Answer:
19
Explanation:
Given that:
wavelength = 600 nm
Distance (d) = 2.02 cm = 2.02 × 10⁻² m
refraction index of air (n) = 1.00028
Pressure = 1.00 atm
∴
The number of bright-dark-bright fringe shifts can be determined by using the formula:
[tex]\Delta m = \dfrac{2d}{\lambda} (n -1 ) \\ \\ \Delta m = \dfrac{2\times2.02 \times 10^{-2}}{600\times 10^{-9}} (1.00028 -1 ) \\ \\ \Delta m = 67333.33 \times 10^{-5}(1.00028 -1) \\ \\ \Delta m = 67333.33 \times 10^{-5}(2.8\times 10^{-4}) \\ \\ \Delta m = 18.853 \\ \\ \mathbf{\Delta m = 19}[/tex]
Newton's third law states that for every action force there is an equal and opposite reaction force. An idiot in your class says, "Wow that means everything cancels and nothing ever moves, it is all an illluussion! Wowwwwww" What statement best proves to him he is an idiot.
a) The equal and opposite forces act on different objects
b) If there is even a slight imbalance in the third law there will be a net force causing acceleration
C) he's right, all forces cancel, any motion I have ever seen is wrong
[tex] \huge \mathfrak{Answer.... }[/tex]
The Correct Answer is :
B. if there is even a slight imbalance in third law there will be a net force causing acceleration.
A slight difference in the forces can result in acceleration of an object.
[tex] \mathrm{✌TeeNForeveR✌}[/tex]
A sack of groceries with a mas of 22 kg is lifted off the floor with a velocity of 6 m/s. What is the kinetic energy of the sack
of groceries?
the answer is 396 joules :D
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 8.2 times the magnitude of the tangential acceleration. What is the angle?
Answer:
The angle is 4.1 rad.
Explanation:
The centripetal acceleration (α) is given by:
[tex] \alpha = \omega^{2} r [/tex] (1)
Where:
ω: is the angular velocity
r: is the radius
And the tangential acceleration (a) is:
[tex] a = \alpha r [/tex] (2)
Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:
[tex] \omega^{2} r = 8.2\alpha r [/tex]
[tex] \omega^{2} = 8.2\alpha [/tex] (3)
Now, we can find the angle with the following equation:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]
Where:
[tex] \omega_{f}[/tex]: is the final angular velocity [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)
[tex]\Delta \theta[/tex]: is the angle
[tex] \omega^{2} = 2\alpha \Delta \theta [/tex] (4)
By entering equation (3) into (4) we can calculate the angle:
[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]
[tex] \Delta \theta = 4.1 rad [/tex]
Therefore, the angle is 4.1 rad.
I hope it helps you!
Ingrid lives in a cold country, that sometimes gets a lot of snow. when that happens people can enjoy lot pf skiing. Ingrid goes outside to see if the snow is fir for skiing. she sinks into the snow, but when she puts her ski on, she can move over it without sinking. Why?
Answer:
Because the surface area of her skis are greater than the surface are of her shoes
Explanation:
the reason for this is that the weight per in is too heavy crushing the snow blower but with the skies the weight is distributed to the point were the snow can support her weight
A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of wire. When running, the solenoid produced a field of 1.3 T in the center. Given this, how large a current does it carry?
Answer:
I = 2172.46 A
Explanation:
Given that,
The length of a solenoid, l = 2.1 m
The inner radius of the solenoid, r = 28 cm = 0.28 m
The number of turns in the wire, N = 1000
The magnetic field in the solenoid, B = 1.3 T
We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :
[tex]B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}[/tex]
Put all the values,
[tex]I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A[/tex]
So, it carry current of 2172.46 A.
What is the source of almost all energy on Earth?
Earth’s hot core
the Sun
stored carbon
moving water
Answer:
the sun
Explanation:
The sun is the source of almost all energy on Earth beacuse both plants and animal on Earth derive their energy from the sun.
Source of energy on EarthThe Earth is one of the planets that make up the solar system. The Sun is the center of the universe and the Earth revolves round the sun.
The source of almost all energy on Earth is from the sun. The energy from the sun is callled solar energy.
This energy from the sun can be used by the planet to manufatcure its own food. The plants are consumed by animals to provide energy their metabolic activities.
