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laiz [17]
3 years ago
9

Each of the four expansion models (recollapsing, critical, coasting, and accelerating) predict different ages for the universe,

given the current expansion rate. Why is this
Physics
1 answer:
cupoosta [38]3 years ago
4 0

Answer:

This is because the age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

Explanation:

The age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

This is due to the fact that the expansion rate in the coasting model is constant and never changes. Because the cosmos is growing faster now than during the old days, recollapsing and critical models give shorter ages. According to the accelerating model, the universe is growing at a slower rate currently than in the past, implying an older age.

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In an effort to decrease the mAs of an exposure, the 15% rule of kVp change may be considered. Changing the original kVp of 84 u
Roman55 [17]

Answer:

Will result in :Greater Compton scatter interaction

Explanation

This is because The 15% rule states that changing the kVp by 15% has the same effect as doubling the mAs, or reducing the mAs by 50%; that is A 15% increase in kVp has the same effect as doubling the mAs and vice versa but in this case reducing mas causes more Compton scatter because Compton scattering occurs at a lower kvp

8 0
3 years ago
A space vehicle accelerates uniformly from 85 m/s at t = 0 to 164 m/s at t = 10.0 s .How far did it move between t = 2.0 s and t
creativ13 [48]

First, we have a change in the velocity from 85 to 164 m/s in 10 sec.

Then, we calculate the <u>acceleration </u>as:

a=\frac{v_{f}-v_{i} }{t} =\frac{164-85}{10}=7.9 m/s^2

Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:

v_{f}=v_{i}+at=85+7.9*2=100.8m/s

Then, using the second equation of motion to calculate the distance:

d=v_{i}  t+\frac{1}{2}at^2

d=100.8*2+\frac{1}{2}*7.9*(2)^2=217.4m

5 0
4 years ago
Read 2 more answers
A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after
vodka [1.7K]

Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s  (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

3 0
4 years ago
Otion
Scorpion4ik [409]

Answer:

SKID

Explanation:

In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.

Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.

Y axis y

     N- W = 0

     N = W

X axis (radial)

        fr = m a

the acceleration in the curve is centripetal

         a = \frac{v^2}{r}

the friction force has the expression

        fr = μ N

we substitute

       μ mg = m v²/r

       v = \sqrt{\mu g r}

we calculate

      v = \sqrt{0.1 \ 9.8 \ 3}

      v = 1,715 m / s

to compare with the cyclist's speed let's reduce to the SI system

        v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s

We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID

5 0
3 years ago
Given the distance between the crest of one wave and the crest of the next wave, you can determine the?
Sonja [21]
Answer: wavelength !!
hope this helped :)
3 0
3 years ago
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