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adelina 88 [10]

3 years ago

7

If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible f

or one to be at rest after the collision.

1 answer:

Ulleksa [173]3 years ago

8 0

Answer:

yes

Explanation:

if the objects a magnetic with a certain amount of force they should be at a stop

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You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1

drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

The given parameters;

<em>Mass of the pendulum, = M </em>
<em>Length of the pendulum, = L</em>
<em>Initial angular speed, </em> $\omega _i$ <em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

$I_i = \frac{1}{3} ML^2$

The moment of inertia of the rod between the middle and the end is calculated as;

$I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}$

Apply the principle of conservation of angular momentum as shown below;

$I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f$

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

3 0

3 years ago

Question: How is the energy in the food we eat an example of chemical potential energy?

Elanso [62]

Answer: D

Explanation: The chemical bonds in the food store energy that is released when we eat it.

3 0

3 years ago

The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d

olya-2409 [2.1K]

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

$|r|=\frac{3\times 2\pi r}{4}=75.408 cm$

$Angle =270^{\circ}$

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

$|r|=\frac{2\pi r}{4}=25.136 cm$

angle turned $=90^{\circ}$

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

Angle turned will also be same i.e. $90 ^{\circ}$

7 0

3 years ago

In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?

kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰ (angle with respect x-axis)

Explanation:

E(t) Electric Field is a vector, so we need to determine module and direction

E(t) = E(q₁) + E (q₂) Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂ respectively.

E(q₁) = K * q₁/ (d₁)² K = 9 *10⁹ [N*m²/C²] d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225 [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) = 9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9 tanα = 2,395 α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0

3 years ago

Read 2 more answers

Someone please help. Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign a

erik [133]

Answer:

The answer is based on the conservation of energy law; something you should really understand by now.

For convenience we can hold one of the two charges still; it becomes the frame of reference. And everything we say is in reference to the designated static charge, call it Q.

So the moving charge, call it q, has total energy TE = PE. It's all potential energy as we start with q not moving.

It has potential energy because in order to separate q from Q, we had to do work, add energy, on q. And from the COE law, that work added is converted into PE.

It's a bit like lifting something off the ground. That's work and it becomes GPE. So there's some work, in separating the two charges in the first place.

But there's more.

Now we let q go. As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.

And there's more work, done by the EMF on charge q. That converts the PE into KE and the q charge smashes into Q with some kinetic energy.

5 0

3 years ago