Answer:
It is example of Closed system
If an electromagnetic wave has a frequency of 6×10^5 hz, what is its wavelength? what is its wavelength? A. 2 x 10^12m, B. 5 x 10^14m, C. 5 x 10^2m, 2 x 10^-3m
Answer:
5*10^2
Explanation:
A p e x
A chemist measures the flow of charged ions through a circuit. Which of these would increase the current? Select all that apply.
Two identical circular, wire loops 35.0 cm in diameter each carry a current of 2.80 A in the same direction. These loops are parallel to each other and are 24.0 cm apart. Line ab is normal to the plane of the loops and passes through their centers. A proton is fired at 2600 m/s perpendicular to line ab from a point midway between the centers of the loops.
Find the magnitude of the magnetic force these loops exert on the proton just after it is fired.
Answer:
The answer is "[tex]4659.2 \times 10^{-24} \ N[/tex]"
Explanation:
The magnetic field at ehe mid point of the coils is,
[tex]\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{\frac{3}{2}}}\\\\[/tex]
Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.
[tex]\to B=\frac{(4\pi\times 10^{-7})(2.80\ A) (\frac{0.35}{2})^2}{( (\frac{0.35}{2})^2+ (\frac{0.24}{2})^2)^{\frac{3}{2}}}\\\\[/tex]
[tex]=\frac{(12.56 \times 10^{-7})(2.80\ A) \times 0.030625}{( 0.030625+ 0.0144)^{\frac{3}{2}}}\\\\=\frac{ 1.07702 \times 10^{-7} }{0.0095538976}\\\\=112.730955 \times 10^{-7}\\\\=1.12\times 10^{-5}\ \ T\\[/tex]
Calculating the force experienced through the protons:
[tex]F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N[/tex]
If a 4 Ohm resistor and a 12 Ohm resistor are connected in parallel, what is the total
resistance?
Rt = 3 ohms
Explanation:
Let R1 = 4-ohm resistor
R2 = 12-ohm resistor
For 2 resistors connected in parallel, the total resistance Rt is given by
1/Rt = 1/R1 + 1/R2
or
Rt = R1R2/(R1 + R2)
= (4 ohms)(12 ohms)/(4 ohms + 12 ohms)
= 48 ohms^2/16 ohms
= 3 ohms
Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard nearly 1000 away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1531 m/s
Answer:
the wavelength of the sound in seawater is 90.1 m.
Explanation:
Given;
frequency of the sound, f = 17 Hz
speed of the sound in seawater, v = 1531 m/s
The wavelength of the wave is calculated as follows;
v = fλ
λ = v / f
where;
λ is the wavelength of the sound
λ = 1531 / 17
λ = 90.1 m
Therefore, the wavelength of the sound in seawater is 90.1 m.
Convex lenses cause light rays to _____.
Answer:
to meet at a single point
(causing rays of light that are initially parallel to meet at a single point)
hope this helps:)
Explanation:
Answer:
focus
Explanation:
see more here
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David is driving a steady 30.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.10 m/s2 at the instant when David passes. Part A How far does Tina drive before passing David
Answer:
Explanation:
Let after time t , Tina catches up David .
Distance travelled by them are equal ,
Distance travelled by Tina
s = ut + 1/2 a t²
= .5 x 2.10 t²
= 1.05 t²
Distance travelled by David
= 30 t ( because of uniform velocity )
1.05 t² = 30t
t = 28.57 s
Distance travelled by Tina
= 1/2 a t²
= .5 x 2.10 x 28.57²
= 857 m approx.
Answer: [tex]857\ m[/tex]
Explanation:
Given
Speed of David car [tex]v=30\ m/s[/tex]
Tina begins to accelerate [tex]2.1\ m/s^2[/tex] after David pass the tina
Suppose it took t time for tina to catch David
Distance traveled by David in t time
[tex]\Rightarrow s_d=30\times t[/tex]
Using the equation of motion to get the distance of Tina is
[tex]s_t=ut+\dfrac{1}{2}at^2\\\\s_t=0+\dfrac{1}{2}\times 2.1t^2[/tex]
now, [tex]s_d=s_t[/tex]
[tex]30t=\dfrac{2.1}{2}t^2\\\\\Rightarrow 2.1t^2-60t=0\\\Rightarrow t(2.1t-60)=0\\\Rightarrow t=0,28.57\ s[/tex]
Neglecting [tex]t=0[/tex]
Distance traveled by tina in [tex]28.57\ s[/tex] is
[tex]s_t=\dfrac{1}{2}\times 2.1\times (28.57)^2\\\\s_t=857.057\approx 857\ m[/tex]
Are surface currents warm or cold?
