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2 years ago

25000 seconds is how many days?

Physics

2 answers:

erastova [34]2 years ago

7 0

The answer is 0.289 day. My dear internet friend

KatRina [158]2 years ago

6 0

0.289352 days is the 25000 seconds.

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A wire delivers 12.0 C of charge in 4.0 s. What is the current in the wire? 3.0 A 8.0 A 16 A 48 A

const2013 [10]

3.0 A i got it off Quizlet and there usually always right lol can't submit tho my answers to short.... Dot dot dot

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3 years ago

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When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

2 years ago

An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the

tamaranim1 [39]

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

Attached is the full solution

5 0

2 years ago

Helllppp mee!

Fed [463]

Given:

m = 0.14 kg

v = 18 m/s

To find:

Kinetic energy= ?

Explanation:

Kinetic energy of the object is possessed when the object is in motion. It is define as the work done required to accelerate a body from rest to motion. Unit of energy is Joule and it is same as unit of work done. Energy is nothing but the capacity to do work.

Solution:

Kinetic energy of the ball is given by,

Kinetic energy = $\frac{1}{2}m v^{2}$

m = 0.14 kg

v = 18 m/s

Kinetic energy = $\frac{1}{2}(0.14) (18)^{2}$

Kinetic energy = 226.8 Joule

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2 years ago

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A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d

Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

$E=U=\frac{1}{2}kA^2$

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

$k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m$

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

$U=\frac{1}{2}kx^2=0$ because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

$K=E=50.9 J$

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

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

where m is the mass

Solving the equation for m, we find the mass:

$m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg$

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

So the kinetic energy is

$K=E-U=50.9 J-31.3 J=19.6 J$

And so we can find the speed through the formula of the kinetic energy:

$K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s$

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

$K=E-U=50.9 J-31.3 J=19.6 J$

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

$U=\frac{1}{2}kx^2$

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant the system passes the equilibrium position, which is zero.

4 0

2 years ago