Thus, we can conclude that the sun is the source of almost all energy on Earth beacuse both plants and animal on Earth derive their energy from the sun.
Learn more about solar energy here: https://brainly.com/question/17711999
URGENTT
Which statement best defines the term "superconductivity"?
Answer:
the ability of certain substances at very low temperatures to conduct electricity with no resistance
A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the bucket is gradually increased by the addition of sand. At some point, the bucket will accumulate enough sand to set the block in motion. The coefficients of static and kinetic friction are 0.60 and 0.50 respectively.
Required:
a. Determine the mass of sand in [kg], including the bucket, needed to start the block moving.
b. Find the blocks acceleration, in [m/s^2] up the plane?
Answer:
a). M = 20.392 kg
b). am = 0.56 [tex]m/s^2[/tex] (block), aM = 0.28 [tex]m/s^2[/tex] (bucket)
Explanation:
a). We got N = mg cos θ,
f = [tex]$\mu_s N$[/tex]
= [tex]$\mu_s mg \cos \theta$[/tex]
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + [tex]$\mu_s mg \cos \theta$[/tex] .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
[tex]$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$[/tex]
[tex]$M=2(m \sin \theta + \mu_s mg \cos \theta)$[/tex]
[tex]$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$[/tex]
M = 20.392 kg
b). [tex]$(h-x_m)+(h-x_M)+(h'+x_M)=l$[/tex] .............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
[tex]$-\ddot{x}-2\ddot x_M=0$[/tex]
[tex]$\ddot x_M=\frac{\ddot x_m}{2}$[/tex]
[tex]$a_M=\frac{a_m}{2}$[/tex] .....................(iv)
We got, N = mg cos θ
[tex]$f_K=\mu_K mg \cos \theta$[/tex]
∴ [tex]$T-(mg \sin \theta + f_K) = ma_m$[/tex]
[tex]$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$[/tex] ................(v)
Mg - 2T = M[tex]a_M[/tex]
[tex]$Mg-Ma_M=2T$[/tex]
[tex]$Mg-\frac{Ma_M}{2} = 2T$[/tex] (from equation (iv))
[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}=T$[/tex] .....................(vi)
Putting (vi) in equation (v),
[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$[/tex]
[tex]$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$[/tex]
[tex]$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$[/tex]
[tex]$a_m= 0.56 \ m/s^2$[/tex]
Using equation (iv), we get,
[tex]a_M= 0.28 \ m/s^2[/tex]
A laser emits a single 3.0-ms pulse of light that has a frequency of 2.83E11 Hz and a total power of 65000 W. How many photons are in the pulse? Please provide all equations and work.
6.0E23
1.0E24
2.4E25
3.6E25
4.8E26
Answer:
The number of photons in the pulse is 1.04 x 10²⁴
Explanation:
Given;
frequency of the emitted photons, f = 2.83 x 10¹¹ Hz
duration of the incident light, t = 3 ms = 3 x 10⁻³ s
power of the incident light, P = 65,000 W
The energy of each photon emitted is calculated as;
E = hf
where;
h is Planck's constant, = 6.626 x 10⁻³⁴ Js
E = 6.626 x 10⁻³⁴ x 2.83 x 10¹¹
E = 1.875 x 10⁻²² J
let the number of photons in the pulse = n
n(E)= Power x time
[tex]n = \frac{Pt}{E} \\\\n = \frac{65,000 \times 3\times 10^{-3}}{1.875 \times 10^{-22}} \\\\n = 1.04 \times 10^{24} \ photons[/tex]
) The rate of submergence is the total change in the elevation of the pier (two m) divided by the total amount of time involved (300 years) and is therefore____ cm/yr. (Remember, 1 m 5 100 cm.)
Answer:
0.67cm/year
Explanation:
Since the rate of submergence is the total change in the elevation of the pier (two m) divided by the total amount of time involved (300)
We have total change in the elevation of the pier => 2m => 200cm
The total amount of time involved (300 years)
Hence, we have 200cm ÷ 300 years. = 0.67cm /yr
Therefore, in this case, the correct answer to the question is 0.67cm/year.