A:war m
B:cold
Answer:
Cold
Explanation:
Im pretty sure im sorry if I am wrong
source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of 10.0 m from this source, the amplitude of the electric field is measured to be 3.50 N>C. What is the electric-field amplitude 20.0 cm from the source
Answer:
[tex]175\ \text{N/C}[/tex]
Explanation:
[tex]E_1[/tex] = Initial electric field = 3.5 N/C
[tex]E_2[/tex] = Final electric field
[tex]r_1[/tex] = Initial distance = 10 m
[tex]r_2[/tex] = Final distance = 20 cm
Electric field is given by
[tex]E=\sqrt{\dfrac{2P}{\pi r^2c\varepsilon_0}}[/tex]
So,
[tex]E\propto \dfrac{1}{r}[/tex]
[tex]\dfrac{E_2}{E_1}=\dfrac{r_1}{r_2}\\\Rightarrow E_2=E_1\dfrac{r_1}{r_2}\\\Rightarrow E_2=3.5\dfrac{10}{0.2}\\\Rightarrow E_2=175\ \text{N/C}[/tex]
The electric field amplitude at the required point is [tex]175\ \text{N/C}[/tex].
g a mass of 1.3 kg is pushed horizontally against a massless spring with a spring constant of 58 n/m until the spring compresses 19.5 cm if the mass is then released what is the kinetic energy of the mass when it is no longer in contact with the spring ignore friction
Answer: [tex]1.102\ J[/tex]
Explanation:
Given
Mass [tex]m=1.3\ kg[/tex]
Spring constant [tex]k=58\ N/m[/tex]
Compression in the spring [tex]x=19.5\ cm\ or\ 0.195\ m[/tex]
When the mass leaves the spring, the elastic potential energy of spring is being converted into kinetic energy of mass i.e.
[tex]\Rightarrow \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}\cdot 58\cdot (0.195)^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}mv^2=1.102\ J[/tex]
The kinetic energy of the mass is 1.102 J.
Using a 2.5 meter long lever, we want to lift a body weighing 15 kilograms. If the lever support so that the pitch of the load is 55 cm, determine the force we must apply to be lift that burden? HURRYYYYY PLEASE ITS A TEST
Answer:
F = 668.9N
Explanation:
Using the formula;
m1d1 = m2d2
m1 and m2 are the masses
d1 and d2 are the distances
Substitute;
15 * 2.5 = x * (55/100) [55cm was converted to metres]
37.5 = 0.55x
Swap
0.55x = 37.5
x = 37.5/0.55
x = 68.18kg
Since F = mg
F = 68.18 * 9.81
F = 668.9N
This gives the required force
Two parallel slits are illuminated by light composed of two wavelengths, one of which is 657 nm. On a viewing screen, the light whose wavelength is known produces its third dark fringe at the same place where the light whose wavelength is unknown produces its fourth-order bright fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown wavelength
Answer:
λ = 5.75 10⁻⁷ mm
Explanation:
This is a slit interference exercise, we analyze each wavelength separately
λ = 657 nm indicate that the third dark pattern
a sin θ = (m + ½) lam
a sin θ = (3 + ½) 657 10⁻⁹
a sin θ = 2299.5 10⁻⁹ nm
for the other wavelength in the same place we have m = 4 bright
a sin θ = m lam
we substitute
2299.5 10⁻⁹ = 4 λ
λ = [tex]\frac{2299.5 \ 10^9 }{ 4}[/tex]
λ = 5.75 10⁻⁷ mm
PHYSICS HELP !! 30 points please answer correctly !! questions attached below
24. A anvil with a mass of 60 kg falls from a height of 9.5 m. How fast is it going right
before it hits the ground?
V= I*R
V = voltage (measured in volts) V
I = current (measured in amperes) A
R = resistance (measured in Ohms) Ω
So they give us this
V=IR
V= 1.8
I=0.4
R=?
So we insert the thing that we know.
1.8=0.4*R
We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.
So we have this.
Lastly we solve.
R=4.5ohms
The formula to find R is V=IR
V/I=R
So the resistance will be the Voltage divided by the Current
The door is 2 m tall. How tall is it in inches? Note: There are 2.54 cm in 1 inch.
