)

F = 25 N (force)

thereofre, the acceleration of the ball is

$a=\frac{F}{m}=\frac{25}{0.50}=50 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

3 years ago

All of the following are involved in the geological process except (2 points) gravity temperature changes thawing and freezing e

ki77a [65]

That us true so I think it is a

5 0

3 years ago