All categories

2 years ago

What happens to the period of a spring-cart system motion when the system is in orbit inside the international space station?

1 answer:

8 0

The period of a spring-cart system motion when the system is in orbit inside the international space station will not change because in conditions of weightlessness it is not possible to measure the mass of a body from its weight and the period depends on the mass of the body.

You might be interested in

The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a

navik [9.2K]

Answer:

230

Explanation:

$\omega$ = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J$

Energy used in one acceleration

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J$

Number of accelerations would be given by

$n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4$

So the number of complete accelerations is 230

8 0

2 years ago

A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho

Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

$80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)$

$80=5(50-x)$

$80=250-5x$

$5x=170$

$x=\frac{170}{5}$

$x=34 cm$

4 0

1 year ago

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri

MaRussiya [10]

Answer:

$dz=19217687.07\ m$

Explanation:

Given:

initial gauge pressure in the container, $P_0=2.02\times 10^{5}\ Pa$

atmospheric pressure at sea level, $P_a=1.01\times 10^5\ Pa$

initial volume, $V_0=4.4\times 10^{-4}\ m^3$

maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

density of the air, $\rho_a=1.2\ kg.m^{-3}$

density of sea water, $\rho_s=1.2\ kg.m^{-3}$

The relation between the change in pressure with height is given as:

$\frac{dP_{max}}{dz} =\rho_a.g_n$

where:

dz = height in the atmosphere

$g_n$ = standard value of gravity

Now putting the respective values:

$\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8$

$dz=19217.687\ km$

$dz=19217687.07\ m$

Is the maximum height above the ground that the container can be lifted before bursting. (Since the density of air and the density of sea water are assumed to be constant.)

7 0

1 year ago

Who pays for Government workers that work on alcohol impaired driving cases?

Shalnov [3]

Answer: Taxpayers

Explanation:

Taking alcohol before driving or while driving is dangerous and has resulted in lots of accidents and deaths. Alcohol tampers with the normal functioning of the brain, and also impairs ones reasoning.

Alcohol impaired driving cases handled by government officials are paid for by the taxpayers. A tax is the levy that the people in the country pays. Those funds are used in handling different government objectives and this is one of such ways.

8 0

1 year ago

Two horses, Thunder and Misty, are accelerating a wagon 1.3 m/s2. The force of friction is 75 N. Thunder is pulling with a force

sasho [114]

Assumption both thunder and misty are pulling in same direction,
Net force= 1000N+800N-75N=1725N
Mass of wagon = 1725N/1.3ms^-2 = 1327kg

7 0

1 year ago

Read 2 more answers