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3 years ago

Answer this question please and you will can’t the brain list

Physics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

We know that velocity equals to difference in distance divided by difference in time.

1st period: velocity is increasing. Where v = (8-0)/(3-0) = 8/3 m/s.

2nd one: velocity decreases. Where

v=(5-8)/(4-3) = (-3) m/s.

3rd one: velocity is constant because no change in distance, so v=0.

4th one: velocity increases again. Where v=(10-5)/(8-6)= 5/2 m/s.

5th one: velocity is constant again, so v=0.

6th one: velocity decreases, where

v=(6-10)/(15-12)= (-4)/3 m/s.

You might be interested in

A) Find the gravitational field strength of an asteroid with the mass of 3.2 * 10^3 kg and an average radius of 30 km when at a

MrMuchimi

a) $1.96\cdot 10^{-16} m/s^2$

The gravitational field strength near the surface of the asteroid is given by:

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the asteroid

R the radius of the asteroid

h is the distance from the surface

Substituting the data of the asteroid:

$M=3.2\cdot 10^3 kg$ is the mass

$R=30 km = 30000 m$ is the radius of the asteroid

$h=3 km = 3000 m$ is the distance from the surface

We find

$g=\frac{(6.67\cdot 10^{-11})(3.2\cdot 10^3)}{(30000+3000)^2}=1.96\cdot 10^{-16} m/s^2$

b) i) $5.53\cdot 10^9 s$

The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):

$g=1.96\cdot 10^{-16} m/s^2$

So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:

$s=ut+\frac{1}{2}gt^2$

where

s = 3 km = 3000 m is his displacement to reach the surface

u = 0 is his initial velocity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(3000)}{1.96\cdot 10^{-16} m/s^2}}=5.53\cdot 10^9 s$

b) ii) $1.08\cdot 10^{-6} m/s$

Again, we can use another suvat equation:

$v=u+gt$

where

v is the final velocity

u is the initial velocity

g is the acceleration of gravity

t is the time

Since we have

u = 0

$t=5.53\cdot 10^9 s$

$g=1.96\cdot 10^{-16} m/s^2$

The velocity of the astronaut at the surface will be

$v=0+(1.96\cdot 10^{-16} m/s^2)(5.53\cdot 10^9)=1.08\cdot 10^{-6} m/s$

b) iii) 175 years

The duration of one year here is

$T=3.16\cdot 10^7 s$

And the time it takes for the astronaut to reach the surface of the asteroid is

$t=5.53\cdot 10^9 s$

Therefore, to find the number of years, we just need to divide the total time by the duration of one year:

$n=\frac{t}{T}=\frac{5.53\cdot 10^9 s}{3.16\cdot 10^7}=175$

So, the astronaut will take 175 years to reach the surface.

8 0

3 years ago

A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the gri

Drupady [299]

Answer:

a) KE = 888.26J

b) N = 294.5 turns

Explanation:

For the kinetic energy:

$KE = I/2*\omega_o^2$

The inertia is:

$I=m/2*R^2=0.072kg.m^2$

So, the kinetic energy will be:

$KE = 888.26J$

Now, friction force is:

Ff = μ*N = 0.80*5N = 4N

The energy balance would be:

Kf - Ko = Wf where Kf=0; Ko = 888.26J; and Wf is the work done by friction force.

Wf = -Ff*d = -Ff*N*2*π*R where N is the amount of turns it gives.

Replacing these values into the energy balance:

0-888.26=-4*N*2*π*0.12

-888.26=-0.96*π*N

N=294.5 turns

6 0

3 years ago

A wave travels through a medium because

Amanda [17]

in case you dont want to read the answer is B

5 0

3 years ago

Read 2 more answers

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

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

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

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

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

Which particle rarely interacts with matter?

Marysya12 [62]

D. Neutrino
Neutrinos are particles that rarely interact with matter.

4 0

3 years ago