In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0342 s, during which time it experiences an acceleration of 186 m/s2. The ball is launched at an angle of 45.9 ° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
Answer:
b) v_y = 4.57 m / s
a) vₓ = 4.43 m / s
Explanation:
This is an exercise in kinematics, where we assume that the acceleration is in the direction of the force and the initial body with zero velocity
v = v₀ + a t
v = 0 + a t
v = 186 0.0342
v = 6.36 m / s
let's use trigonometry to decompose this velocity
sin 45.9 = v_y / v
cos 45.9 = vₓ / v
v_y = v sin 45.9
vₓ = v cos 45.9
v_y = 6.36 sin 45.9
vₓ = 6.36 cos 45.9
v_y = 4.57 m / s
vₓ = 4.43 m / s
ₓ
Which element makes up most of the Sun?
A. Sodium
B. Carbon
C. Lithium
D. Hydrogen
Answer:
D. Hydrogen
Explanation:
The sun is a big ball of gas and plasma. Most of the gas — 91 percent — is hydrogen.
Answer:
D. Hydrogen
Explanation:
Hydrogen makes up most of the Sun. It is nearly 91 percent.
Question 2 of 32
A water-skier with a mass of 68 kg is pulled with a constant force of 980 N by
a speedboat. A wave launches him in such a way that he is temporarily
airbome while still being pulled by the boat, as shown in the image below.
Assuming that air resistance can be ignored, what is the vertical acceleration
that the water-skier experiences on his return to the water surface? (Recall
that g = 9.8 m/s2)
Rope Force
ODON
Weight
O A. - 18.1 m/s2
OB. - 15.6 m/s2
O C. -11.2 m/s2
OD. -9.8 m/s2
Answer:
OD. -9.8 m/s2
Explanation:
The only force vertical force that is acting on the skier is gravity and since its pulling him back it's a negative force down the y axis.
Two identical conductors have charge -1.8 C and 5.5 C on them, respectively. They are connected by a conducting wire for a short period of time and then disconnected. What is the net charge on each of the conductors after the interaction? g
Answer: 1.85 C
Explanation:
Given
charges on the conductors are [tex]-1.8\ C[/tex] and [tex]5.5\ C[/tex]
They are connected by a conducting wire for a short period of time and then disconnected. During this time charge flow from the wire and net charge becomes [tex]5.5-1.8=3.7\ C[/tex]
This charge will be equally distribute among the two conductors i.e. 1.85 C on each conductor.
A canoe has a velocity of 0.330 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.540 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.
Answer:
The velocity of the canoe relative to the river is 0.385 m/s, S37.26⁰W
Explanation:
Given;
velocity of the canoe relative to the earth, [tex]V_{r/e} = 0.33 \ m/s[/tex]
velocity of the river relative to the earth, [tex]V_{r/e} = 0.54 \ m/s[/tex]
The velocity of the canoe relative to the river is calculated as;
[tex]V_{(c/r)x} = V_{(c/e)x}- V_{(r/e)x} \ \ ----(1)\\\\V_{(c/r)y} = V_{(c/e)y}- V_{(r/e)y} \ \ ----(2)[/tex]
The x - component of the velocity of the canoe relative to the earth;
[tex]V_{(c/e)x} = 0.33 \times cos \ 45^0\\\\V_{(c/e)x} = 0.2333 \ m/s[/tex]
The y-component of the velocity of the canoe relative to the earth;
[tex]V_{(c/e)y} = 0.33 \times sin \ 45^0\\\\V_{(c/e)y} = 0.2333 \ m/s[/tex]
Note: velocity of the river relative to the earth has only x-component = 0.54 m/s
Apply equation (1) and (2) to calculate the velocity of the canoe relative to the river;
[tex]V_{(c/r)}x = 0.2333 - 0.54 = -0.3067 \ m/s\\\\V_{(c/r)}y = 0.2333 - 0 = 0.2333 \ m/s\\\\The \ resultant \ velocity;\\\\V_{c/r} = \sqrt{(-0.3067)^2 + (0.2333)^2} \\\\V_{c/r} = 0.385 \ ms/\\\\The \ direction:\\\\\theta = tan^{-1} (\frac{0.2333}{0.3067} ) = 37.26^0 \ south \ west \ of \ the \ river[/tex]
According to Newton’s law of universal gravitation, which statements are true?
List the 5 theoretical perspectives that underlie much of the research on human development. Also, name an individual strongly associated with each perspective.
