Answer:
Weather is the conditions (temperature, wind etc.) at a given time, like on that day. Climate, which is what his data would show, is the conditions over an extended period of time like the 3 months he collected data
The second-order dark fringe in a single-slit diffraction pattern is 1.40 mm from the center of the central maximum. Assuming the screen is 89.0 cm from a slit of width 0.710 mm and assuming monochromatic incident light, calculate the wavelength of the incident light.
We know, for single slit :
[tex]y =\dfrac{ n\lambda L}{a}\\\\\lambda = \dfrac{ya}{nL}[/tex] ...1)
[tex]y = 1.4\ mm = 1.4 \times 10^{-3}\ m[/tex]
n = 2
L = 89 cm = 0.89 m
[tex]a=7.1\times 10^{-4}\ m[/tex]
Putting all these in equation 1), we get :
[tex]\lambda = \dfrac{ya}{nL}\\\\\lambda = \dfrac{1.4\times 10^{-3}\times 7.1\times 10^{-4}}{2\times 0.89 }\\\\\lambda = 5.584 \times 10^{-7}\ m[/tex]
Therefore, wavelength of the incident light is [tex]5.584 \times 10^{-7}\ m[/tex] or 558.4 nm.
Hence, this is the required solution.
Which statement is part of Dalton's atomic theory?
Matter is composed of small particles called atoms.
Atoms can be divided into their subatomic particles.
Atoms are able to be seen with proper spectroscopy equipment.
Chemical reactions can change atoms from one type to another.
Answer:
Dalton’s atomic theory was a scientific theory on the nature of matter put forward by the English physicist and chemist John Dalton in the year 1808. It stated that all matter was made up of small, indivisible particles known as ‘atoms’.
Explanation:
Atoms can be divided into their subatomic particles.
The statement that is part of Dalton's atomic theory is as follows: Matter is composed of small particles called atoms.
What is Dalton's atomic theory?John Dalton is a scientist that first stated the theory of chemical combination in 1803.
The components of these theory are as follows:
Elements consist of indivisible small particles called atoms.All atoms of the same element are identical i.e. different elements have different types of atom. Atoms can neither be created nor destroyedTherefore, according to this question, the statement that is part of Dalton's atomic theory is as follows: Matter is composed of small particles called atoms.
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#SPJ2
The metric unit of power is _____.
Answer:
Watt
Explanation:
Power is defined as the work done by an object per unit time. Its mathematical form is given by :
[tex]P=\dfrac{W}{t}[/tex]
The SI unit of work done is Joules (J) and that of time is seconds (s).
J/s is equal to watts. Watt is the metric unit of power.
How many stars are in the universe (approximately)? O 40 sixtillion 0 365 billion O 86.4 million O one
Answer:
a i belive
Explanation:
the univerce is VERY large so a, if im wrong i apologise :(
2. Heather and Matthew walk with an average velocity of +0.87 m/s eastward.
If it takes them 27 minutes to walk to the store, what is their
displacement? (include direction)
(5 points)
The CERN particle accelerator is circular with a circumference of 7.0 km.
Required:
a. What is the acceleration of the protons (m=1.67×10^−27kg) that move around the accelerator at of the speed of light? (The speed of light is v=3.00×10^8m/s.)
b. What is the force on the protons?
Answer:
Explanation:
a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;
a = mv²/r
m is the mass of proton = 1.67×10^−27kg
v is the velocity = 3.00×10^8m/s
r is the radius
Since C = 2πr
7000m = 2πr
r = 7000/2π
r = 1114.08m
Substitute
a = 1.67×10^−27 (3.00×10^8)²/1114.08
a = 1.67×10^−27 * 9×10^16/1114.08
a = 15.03*10^-11/1114.08
a = 0.001346*10^-11
a = 1.346*10^-14m/s²
b) Force on the proton = mass * acceleration
Force = 1.67×10^−27kg * 1.346*10^-14
Force = 2.246*10^-41N
hence the force on the proton is 2.246*10^-41N
* WILL GIVE BRAINLIEST TO CORRECT ANSWER *
Name a product that people commonly purchase by mass and not by weight.
I would say food but like they take weight too
Hmmm... Who can answer this question?
How did life begin?
Just a practice
Answer:
The earliest known life-forms are putative fossilized microorganisms, found in hydrothermal vent precipitates, that may have lived as early as 4.28 Gya (billion years ago), relatively soon after the oceans formed 4.41 Gya, and not long after the formation of the Earth 4.54 Gya.
Explanation:
Is this what your looking for?
Catching a wave, a 77 kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.65 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?
Answer:
Explanation:
The total work done by the wave is expressed as;
Workdone = Potential energy + Kinetic energy
Workdone = mgh + 1/2mv²
m is the mass = 77kg
g is the acceleration due to gravity = 9.8m/s²
v is the velocity = 8.2m/s
h is the height = 1.65m
Substitute into the formula;
Workdone = 77(9.8)(1.65) + 1/2(77)8.2²
Workdone = 1245.09 + 2588.74
Workdone = 3833.83Joules
Hence the amount of non conservative work done on the sofa is 3833.83Joules
Given:
Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kgWe know,
→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]
or,
[tex]= mgh +\frac{1}{2} mv^2[/tex]
By putting the values,
[tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]
[tex]= 1245.09+2588.74[/tex]
[tex]= 3833.83 \ Joules[/tex]
Thus the above approach is right.
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A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it looks like this group varied the amount of mass sitting on the block with each trial - this is not recommended). Nonetheless, what is their average coefficient of static friction?
Trial Mass of block (g) Hanging mass (kg)
1 105 0.053
2 165 0.081
3 220 0.118
4 280 0.149
5 315 0.180
6 385 0.198
Answer:
0.130
Explanation:
From the given data, the coefficient of static friction for each trial are:
1. 0.053
2. 0.081
3. 0.118
4. 0.149
5. 0.180
6. 0.198
The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
So that;
the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]
= [tex]\frac{0.779}{6}[/tex]
= 0.12983
The average coefficient of static friction is 0.130
The average coefficient of static friction is 0.13.
The coefficient of static friction is obtained using the formula; μ = F/R
Where;
F = force acting on the body
R = reaction
μ = coefficient of static friction
The average of measurements is given as; ∑summation of measurements/number of measurements
We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;
0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198/6
= 0.13
The average coefficient of static friction is 0.13
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significance of practicals in the discipline of geography
A golf ball (m=26.7g) is struck a blow that makes an angle of 33.6 degrees with the horizontal. The drive lands 190m away on a flat fairway. The acceleration of gravity is 9.8 m/s^2 . If the golf club and ball are in contact for 7.13 ms, what is the average force of impact?
Answer:
Th average force impact is [tex]F = 168.298 \ N[/tex]
Explanation:
From the question we are told that
The mass of the golf ball is [tex]m_g = 26.7 \ g = 0.0267 \ kg[/tex]
The angle made is [tex]\theta = 33.6 ^o[/tex]
The range of the golf ball is [tex]R = 190 \ m[/tex]
The duration of contact is [tex]\Delta t = 7.13 \ ms = 7.13 *10^{-3} \ s[/tex]
Generally the range of the golf ball is mathematically represented as
[tex]R = \frac{v^2 sin2(\theta)}{g}[/tex]
Here v is the velocity with which the golf club propelled it with, making v the subject
[tex]v = \sqrt{\frac{R * g}{sin 2 (\theta)} }[/tex]
=> [tex]v = \sqrt{\frac{190 * 9.8}{sin 2 (33.6)} }[/tex]
=> [tex]v = 44.94 \ m/s[/tex]
Generally the change in momentum of the golf ball is mathematically represented as
[tex]\Delta p = m * (v - u )[/tex]
here u is the initial velocity of the ball before being stroked and the value is 0 m/s
[tex]\Delta p = 0.0267 * ( 44.94 - 0 )[/tex]
=> [tex]\Delta p = 1.19996 \ kg \cdot m/s[/tex]
Generally the average force of impact is mathematically represented as
[tex]F = \frac{\Delta p }{\Delta t}[/tex]
=> [tex]F = \frac{1.19996 }{7.13 *10^{-3}}[/tex]
=> [tex]F = 168.298 \ N[/tex]
In football we see unbalanced forces. When 1 player exerts an unbalanced force on another player and causes a player to
Answer:
Fall
Explanation:
In ultimate the disc may be passed in any direction.
True
False
I think is True: Because...
when you throw a disc you can throw it in any direction, many people call it "Flying Saucer" I'll give you an example ... When you throw something, for example a paper, you want to throw it at your classmate. You already know what address you want to send it to, then I say it is: True ...
Sorry if it's wrong :(
from a flying aeroplane abody should be dropped in advance to hit the target why
From a flying plane a body should dropped in advance to hit the target,Why? ... The body should be dropped in advance as when the body is dropped it has the velocity of the plane. So, in air the body moves forward which we have to take into consideration in order to hit the target.
we feel cold in winter when we come out from the quilt but the same room becomes warmer after coming back from outside the room
Answer:
Yes
Explanation:
I think this is because when you go out of the room and going to a hotter room you then get the heat from that room. It then becomes warmer in the room you are coming from because your body got the heat from the outside the room. I think it is because of body temperature.
HOPE THIS HELPED
Zookeepers carry a stretcher that holds a sleeping lion. The total mass of the stretcher and lion is 175 kg. The lion's forward acceleration is 2 m/s2. What is the force necessary to produce this acceleration?
Answer:
350 N
Explanation:
Newton's second law:
∑F = ma
∑F = (175 kg) (2 m/s²)
∑F = 350 N
Morgan does 50 J of work with a lever
that has an efficiency of 92%. What is
the output work of the lever?
Answer:
46 J
Explanation:
Simply calculate the 92% of 50 Joules as:
0.92 * 50 J = 46 J
C4. A 50.0 kg boy runs at 10.0 m/s, jumps on a cart and rolls off at 2.50 m/s. What is the mass of the cart
Answer:
The mass of the cart is 150 kg.
Explanation:
Given that,
Mass of a boy, m₁ = 50 kg
Initial speed of boy, u₁ = 10 m/s
Initial speed of car, u₂ = 0 (at rest)
The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s
Let m₂ is the mass of the cart. Using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg[/tex]
So, the mass of the cart is 150 kg.
Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 Newtons on earth, which has more mass?
a. Object a
b. Object b
c. Same mass
d. Other
Answer:
a. Object A
Explanation:
The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.
The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.
So that,
For object A on the moon, mass = 25 lbs
For object B on the earth, W = 25 N,
W = m x g
25 = m x 10 (g = 10 m/[tex]s^{2}[/tex])
m = [tex]\frac{25}{10}[/tex]
= 2.5 lbs
Mass of object B is 2.5 lbs.
Therefore, the mass of the object A is more than that of B.
A force of 1.50 N acts on a 0.20 kg trolley so as to accelerate it along an air track.
The track and force are horizontal and in line. How fast is the trolley going after acceleration from rest through 30 cm, if friction is negligible?
Answer: The trolley is moving at 2.12m/s
Explanation:
Given from the question that the track and force are horizontal and inline, we have that
F= ma
where F= force= 1.50 N
m= mass = 0.2kg
therefore Acceleration , a = F/ m= 1.50/0.2 =7.5 m/s^2
To find how fast the trolley is going ie the Velocity, v
Having that
initial velocity at rest , u = 0,
acceleration a = 7.5 and
and distance, s = 30 cm = 30/100 = 0.30 m
we use motion equation that
v² = u² + 2 a s
v² = 0² + 2 x 7.5 x 0.30
v² = 4.5
v = [tex]\sqrt{4.5}[/tex]
v = 2.12 m/s
A 4.80 g bullet moves with a speed of 170 m/s perpendicular to the Earth's magnetic field of 5.00×10−5T.
