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3 years ago

A cat falls from a table of height 1.3m. What is the impact speed of the cat

Physics

1 answer:

Lorico [155]3 years ago

6 0

Answer: 12.753

Explanation: Multiply 1.3 and 9.81 (the speed that everything drops at.)

Also, i feel bad for the cat

it might be wrong sorry

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Because the earth's orbit is slightly elliptical, the earth actually gets closer to the sun during part of the year. When the ea

BabaBlast [244]

Closer to the sun . . . orbital speed is faster.

Farther from the sun . . . orbital speed is slower.

Flag answer: Answer 13 Answer 13

5 0

2 years ago

If gravity on the earth increased, what affect would it have on the moon

Rufina [12.5K]

Answer:

If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.

6 0

3 years ago

A wave With wavelength 20 m has a frequency of 12 Hz what is the waves speed

pentagon [3]

Answer:

240m/s

Explanation:

The equation to calculate is wavelength= velocity/ frequency so to find the velocity you would have to multiply frequency by wavelength.

7 0

3 years ago

A cord of mass 0.65 kg is stretched between two supports 28 m apart. if the tension in the cord is 150 n, 2 how long will it tak

MaRussiya [10]

Ans: Time <span>taken by a pulse to travel from one support to the other = 0.348s
</span>
Explanation:
First you need to find out the speed of the wave.

Since
Speed = v = $\sqrt{ \frac{T}{\mu} }$

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m

So
v = $\sqrt{ \frac{150}{0.0232} }$ = 80.41 m/s

Now the time-taken by the wave = t = Length/speed = 28/80.41=0.348s

5 0

3 years ago

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

3 years ago