A. 78.7 in
B. 500 in
C. 787.4 in
D. 201.4 in
Answer:
Height of the door = 2m = 2000 cm
1 in = 2.54 cm
So 1 cm = 1/2.54 in
2000 cm = 200000/ 254
=
787.401574803
So no.c is correct
The door is 78.7 inch tall. Hence, option (A) is correct.
What is unit of length?Any arbitrarily selected and widely used reference standard for length measurement is referred to as a unit of length. The metric system, which is adopted by every nation on earth, is the most widely utilized in modern times.
The American customary units are also in use in the United States. In the UK and several other nations, British Imperial units are still used sometimes. There are SI units and non-SI units in the metric system.
Given that: the height of the door is = 2 meter
= 2*100 centimeter
= 200 centimeter.
There are 2.54 centimeter in 1 inch.
Hence, the height of the door is = 2 meter = 200 centimeter
= (200/2.54) inch
= 78.7 inch.
The door is 78.7 inch tall.
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Help me with this please
Answer:
check out of phase
Explanation:
this is my answer
I'm a little bit unsure about this question.
Answer:
Option C. 4 Hz
Explanation:
To know the correct answer to the question given above, it is important we know the definition of frequency.
Frequency can simply be defined as the number of complete oscillations or circles made in one second.
Considering the diagram given above, the wave passes through the medium over a period of one second.
Thus, we can obtain the frequency by simply counting the numbers of complete circles made during the period.
From the diagram given above,
The number of circles = 4
Thus,
The frequency is 4 Hz
You purchased 1.9 kg of apples from Wollaston. You noticed that they used a spring scale with the smallest division of 2.1 g to weigh them. What is the relative error in this weight measurement as a percentage
The relative error in this weight measurement as percentage is
[tex]0.1105\%[/tex]What is relative error?is the ratio of the absolute error of a measurement to the measurement being taken. In other words, this type of error is relative to the size of the item being measured.
Therefore,
[tex]\% relative error = \frac{smallest division}{mass of apples}*100\\\\\% relative error = \frac{2.1g}{1900}*100\\\\\% relative error = 0.1105\%[/tex]
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Answer:
The relative error in this weight measurement is approximately 0.1105 %.
Explanation:
We know that,
Relative Error is the ratio of the absolute error of a measurement to the actual measurement . It can be mathematically represented as,
Relative Error = (Measured Value - Actual Value) / Actual Value * 100
In this case, the actual weight is 1.9 kg, and the smallest division of the spring scale is 2.1 g.
Actual Weight = 1.9 * 1000 = 1900 (As 1 kg = 1000g)
We know that the absolute error will be equal to the smallest division hence,
Measured Value - Actual Value = 2.1 g
By replacing this in the general formula we get,
Relative Error = (Measured Value - Actual Value) / Actual Value * 100
= [tex]\frac{2.1}{1900} *100[/tex] %
= [tex]\frac{2.1}{19}[/tex] %
≈ 0.1105 %
Therefore, the relative error in this weight measurement is approximately 0.1105 %.
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A 0.413 kg block requires 1.09 N
of force to overcome static
friction. What is the coefficient
of static friction?
(No unit)
PLEASE HELP!
Answer:
static friction=0.126
A 35.0 g bullet strikes a 5.3 kg stationary wooden block and embeds itself in the block. The block and bullet fly off together at 7.1 m/s. What was the original speed of the bullet? (WILL GIVE BRAINLIEST)
Answer:
= 1200m/s or 1.2 x [tex]10^{3}[/tex] m/s
Explanation:
Light travels at 300,000,000 m/s. This is an example
Answer:
ook soooooo
Explanation:
Moving current has electrical energy.
. A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The
centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car.
four times
twice
half
equal to
Complete question is;
A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The
centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car. How does the Force of the small car FS compare to the force of the large car FL as they round the curve.
