Ask question

Ask question

All categories

3 years ago

9

A grandfather clock is "losing" time because its pendulum moves too slowly. Assume that the pendulum is a massive bob at the end

of a string.A grandfather clock is "losing" time because its pendulum moves too slowly. Assume that the pendulum is a massive bob at the end of a string.The motion of this pendulum can be sped up by:- shortening the string.- lengthening the string.- increasing the mass of the bob.- decreasing the mass of the bob.

1 answer:

nignag [31]3 years ago

4 0

Answer:

shortening the string

Explanation:

L = Length of pendulum

g = Acceleration due to gravity = 9.81 m/s²

Time period of a pendulum is given by

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

$T\propto \sqrt{L}$

It can be seen that time period is proportional to the length of the pendulum.

The time period is getting longer.

So, if the length of the string is reduced this will speed up the time period.

You might be interested in

A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The

qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0

2 years ago

A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force

salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

$KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J$

Energy lost due to friction

$W=Fd=0.031\times 0.158=0.0048J$

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

$\frac{1}{2}mv^2=0.0052$

$\frac{1}{2}\times 0.00524\times v^2=0.0052$

v = 1.40 m/sec

7 0

3 years ago

I beggggg youu to solve this for me, I'm giving you 60 POINTS !!!

djyliett [7]

Answer:

but where is the question ?

Explanation:

<em>hope</em><em> it</em><em> </em><em>works</em><em> out</em>

6 0

2 years ago

Read 2 more answers

A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w

FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

$a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2$

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

$E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C$

⇒ E = 1.04*10⁻¹ N/C

5 0

3 years ago

John’s mass is 95.6 kg, and Barbara’s is 55.3 kg. He is standing on the x axis at xJ = +10.9 m, while she is standing on the x a

Anna11 [10]

Answer:Shifted towards Left by distance of 2.243 m

Explanation:

Given

Mass of john $m_1=95.6 kg$

Mass of barbara $m_2=55.3 kg$

John is standing at $x=10.9 m$

Barbara is standing at $x=2.50 m$

$x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$x_{com}=\frac{95.6\times 10.9+55.3\times 2.5}{95.6+55.3}$

$x_{com}=\frac{1180.29}{150.9}$

$x_{com}=7.821 m$

Now if they change their Position then

$x'_{com}=\frac{95.6\times 2.5+55.3\times 10.9}{95.6+55.3}$

$x'_{com}=\frac{841.77}{150.9}$

$x'_{com}=5.578$

Thus we can see that center of mass shifted towards left by a distance of $2.243 m$ because heavier is shifted towards left

8 0

3 years ago