Answer:
25
Explanation:
5 theroical name indvivdual perspective asssssoitive each persp
What country first colonised Ghana
Answer: Colonialism is a big topic, but it can only be understood by looking at human experiences. Formal colonialism first came to the region we today call Ghana in 1874, and British rule spread through the region into the early twentieth century. The British called the territory the “Gold Coast Colony”.
Explanation: hey, hope this hlps! oh, btw you picked the wrong subject for this question it should have been history insteat of phiscics!
Which shows the formula for converting from degrees Celsius to degrees Fahrenheit?
°F = (9/5 × °C) +32
°F = 5/9 × (°C – 32)
°F = °C – 273
°F = °C + 273
Answer:
the first answer
Explanation:
(32°F − 32) × 5/9 = 0°C
Answer:
Answer: A
Explanation:
A lumberjack is trying to drag a small tree that he cut down. If the static
coefficient of friction of the tree on the ground is 0.5 and the tree weighs 430
N, what is the minimum amount of horizontal force that he will need to apply
so that the tree will start moving?
A. 215 N
B. 430 N
C. 365 N
D. 500 N
Answer:
A
Explanation:
weight of the tree =normal force
Horizontal force =coefficient of friction x Fnormal
0.5×430=215
The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.
Explanation:
The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum
What is the energy equivalent of an object with a mass of 1.05g?
Answer:
The equivalent energy of an object given its mass is calculated through the equation,
E = mc²
where c is the speed of light (3 x 10^8 m/s)
Substituting the known values,
E = (1.05 g/ 1000) (3 x 10^8 m/s)²
E = 9.45x10^13 J
Explanation:
semiconductor have negative temperature coefficient of resistance why
Answer:
As the number of free electrons increases, the resistance of this type of non-metallic material decreases with increasing temperature.
Explanation:
A spring of spring constant k=8.25N/m is displaced from equilibrium by a distance of 0.150m. What is the stored energy in the form of spring potential energy? Show your work.
Answer:
0.0928J
Explanation:
the pulling force of spring F=-kx
where x is the displacement from equilibrium position.
energy stored:
[tex]\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}[/tex]
*** Its fine if you know nothing about calculus. Just apply the equation
[tex]U=\frac{kx^{2} }{2}[/tex]
where U is the potential energy of the spring***
put x=0.150, [tex]U=\frac{8.25}{2}[/tex]×[tex]0.150^{2}[/tex] = 0.0928J (corr. to 3 sig. fig.)
A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutral. When the ruler was brought closer, the paper is attracted to the ruler. The surface of the paper became charged through:_________
Answer: static electricity
Explanation:
When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the positive charge in the piece of paper.
the two factors that affect the amount of heat
Answer:
The two important factors that affect heat energy are specific heat and temperature. Specific heat is a heat-constant of a material per unit mass per degree of temperature change (in units of energy per mass and temperature), like Joules/Kg-°C .
Thank you.....
Have a good day.....
. A wave moves at a constant speed along a string. Which one of the following statements is false concerning the motion of particles in the string?
Answer:
The particle speed is constant.
Explanation:
Particles in gases travel quickly in all directions, frequently clashing with each other and the container's edge. The particles gather kinetic energy and travel faster as the temperature rises. The true average speed of the particles is determined by their mass and temperature; larger particles travel more slower around the same temperature than lighter particles.
Thus, the false statement about a wave moving through a constant speed is that:
The particle speed is constant.
If a sprinter ran a distance of 100 meters starting at his top speed of 11 m/s and running with constant spreed throughout. How long would it take him to cover the distance?
Answer:
9.09 s
Explanation:
If the sprinter ran the 100 meters at the constant speed of 11 m/s it would take him 9.09 s to cover the full distance.
We can find this number by dividing 100 meters (the distance covered) by 11 meters per second (the speed)
[tex]\frac{100}{11} =9.09[/tex]
An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth
Answer:
0 N
Explanation:
This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g
Answer:
a = 5.53 g , a = -15g
Explanation:
This is an exercise in kinematics.
a) Let's look for the acceleration
as part of rest v₀ = 0
v = v₀ + a t
a = v / t
a = 282 / 5.2
a = 54.23 m / s²
in relation to the acceleration of gravity
a / g = 54.23 / 9.8
a = 5.53 g
b) let's look at the acceleration to stop
va = 0
0 = v₀ -2 a y
a = vi / y
a = 282/2 1
a = 141 m /s²
a / G = 141 / 9.8
a = -15g