four times
twice
half
equal to
Answer:
Half
Explanation:
Formula for centripetal force is given as;
F = mv²/R
Where;
v is velocity
R is radius
Now, centripetal acceleration is given by;
a = v²/R
Since they both travel with the same velocity V and radius remains the same, we can say that;
F = ma
For the small car;
FS = ma
For the big car;
FL = 2ma
This means the force of the small car is half of that of the Large car
Thus;
FS = ½FL
Problem 3:A starship voyages to a distant planet 10 ly away. The astronauts in the starship reach theplanet and then they immediately return to the Earth at the same speed. The round trip from the Earthto the planet and back to the Earth takes 25 years.(a) What is the speed of the starship
Answer:
speed = 0.8c
Explanation:
Given :
Distance from earth to the distant planet = 10 ly
Time taken by the astronauts for the entire journey = 25 years
The time taken to reach the planet is [tex]$t_1=\frac{25}{2}$[/tex]
= 12.5 years
Therefore, speed of the starship can be calculated by :
[tex]$\text{Speed} = \frac{\text{distance}}{\text{time}}$[/tex]
[tex]$v=\frac{10 \times c \times 3.15 \times 10^7}{12.5 \times 3.15 \times 10^7}$[/tex]
[tex]$=0.8c$[/tex]
Therefore the speed of the starship is 0.8c
9. Cellular respiration occurs in what types of cells?
Answer:
Cellular respiration takes place in the cells of all organisms. It occurs in autotrophs such as plants as well as heterotrophs such as animals. Cellular respiration begins in the cytoplasm of cells. It is completed in mitochondria
Explanation:
Cellular respiration takes place in the cells of all organisms. It happening in autotrophs such as plantas as well as heterotrophs such as animals. Cellular respiration starts in the cytoplasm of cells.
It is finished in mitochondria.
g Monochromatic light with wavelength 633 nn passes through a narrow slit and a patternappears on a screen 6.0 m away. The distance on the screen between the centers of thefirst minima on either side of the screen is 32 mm. How wide (in mm) is the slit
Answer:
d = 1.19 x 10⁻⁴ m = 0.119 mm
Explanation:
This problem can be solved by using Young's double-slit experiment formula:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = fringe spacing = 32 mm = 0.032 m
L = slit to screen distance = 6 m
λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m
d = slit width = ?
Therefore,
[tex]0.032\ m = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{d}\\\\d = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{0.032\ m}[/tex]
d = 1.19 x 10⁻⁴ m = 0.119 mm
1. Pam has a mass of 48.3 kg and she is at rest on
smooth, level, frictionless ice. Pam straps on
a rocket pack. The rocket supplies a constant
force for 27.3 m and Pam acquires a speed of
62 m/s.
What is the magnitude of the force?
Answer in units of N.
2. What is Pam’s final kinetic energy?
Answer in units of J.
3. A child and sled with a combined mass of 55.7
kg slide down a frictionless hill that is 11.3 m
high at an angle of 29 ◦
from horizontal.
The acceleration of gravity is 9.81 m/s
3. If the sled starts from rest, what is its speed
at the bottom of the hill?
Answer in units of m/s
Answer:
1. F = 3400 N = 3.4 KN
2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]
3. v = 14.9 m/s
Explanation:
1.
First, we will calculate the acceleration of Pam by using the third equation of motion:
[tex]2as = v_f^2-v_i^2[/tex]
where,
a = acceleration = ?
s = distance = 27.3 m
vf = final speed = 62 m/s
vi = initial speed = 0 m/s
Therefore,
[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]
Now, we will calculate the force by using Newton's Second Law of Motion:
F = ma
F = (48.3 kg)(70.4 m/s²)
F = 3400 N = 3.4 KN
2.
Final kinetic energy is given as:
[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]
[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]
3.
According to the law of conservation of energy:
[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]
where,
v = speed at bottom = ?
g = acceleration due to gravity = 9.81 m/s²
h = height at top = 11.3 m
Therefore,
[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]
v = 14.9 m/s
A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of 0.600c and 0.600c. What is the relative speed of the two ships as measured by a passenger on either one of the spaceships
Answer:
If we use the equation for the transformation of velocities for moving frames:
v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'
v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)
or v' = 1.2 c / (1 + .36) = .88 c
v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u (-.6 c) and we measure the speed of v as seen in the frame moving to the left
What is Hooke's law? what is meant by elastic limit?
please answer me
Answer:
Hooke's law describes the elastic properties of materials only in the range in which the force and displacement are proportional. Hooke's law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. the maximum extent to which a solid may be stretched without permanent alteration of size or shape, is called elastic limit
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Mass, volume and density are all properties of
Answer:
Properties of matter
Explanation:
All properties of matter are either extensive or intensive and either physical or chemical. Extensive properties, such as mass and volume, depend on the amount of matter that is being measured. Intensive properties, such as density and color, do not depend on the amount of matter.
Answer:
The pythagoream theorem
Explanation: