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3 years ago

Suppose that you take a 10 kg mass on the surface of the earth and then place it on the moon. What will

Physics

1 answer:

sergey [27]3 years ago

5 0

Answer:

10 kg

Explanation:

The mass of an object does not change even if the amount of gravtiy changes.

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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

Find the current if 55 C of charge pass a particular point in a circuit in 5 seconds.

uranmaximum [27]

Answer:

The current is 11 Amperes

Explanation:

Electric Current

The electric current is defined as a stream of charged particles that move through a conductive path.

The current intensity can be calculated as:

$\displaystyle I=\frac{Q}{t}$

Where:

Q = Electric charge

t = Time taken by the charge to move through the conductor

The current intensity is often measured in Amperes.

The charge passing through a point in a circuit is Q= 55 c during t=5 seconds, thus the current intensity is:

$\displaystyle I=\frac{55}{5}$

I = 11 Amp

The current is 11 Amperes

4 0

3 years ago

A 800-gram grinding wheel 27.0 cm in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the cent

SSSSS [86.1K]

Explanation:

d = Diameter of wheel = 27 cm

r = Radius = $\frac{d}{2}=\frac{27}{2}=13.5\ cm$

m = Mass of wheel = 800 g

$\omega_i$ = Initial angular velocity = $245\times \frac{2\pi}{60}\ rad/s$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-245\times \frac{2\pi}{60}}{50}\\\Rightarrow \alpha=-0.51312\ rad/s^2$

Moment of inertia is given by

$M=\frac{1}{2}mr^2\\\Rightarrow M=\frac{1}{2}\times 0.8\times 0.135^2\\\Rightarrow M=0.00729\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.00729\times -0.51312\\\Rightarrow \tau=-0.0037406448\ Nm$

The torque the friction exerts is -0.0037406448 Nm

For more information on torque and moment of inertia refer

brainly.com/question/13936874

brainly.com/question/3406242

7 0

3 years ago

A rocket for use in deep space is to have the capability of boosting a total load (payload plus the rocket frame and engine) of

choli [55]

Answer:

13.54 tons

Explanation:

Let f be the amount of fuel oxidizer needed

v be the speed

The relationship between them is inverse in nature i.e

f ∝ 1/v

f = k/v

If a rocket for use in deep space is to have the capability of boosting a total load (payload plus the rocket frame and engine) of 3.25 metric tons to a speed of 10,000 m/s, then f = 3.25 when v = 10,000

Substitute and get k

k = fv

k = 3.25 * 10,000

k = 32500

To get the amount of fuel oxidizer required to produce a speed of 2400m/s, we will find f when v = 2400m/s

Recall that f = k/v

f = 32500/2400

f = 13.54 metric tons

Hence the fuel plus oxidizer that will be required is 13.54 tons

4 0

3 years ago

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by

OlgaM077 [116]

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$

Note: Refer the image attached

8 0

3 years ago