How does an object move when it is subject to a steady centripetal force

How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the o

melisa1 [442]

Question:

A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes

How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.

Answer:

S = 5.508 × 10¹¹m

V = 2.62 × 10⁸ m/s

Explanation:

The radius of the orbit of Jupiter, Rj is 43.2 light-minutes

radius of the orbit of Mars, Rm is 12.6 light-minutes

Distance travelled S = (Rj - Rm)

= 43.2 - 12.6 = 30.6 light- minutes

= 30.6 × (3 ×10⁸m/s) × 60 s

= 5.508 × 10¹¹m

time = 35mins = (35 × 60 secs)

= 2100 secs

speed = distance/time

V = 5.508 × 10¹¹m / 2100 s

V = 2.62 × 10⁸ m/s

7 0

3 years ago

An outfielder throws a baseball with an initial speed of 76.9 mi/h. Just before an infielder catches the ball at the same level,

Lyrx [107]

Answer:

-9.62ft-lb

Explanation:

see the attached file

7 0

3 years ago

Read 2 more answers

Consider the possibility of using two rotating cylinders to replace the conventional wings on an airplane for lift. Consider an

EleoNora [17]

Answer:

27.35m

Explanation:

For the calculation of the Support Force we rely on the formula for obtaining the force in a cylinder of a certain length l,

$F_y = - \rho Ul\Gamma$

Here each term is,

$F_y$ = Lift force

$\rho$ = density of air

$\Gamma$ = vortex strength

For this last equation, its mathematical representation is given by,

$\Gamma = 2\pi av_{\theta}$

Here each term is,

a= 1m, radios of cylinder

$v_{\theta}= 20 Km/hr=5.5m/s$ , the velocity of cylinder surface.

$\Gamma = 2\pi (1)(5.5) = 34.90m^2/s$

In order to find the density of the area at 2000m we will refer to the table of Standard Atmosphere of the United States, that is $1.007kg/m^3,$

$U= 150Km/hr = 41.6m/s, F_y = 40000N, \Gamma = 34.90m^2/s$

Replacing the values,

$40000 = -(1.007)(41.6)l(34.90)$

Clearing l and solving for it we have,

$l=-27.35m$

<em>In this way we can conclude that the length of the cylinder must be 27.35m</em>

7 0

4 years ago

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2

Leno4ka [110]

Answer:

A. T=15.54 °C

B. Q/A= 0.119 W/m2

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

$Q= kA\frac{dT}{dx}$

Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

$\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C$

Then, to find the rate of heat flow per square meter, we have:

$\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}$

$T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness$

7 0

3 years ago

Think about your displacement at three different times throughout your day and compare it with the distance you traveled.

Paul [167]

-- The first thing I do when I wake up is go STRAIGHT to the bathroom. Up to that time, my displacement is equal to distance I traveled from my bed.

-- Once I'm relaxed and back in my room, dithering around and getting dressed, the distance I've traveled since I woke up is growing and growing, but my displacement is staying pretty steady, because I'm still hanging right around my bed.

-- I walk to school, walk between classes, maybe run around the track a couple times, walk to the lunchroom and back to classes, then walk home. By dinner time, my distance traveled during the day might be 3 or 4 MILES, but my displacement is only one floor down from my bedroom to the kitchen.

-- After my homework is done, I slide back into my warm bed and turn out the light. My displacement for the day is now zero ! The straight-line distance from the place I started to the place I finished is zero.

6 0

4 years ago