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3 years ago

Cosmologists study ______.

Physics

2 answers:

____ [38]3 years ago

6 0

Your answer is:

<em>The origin, structure, and future of the universe</em>

You can just search up the definition of Cosmology and receive your answer. Hopefully this helps at least a little :3

Irina-Kira [14]3 years ago

4 0

Answer: the origin, structure, and future of the universe

Explanation:

Cosmology is a branch of science (astronomy) that deals with the study of the origin, structure, and future of the universe. Cosmologists are the people who study cosmology.

Cosmologist study the universe on large scale. They study the universe as a whole from events - Big bang till today and possibilities in future.

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Find the minimum radius of a helium balloon that will lift her off the ground. the density of helium gas is 0.178 kg/m3, and the

marysya [2.9K]

$Volume of balloon = \frac{4}{3} \pi r^{3}$

Where r is the radius of balloon.

Here mass of woman = 68 kg

Mass of air displaced by a balloon with volume V = 1.29*V

Mass of helium inside balloon = 0.178*V

Total mass to be lifted by balloon = 68 +0.178*V

Buoyant force = 1.29V-0.178V=1.112V

So we have 1.112 V = 68+ 0.178*V

0.934 V = 68

V = 72.81 $m^3$

\frac{4}{3} \pi r^{3}[/tex]= 72.81

r = 2.59 m

So radius of helium balloon = 2.59 m

4 0

3 years ago

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$

The number of revolutions is 29.57239

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s$

The time it takes for the car to stop is 3.91176 seconds

Linear distance

$s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m$

The distance the car travels is 52.026478 m

8 0

2 years ago

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

slega [8]

Answer:

a) v_average = 11 m / s, b) t = 0.0627 s

, c) F = 7.37 10⁵ N

, d) F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

v² = v₀² - 2 a x

The fine speed zeroes them

a = v₀² / 2x

a = 22²/2 0.69

a = 350.72 m / s²

The average speed is

v_average = (v + v₀) / 2

v_average = (22 + 0) / 2

v_average = 11 m / s

b) The average time

v = v₀ - a t

t = v₀ / a

t = 22 / 350.72

t = 0.0627 s

c) The force can be found with Newton's second law

F = m a

F = 2100 350.72

F = 7.37 10⁵ N

.d) the ratio of this force to weight

F / W = 7.37 10⁵ / (2100 9.8)

F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

5 0

2 years ago

Which of the following is an example of acceleration? I. A car speeds up. II. A car slows down. III. A car travels in a straight

svetlana [45]

I., II., and IV. are examples of acceleration. III. isn't.

6 0

3 years ago

A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel

trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2$

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s$

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

$1\ mile=1609.34\ m$

$1\ h=3600\ seconds$

$18.03\times \frac{3600}{1609.34}=40.33\ mph$

Speed of the boat by when it will hit the dock is 40.33 mph

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m$

The distance at which the boat will have to start decelerating is 312.5 m

5 0

2 years ago