Part A
If the bullet possesses a net charge of 1.06×10−8 C , by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.00 km ?
Answer:
[tex]3.24\times 10^{-7}\ \text{m}[/tex]
Explanation:
m = Mass of bullet = 4.8 g
v = Velocity of bullet = 170 m/s
B = Magnetic field of Earth = [tex]5\times 10^{-5}\ \text{T}[/tex]
q = Charge of bullet = [tex]1.06\times 10^{-8}\ \text{C}[/tex]
a = Acceleration
Time the bullet will be in the air for is [tex]t=\dfrac{1000}{170}=5.88\ \text{s}[/tex]
Force is given by
[tex]F=ma[/tex]
Magnetic force is given by
[tex]F=qvB[/tex]
So
[tex]ma=qvB\\\Rightarrow a=\dfrac{qvB}{m}\\\Rightarrow a=\dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\ \text{m/s}^2[/tex]
From the linear equations of motion we have
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0+\dfrac{1}{2}\times \dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\times 5.88^2\\\Rightarrow s=3.24\times 10^{-7}\ \text{m}[/tex]
The defelection of the bullet is [tex]3.24\times 10^{-7}\ \text{m}[/tex]
If you are riding your bike west to your friend’s house, and you ride the 1.25 miles in 5 minutes, what is your velocity, in miles per hour?
Answer:
velocity = 15 miles / hourExplanation:
distance = 1.25 mile
time traveled = 5 min.
find velocity in miles / hour
solution:
use the formula: velocity = distance / time
velocity = 1.25 mile x 60 min
5 min 1 hour
velocity = 15 miles / hour
Why does a watched pot never boil?
Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 970 km/hr above East Lansing. The downward component of the earth's magnetic field at this place is
This question is incomplete, the complete question is;
Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 970 km/hr above East Lansing. The downward component of the earth's magnetic field at this place is 0.7 × 10⁻⁴ T. (DATA: Assume that the wingspan is 60 meters.)
Answer:
the average induced voltage between the tips of the wings of a Boeing 747 flying is 1.132 Volts
Explanation:
Given that;
Speed S = 970 KM/hr
downward component of the earth's magnetic field B = 0.7 × 10⁻⁴
wingspan 1 = 60min
Velocity V = (970 × 10³) / 3600 = 269.44 m/s
So Average Induced Voltage E = VBI
we substitute
E = 269.44 × (0.7 × 10⁻⁴) × 60
E = 1.132 Volts
Therefore the average induced voltage between the tips of the wings of a Boeing 747 flying is 1.132 Volts
a heavty downpour and thundrstorm is probably caused by
1.) There was an earthquake in Salt Lake City, Utah, on March 18, 2020, in the morning at 9 hours, 9 minutes, and 45 seconds Mountain Standard Time (9:9:45 MST). If the velocity of the p-wave is 7.3 km/sec and the velocity of the s-wave is 5.1 km/sec and the s-p time lag is 16 seconds, what is the distance in kilometers from Salt Lake City to the focus of the earthquake? Explain how you calculated the answer.
Answer:
7 because salt lake and Southis weat
A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
Answer:
Explanation:
To get the focal length, we will use the lens formula;
1/f = 1/u + 1/v
f is the focal length
u is the object distance
v is the image distance
Given
since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.
u = 35cm
v = -2.0
1/f = 1/35-1/2
1/f = 2-35/70
1/f = -33/70
f = -70/33
f = -2.12 cm
f = -0.0212m
Power of a lend is the reciprocal of its focal length
Power of the lens = 1/f
P = 1/-0.0212
P = -47.17dioptres
The power of the lens is -47.17D
Assume you are in the car and the car is moving at a certain speed to
school. Are you at rest or in motion with respect to the school? With
respect to the car?
A spring gun fires a ball horizontally at 15 m/s. It is mounted on a flat car moving in a straight line at 25 m/s. relative to the ground, what is the horizontal speed of the ball when the gun is aimed forward?
Answer:
15 m/s
Explanation:
The Speed Of The Car Does Not Add To The Speed Of The Bullet