To an observer on Earth, the time on a moving spaceship is answer: slower than the time on Earth.

An astronaut is rotated in a horizontal centrifuge at a radius of 5.0 m. (a) What is the astronaut’s speed if the centripetal ac

sergeinik [125]

Explanation:

Given that,

Radius of circular path, r = 5 m

Centripetal acceleration, $a=7g\ m/s^2$

(a) Let v is the astronaut’s speed. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$v=\sqrt{ar}$

$v=\sqrt{7\times 9.8\times 5}$

v = 18.5 m/s

(b) Let T denotes the time period. It is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi \times 5}{18.5}$

T = 1.69 s

Let N is the number of revolutions. So,

$N=\dfrac{60}{1.69}=35.5\ rev/min$

So, the number of revolutions per minute is 35.5

(c) T = 1.69 seconds

Hence, this is the required solution.

8 0

4 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time

Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

4 0

4 years ago

If you remove large amounts of heat from a liquid, what could happen

snow_lady [41]

A liquid becomes a solid when energy is removed. The energy content decreases, and the speed of the particles decrease.

3 0

3 years ago

Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i

spayn [35]

Answer:

$v = 15.56 m/s$

$v = 56 km/h$

Explanation:

When coefficient of friction is approximately zero then we have

$F_ncos\theta = mg$

$F_n sin\theta = \frac{mv^2}{R}$

$tan\theta = \frac{v^2}{Rg}$

here we know that

$v = 40 km/h = 11.11 m/s$

R = 30 m

$tan\theta = \frac{11.11^2}{30\times 9.81}$

$\theta = 22.75 degree$

now when friction coefficient is 0.30 then we have

$F_n cos\theta = mg + F_f sin\theta$

$F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}$

now we have

$v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}$

$v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}$

$v = 15.56 m/s$

$v = 56 km/h$

3 0

3 years ago

What is the equation of instantaneous velocity<br>

Oksi-84 [34.3K]

Answer:

it's possible to calculate an object's velocity at any moment along its path. This is called instantaneous velocity and it is defined by the equation v = (ds)/(dt),in other words, the derivative of the object's average velocity equation.

7 0

3 years ago