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german
2 years ago
14

A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks

to see the kind of star that our Sun will ultimately become, long, long after it has turned into a white dwarf. Why is the astronomer on duty going to have a bit of a problem satisfying her request? a. All the old stars in our Galaxy are located in globular clusters and all of these are too far away to be seen with the kind of telescope a college or university campus would have. b. After being a white dwarf, the Sun will explode, and there will be nothing left to see. c. The universe is not even old enough to have produced any white dwarfs yet d. Astronomers only let people with PhD's look at these stellar corpses; it's like an initiation rite for those who become astronomers. e. After a white dwarf cools off it becomes too cold and dark to emit visible light
Physics
1 answer:
lidiya [134]2 years ago
3 0

Answer:

b

Explanation:

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Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hourmiles/hour
slavikrds [6]

Answer:

The acceleration of the cheetahs is 10.1 m/s²

Explanation:

Hi there!

The equation of velocity of an object moving along a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time t.

v0 = initial velocity.

a = acceleration.

t = time

We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.

Let's convert mi/h into m/s:

50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s

Then, using the equation:

v = v0 + a · t

22.4 m/s = 0 m/s + a · 2.22 s

Solving for a:

22.4 m/s / 2.22 s = a

a = 10.1 m/s²

The acceleration of the cheetahs is 10.1 m/s²

5 0
2 years ago
A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a
vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451

tan\alpha = 0.451

\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0

3 0
2 years ago
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel
Svetradugi [14.3K]

Answer:

a)693.821N/m

b)17.5g

Explanation:

We the Period T we can find the constant k,

That is

T = 2 \pi \sqrt{\frac{m}{k}}

squaring on both sides,

T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}

where,

M=hanging mass, m = spring mass,

k =spring constant

T =time period

a) So for the equation we can compare, that is,

y=T^2=0.0569x+0.0010

the hanging mass M is x here, so comparing the equation we know that

\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m

b) In order to find the mass of the spring we make similar process, so comparing,

\frac{4\pi^2}{k}m =0.001\\m=\frac{0.004k}{4\pi^2} =\frac{0.001*693.821}{4\pi^2}\\m=0.0175kg\\m=17.5g

3 0
2 years ago
You need to determine the density of an unknown liquid and decide to perform an experiment. You notice that a wooden block float
Allushta [10]

Answer:

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

Explanation:

The density of a liquid is inversely proportional to the volume (height) of object submerged in it.

High density liquid possess higher buoyant force preventing objects from submerging.

p ∝ 1/V ∝ 1/h

since V = Ah

pu/pw = hw/hu

pu = pwhw/hu

Where;

p = density

h = height submerged

pu and pw is the density of unknown liquid and water respectively

hu and hw is the height of object submerged in unknown liquid and water respectively

pw = 1000kg/m^3

hu = 4.6cm = 0.046m

hw = 5.8cm = 0.058m

Substituting the given values;

pu = 1000×0.058/0.046

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

5 0
2 years ago
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