Answer:hi
Explanation:
A chemist measures the flow of charged ions through a circuit. Which of these would increase the current? Select all that apply.
If the output work of a simple machine equals the input work, the machine is said to have ______ efficiency.
If the output work of a simple machine equals the input work, the machine is said to have 100% efficiency.
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Could anyone help me on this question?
You really don't need any help on the question. It's all right there in the picture. What you need help with is the answer.
The number of times the same thing happens each second is called its "frequency". The frequency of the dragonfly's flaps is 477 Hz. (If you're close enough to the dragonfly, you can hear the wings flapping. It sounds like a raspy tone with a frequency of 477 Hz.)
The "period" is just the length of time it takes to happen once. That length of time is just (1 / frequency) .
The dragonfly flaps its wings once every (1 / 477 Hz) = 0.0021 second (C)
the caste system is an example of
Answer:
It is example of Closed system
A French submarine and a U.S. submarine move toward each other during maneuvers in motionless water in the North Atlantic. The French sub moves at 43.00 km/hkm/h , and the U.S. sub at 64.00 km/hkm/h . The French sub sends out a sonar signal (sound wave in water) at 1000 Hz. Sonar waves travel at 5470 km/hkm/h . What is the signal's frequency as detected by the U.S. sub
Answer:
[tex]f_{U}= 1019.72hz[/tex]
Explanation:
From the equation we are told that:
Velocity of French sub [tex]V_{U}= 43.00km/h[/tex]
Velocity of U.S. sub at [tex]V_{F}|=64.00 km//h[/tex]
French Wave Frequency [tex]F_{F}=1000Hz[/tex]
Velocity of wave [tex]V_{s}=5470 km/h[/tex]
Generally the equation for Signal's frequency as detected by the U.S. is mathematically given by
Doppler effect
[tex]f_{U} = \frac{ f_F (vs + v_{U})}{(v_s - v_F) }[/tex]
[tex]f_{U}= \frac{ 1000 x ( 5470+64)}{(5470-43)}[/tex]
[tex]f_{U}= 1019.72hz[/tex]
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
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]
Two long anti-parallel current-carrying wires carry currents of 20 A and 5.0 A. If the wires are separated by 20 cm. Find the magnitude of the magnetic field a point equally separated between both wires.
Answer:
5 x 10⁻⁵ T
Explanation:
Magnetic field at a point d distance away from a current carrying wire having current of I
B = 10⁻⁷ x 2I / d
Here magnetic field due to current of 20 A
B₁ = 10⁻⁷ x 2 x 20 / 10 x 10⁻²
= 4 x 10⁻⁵ T .
Here magnetic field due to current of 5 A
B₂ = 10⁻⁷ x 2 x 5 / 10 x 10⁻²
= 10⁻⁵ T .
These magnetic fields are acting in the same direction
Total magnetic field
B = B₁ + B₂
= 4 x 10⁻⁵ + 10⁻⁵ = 5 x 10⁻⁵ T .
Light travels at 300,000,000 m/s. This is an example
Answer:
ook soooooo
Explanation:
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
Moving current has electrical energy.
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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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.
Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillate up and down, making one complete cycle every 10 seconds. When Boat A is at its peak, Boat B is at its low point and there is a crest in between the two boats. The vertical distance between Boat A and Boat B at their extreme is 8 meters. The wavelength is ___ m, the period is ___ s, the frequency is ___ Hz, and the amplitude is ___ m.
Answer:
wavelength = 24 m
Period = 10 s
f = 0.1 Hz
Amplitude = 4 m
Explanation:
Wavelength:
Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:
wavelength = 24 m
Period:
The period is given as:
[tex]Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\[/tex]
Period = 10 s
Frequency:
The frequency is given as:
[tex]f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\[/tex]
f = 0.1 Hz
Amplitude:
Amplitude will be half the distance between extreme points, that is, crest and trough:
Amplitude = 8 m/2
Amplitude = 4 m
PHYSICS HELP !! 30 points please answer correctly !! questions attached below
cellus
Find the x-component of this
vector:
92.5 m
32.0
Explanation:
x-component:
Vx = Vcos(theta)
= (92.5 m)cos(32.0)
= 78.4 m
Answer:
-78.4
Explanation:
For acellus students
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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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.
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]
In an old Sesame Street skit, Kermit the Frog interviewed a local resident on the planet Koozebane, who measures time in gleeps and distance in glorps. One glorp is defined as the distance a rock will fall from rest in one gleep. How far will a rock fall from rest during the second gleep
Answer:
four glorps
Explanation:
We know :
[tex]$y=v_{0y}t + \frac{1}{2}a_yt^2$[/tex]
[tex]$\Rightarrow -1 \text{glorp} = 0 - \frac{g}{2} \times (1 g\text{ gleep})^2$[/tex]
[tex]$\Rightarrow 1 \text{ glorp}= \frac{g}{2} (1 \text{ gleep})^2$[/tex] .............(i)
Now, t' = 2 gleep
[tex]$y=v_{0y}t + \frac{1}{2}a_yt^2$[/tex]
[tex]$=0+ \frac{-g}{2} (2 \text{ gleep})^2$[/tex]
[tex]$=-\frac{4g}{2}(2 \text{ gleep})^2$[/tex]
[tex]$=4\left[\frac{-g}{2} (\text{gleep})^2\right]$[/tex]
= 4 (-1 gleep) (From (i))
So, |y| = 4 glorp
Copper wires are used as connecting wires because
(1 Point)
Copper has very high melting point
Copper wires are very thick wires
Copper wire offers a lower resistance
Copper wire has an attractive colour
[tex]\huge{ \mathfrak{ \underline{ Answer }\: \: ✓ }}[/tex]
Copper wires are used as connecting wires in circuits because they offer very less resistance.
__________________________________________________________
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. 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
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
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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Help me with this please
Answer:
yong may check po ayon yong sagot
An individual's belief that they can master a situation and produce positive outcomes is
Answer:
optimism
Explanation:
just an idea
....
A wind turbine is rotating 5.98 rad/s when the wind abruptly stops blowing. It takes the turbine 27.5s to stop rotating. how many revolutions does his turbine make before it comes to a stop
Answer:
S = Vo t * 1/2 a t^2 equation for distance traveled
w = w0 t + 1/2 a t^2 equivalent circular equation where w equals omega
w = 5.98 * 27.5 + a (27.5)^2/ 2
a = (w2 - w1) / t = -5.98 / 27.5 = -.217
w = 5.98 * 27.5 - 1/2 * .217 * 27.5^2 = 82.4
Since 1 Rev = 2 pi radians
Rev = 82.4 / 2 * pi = 13.1 Rev
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.
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
Help me with this please
Answer:
check out of phase
Explanation:
this is my answer