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3 years ago

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Suppose you took a trip to the moon. Write a paragraph describing how and why your weight would change. Would your mass change t

oo?

1 answer:

noname [10]3 years ago

5 0

Your weight would change but not your mass, the moon has less gravity so therefore you are going to be lighter :-)

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Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher

FrozenT [24]

Answer: Option (C) is the correct answer.

Explanation:

As we know that metals are able to conduct electricity so, when a negatively charges rod is kept closer to the left sphere then electrons will enter the sphere.

Since, like charges repel each other. Hence, some of the negative changes from the rod will repel the negative charges of left sphere.

As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.

Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.

7 0

3 years ago

Read 2 more answers

2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg & object B’s mass is 11kg. After the colli

ss7ja [257]

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5 x v + 11 x 15

where v is velocity of A after the collision .

5 x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

3 0

3 years ago

A ball is rolling down a hill. Wich action would slow the ball down?

Feliz [49]

Answer:

Friction Force

hope this helped :)

Explanation:

4 0

3 years ago

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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

Polonium, the Period 6 member of Group 6A(16), is a rare radioactive metal that is the only element with a crystal structure bas

Burka [1]

Answer:

Approximate atomic radius for polonium-209 is 167.5 pm .

Explanation:

Number of atom in simple cubic unit cell = Z = 1

Density of platinum = $9.232 g/cm^3$

Edge length of cubic unit cell= a = ?

Atomic mass of Po (M) = 209 g/mol

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

ρ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$9.232 g/cm3=\frac{1 \times 209 g/mol}{6.022\times 10^{23} mol^{-1}\times (a)^{3}}$

$a = 3.35\times 10^{-8} cm$

Atomic radius of the polonium in unit cell = r

r = 0.5a

$r=0.5\times 3.35\times 10^{-8} cm=1.675\times 10^{-8} cm$

$1 cm = 10^{10} pm$

$1.675\times 10^{-8} cm=1.675\times 10^{-8}\times 10^{10}=167.5 pm$

Approximate atomic radius for polonium-209 is 167.5 pm.

7 0

3 years ago