Answer:
The lid becomes tighter
It becomes tighter because metals have a lower heat capacity than glass meaning their temperature drops (or increases) much faster than glass for the same energy change. So in this example, the metal will contract faster than the glass causing it to be more tighter around the glass.
Complete Question
If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don’t have time to expand to compensate for the blood pressure drop. If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart? The density of blood plasma is about 1025 kg/m3 and a typical maximum (systolic) pressure of the blood at the heart is 120 mm of Hg (= 0.16 atm = 16 kP = 1.6 × 104 N/m2).
Answer:
The pressure at the brain is
Explanation:
Generally is mathematically denoted as
Substituting for (the density) , for g (acceleration due to gravity) , 0.4m for h (the height )
We have that the pressure difference between the heart and the brain is
But the pressure of blood at the heart is given as
Now the pressure at the brain is mathematically evaluated as
When you stand up quickly, the blood pressure at your brain is lower than at your heart. The decrease in blood pressure can be calculated using the equation ΔP = ρgh, where ΔP is the change in pressure, ρ is the density of the blood, g is the acceleration due to gravity, and h is the height difference between the two points. In this case, the blood pressure at the brain is approximately 416.32 Pa lower than at the heart.
When you stand up quickly, your blood pressure drops because the blood vessels don't have enough time to expand and compensate for the change in posture. The brain, which is 0.4 m higher than the heart when standing, experiences a decrease in blood pressure. To calculate how much lower the blood pressure is at the brain compared to the heart, we need to use the equation: ΔP = ρgh, where ΔP is the change in pressure, ρ is the density of the blood, g is the acceleration due to gravity, and h is the height difference between the two points. In this case, we can use the height difference of 0.4 m and the density of blood to find the change in pressure.
Using the equation, ΔP = ρgh, we can calculate the change in pressure:
Plugging in the values into the equation, we get:
ΔP = (1060 kg/m³)(9.8 m/s²)(0.4 m) = 416.32 Pa
Therefore, the blood pressure at the brain is approximately 416.32 Pa lower than at the heart when standing up quickly.
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A.0.41 sec
B.41 sec
C.4.1 sec
D.4 sec
A horizontal baseball pitch is launched at 44 m/s. The ball will stay for 4.1 sec (approx) in the air. Hence, option C is correct.
The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.
Its SI unit is represented as m/s, and it is a vector quantity, it means that it has both magnitude and direction.
According to the question, the given values are :
Initial Velocity, u = 44 m/s,
Distance travelled, s = 18 m and,
Final velocity, v = 0.
Use equation of motion :
v = u + at
0 = 44 + (-9.8)t
t = 44 / 9.8
t = 4.3 (approx)
Hence, the time for which the ball stay in the air is 4.1 sec (approx).
To get more information about velocity :
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Answer:
a 0.41
plug number into equation
The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.
Through these waves, you can know that there is beyond the clouds.
Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.
Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.
It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.
Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.
Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.
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Answer:
θ=π/2
Explanation:
The definition of work is W = → F ⋅ → d = q E c o s θ d W=F→⋅d→=qEcosθd. So if no work is done, the displacement must be in the direction perpendicular to the force ie c o s θ = 0 → θ = π / 2 cosθ=0→θ=π/2
A charged particle can be displaced without any external work done on it in a uniform electric field when its movement is perpendicular to the direction of the electric field.
In a uniform electric field, the electric force is the same in every direction. Therefore, if a charge were to be displaced perpendicular to the original direction of the electric field (i.e., in the y or z direction), it would not encounter any extra electric forces. This means there would be no external work being done on the charge. When a charge is moved perpendicular to an electric field, the field does not affect it, and hence, no work is done by the field.
In other words, a charge can be displaced in this field without any external work being done on it when it is moved in a direction perpendicular to the uniform electric field, either in y-axis or z-axis, assuming the electric field is constant in the x-axis direction.
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Answer:
5N
Explanation:
Given parameters:
Original length = 22cm
Spring constant, K = 50N/m
New length = 32cm
Unknown
Force applied = ?
Solution:
The force applied on a spring can be derived using the expression below;
Force = KE
k is the spring constant
E is the extension
extension = new length - original length
extension = 32cm - 22cm = 10cm
convert the extension from cm to m;
100cm = 1m;
10cm will give 0.1m
So;
Force = 50N/m x 0.1m = 5N
To calculate the force used to stretch the spring, Hooke's Law is utilized, which leads to the conclusion that a force of 5 N was exerted to stretch the spring from its original length of 22 cm to a final length of 32 cm.
The force exerted by a spring is governed by Hooke's Law, which states that the force required to stretch or compress a spring by a certain distance is proportional to that distance. In this case, the spring constant, k, is given as 50 N/m and the spring is stretched from its original length of 22 cm to a final length of 32 cm. This represents a stretch, or displacement, of 10 cm (or 0.1 m when converted to the standard unit).
The force (F) can be calculated using Hooke's law: F = kx, where x is the displacement of the spring. Substituting the given values, the force amounts to F = (50 N/m) * (0.1 m) = 5 N. Therefore, the force used to stretch the spring to its final length of 32 cm is 5 N.
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Answer:
To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.
Explanation:
First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:
Here h is the Planck's constant, c is the speed of light, is the wavelength of the light and W the work function of the element:
Now, we calculate the wavelength for the new maximum kinetic energy:
This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.