All categories

3 years ago

The answer would be ?

Physics

1 answer:

Cerrena [4.2K]3 years ago

4 0

Answer:

D. demand; increased

Explanation:

Demand is how much people want it.

You might be interested in

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

3 0

3 years ago

Determine the total amount of heat, in joules, required to completely vaporize a 50.0-gram sample of h2o(?) at its boiling point

Tanzania [10]

In order to calculate the amount of energy required, we must first check the latent heat of vaporization of water from literature. The latent heat of vaporization of any substance is the amount of energy required per unit mass to convert that substance from a solid to a liquid. For water this is 2,260 J/g. We now use the formula:
Energy = mass * latent heat
Q = 50 * 2,260
Q = 113,000 J

113,000 Joules of heat energy are required.

3 0

3 years ago

The midpoint M of a guitar string is pulled a distance d = 1.7 mm from equilibrium and released. Point M is observed to undergo

nlexa [21]

Answer:

ω = 380π rad/s

Explanation:

The formula for the angular frequency is the oscillation frequency f (hertz) multiplied by 2π

ω = 2πf

then

ω = 2π(190)

ω = 380π rad/s

8 0

2 years ago

On the Moon the acceleration due to gravity is about one sixth that on Earth. If a golfer on the Moon imparted the same initial

swat32

Explanation:

We know that that the range of the ball on the earth

$R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}$

therefore, range of the ball on moon

$R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}$

$R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}$

therefore,

$R_{moon}=6R_{earth}$

Therefore, the range of ball will be 6 times on the moon than that on earth

6 0

2 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago