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3 years ago

Measuring the amount of deuterium in the universe allows us to set a limit on _________.

Physics

1 answer:

Mademuasel [1]3 years ago

4 0

Answer:

the density of ordinary (baryonic) matter in the universe

Explanation:

Deuterium detection is of interest because the amount of it may be related to the amount of dark matter in the universe, but precise measurements have been difficult to obtain. Due to the way in which deuterium was created in the Big Bang, an exact measurement of the amount of deuterium would allow scientists to set limits on the models of the great explosion.

Also, an exact measure of deuterium would be an indicator of the cosmic density of barions (ordinary matter), and that density of barions would indicate whether ordinary matter is dark and is found in regions such as black holes, gas clouds or brown dwarfs, or it is bright and can be found in the stars. This information will help scientists who try to understand the very beginning of our universe.

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A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan

ASHA 777 [7]

Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of $a \;\;m/s^2$ from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, $a=\frac{v-u}{t}=\frac{5-4.2}{t}$

$\Rightarrow at=0.8\cdots(i)$

Now, from the equation of motion

$s=ut+\frac 12 at^2$

$\Rightarrow s=ut+\frac 12 at(t)$

$\Rightarrow 115=4.2t+\frac 12 \times 0.8 t$ [ from equation (i)]

$\Rightarrow 115=(4.2+0.4)t$

$\Rightarrow t= 115/4.6 = 25$ seconds.

Hence, she takes 25 seconds to walk the distance.

5 0

2 years ago

Two objects are moving at equal speed along a level, frictionless surface. the second object has twice the mass of the first obj

lbvjy [14]

Answer:

They both rises to same height.

Explanation:

When an object is sliding up in friction less surface than according to conservation of energy its potential energy will be converted into kinetic energy.

$mgH=\frac{1}{2}mv^{2}\\ v=\sqrt{2gH}$

Here, m is the mass, v is the velocity, g is the acceleration due to gravity and H is the height.

Here the height is independent on the mass of an object and its only depend on velocity.

Now according to the question, two objects have same velocity but they have different masses.

Therefore, they rises to the same height because height will not change with mass.

8 0

3 years ago

Read 2 more answers

True or False: Exercise is an underrated stress reliever and can be used to 25

Andreyy89

Answer:

I think it is true I'm not saying it is but if you get another person who says its true say true

Explanation:

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3 years ago

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A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/

exis [7]

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2 years ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

pickupchik [31]

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

In that particular situation, you can prove it like this: 

initial velocity is Vo 
launch angle is α 

initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity is 
Vh = Vo×cos(α) 

total time in the air is the the time it needs to fall back to a height of 0 m, so 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time = ? 
a = acceleration by gravity = g (= -9.8 m/s²) 
so 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now look at the horizontal range. 
r = v × t 
where 
r = horizontal range = ? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
so 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

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2 years ago