Answer:
144.6ohms
Explanation:
v=IR
R =V/I=120/0.83
R=144.6
1) A rock of mass 1.50kg is released from rest at a height of 30m. Ignore air resistance and
calculate;
(a) its speed half way down
(b) its speed on reaching the ground
(c) its total mechanical energy
(i) at a height of 30m
(ii) its kinetic energy half way through
(iii) its potential energy on striking the ground
Here's link to the answer:
bit.[tex]^{}[/tex]ly/3a8Nt8n
Which function represents g(x), a reflection of f(x) = 6(one-third) Superscript x across the y-axis?
Answer:
g(x)=6(3)°x
[tex]g(x) = 6(3){x} [/tex]
Which two mixtures are homogeneous?
Answer:
sand and air
Explanation:
air is a mix of carbon dioxide and oxygen and sand is a mix of rocks and stuff
11) A tank of kerosene with density of 750 kg/m3 has a syphon used to remove the fluid that then exits into the local atmosphere, with pressure of 101 kPa. The pressure above the kerosene in the tank is 120 kPa absolute. The syphon tube has a diameter of 2 cm, exits the tank rising to 10 cm above the level of the kerosene and then drops down to 15 cm below the level of the kerosene where it exits into the atmospheric pressure. Calculate the exit velocity from the tube.
Answer:
[tex]7.32\ \text{m/s}[/tex]
Explanation:
[tex]v_1[/tex] = Velocity at initial point = 0
[tex]P_1[/tex] = Pressure in tank = 120 kPa
[tex]P_2[/tex] = Pressure at outlet = 101 kPa
[tex]\rho[/tex] = Density of kerosene = [tex]750\ \text{kg/m}^3[/tex]
[tex]Z_1[/tex] = Tank height = 15 cm
[tex]Z_2[/tex] = Height of pipe exit = 0
[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From Bernoulli's equation we have
[tex]\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+Z_2\\\Rightarrow \dfrac{P_1}{\rho g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}\\\Rightarrow v_2=\sqrt{2g(\dfrac{P_1}{\rho g}+Z_1-\dfrac{P_2}{\rho g})}\\\Rightarrow v_2=\sqrt{2\times 9.81(\dfrac{120\times 10^3}{750\times 9.81}+0.15-\dfrac{101\times 10^3}{750\times 9.81})}\\\Rightarrow v_2=7.32\ \text{m/s}[/tex]
The exit velocity from the tube is [tex]7.32\ \text{m/s}[/tex].
What is the mass of 1.000 L of seawater? kg
Answer:
1000 L= 1000 kg
1.000 L= 1.000 kg
Explanation:
It will be the same because L and kg have the same mass
Polarizing windows, filters, etc. are often used to reduce the amount of light that enters the lens of a camera or into a room or a car. A library atrium has an overhead skylight that lets in too much light during the day which heats up the interior of the library far too much. The building engineer installs new double paned polarizing sky lights to reduce the intensity. If sunlight, which is unpolarized, has an average intensity of 1250 W/m^2.
Required:
What angle should the polarizing axis of the second pane of the window make with the polarizing axis of the first pane of the window in order to reduce the intensity of the sunlight to 33% of the original value?
Answer:
The answer is "[tex]35.6^{\circ}[/tex]"
Explanation:
The sunlight level of the first panel:
[tex]I_1 = \frac{I_o}{2}[/tex]
When the light of this intensity passes through the second window:
[tex]I_2 = I_1 \cos^2 \theta\\\\I_2 = \frac{I_o}{2} \cos^2 \theta[/tex]
[tex]\frac{I_2}{I_o} = 0.33 (33\%) \\\\[/tex]
therefore,
[tex]0.33 = \frac{1}{2} \cos^2 \theta\\\\\cos^2 \theta = 0.66\\\\\cos \theta = \sqrt{0.66} = 0.8124\\\\\theta = \cos^{-1}( 0.8124) = 35.6^{\circ}\\\\[/tex]
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
Answer:
Wavelength = 1.36 * 10^{-34} meters
Explanation:
Given the following data;
Mass = 0.113 kg
Velocity = 43 m/s
To find the wavelength, we would use the De Broglie's wave equation.
Mathematically, it is given by the formula;
[tex] Wavelength = \frac {h}{mv} [/tex]
Where;
h represents Planck’s constant.
m represents the mass of the particle.
v represents the velocity of the particle.
We know that Planck’s constant = 6.6262 * 10^{-34} Js
Substituting into the formula, we have;
[tex] Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43} [/tex]
[tex] Wavelength = \frac {6.6262 * 10^{-34}}{4.859} [/tex]
Wavelength = 1.36 * 10^{-34} meters
solve this with figure.help me ......
Answer:
[tex] \huge\mathfrak\pink{Hope \: it \: helps}\ [/tex]
places where computers is used
Answer:
Banks and financial.
Business.
Communication.
Defense and military.
Education.
Internet.
Medical.
Transportation.
etc..
Answer:
Super markets
Hospitals
Industries
Explanation:
In supermarket's computers helps them manage and organise data .
5. How does a jack make changing a tire easier?
Answer: An jack makes changing a tire easier because it lifts up the car to get the tire off of the ground.
Explanation:
A honey bee's wings beat at 230 beats per second. If the speed of sound in air is 340 m/s, what is the wavelength of
the sound waves?
Answer:
[tex]from \: the \: wave \: equation \\ velocity = frequency \times wavelength \\ 340 = 230 \times \lambda \\ \lambda = \frac{340}{230} \\ \lambda = 1.5 \: m[/tex]
A trumpeter plays at a sound level of 75dB. three equally loud trumpet players join in. what is the new sound level?
The energy an object possesses due to its motion
Difference between corpuscular theory and wave theory
Answer:
Explanation:
Isaac Newton argued that the geometric nature of reflection and refraction of light could only be explained if light were made of particles, referred to as corpuscles, because waves do not tend to travel in straight lines.
Someone tryna help a homie out
Answer: thats hard...
Explanation:1 c
2 a
3 b
4 d
6 b
7 a
8 b
i tried
If a ball rolls down a hill at a speed of 6.5m/s for 7s, how far did it travel?
Answer:
DABABY CONVERTIBLE LESS GOOO HA YE YE
move the compass around the edge of the black screen far away from the bar magnet does the needle still interact with the poles of the bar magnet?
Answer: Yes
Explanation:
Answer:
Yes. The behavior of the bar magnet and the needle is similar to that in part A.
Explanation:
plato
Choose the statement(s) that are FALSE. A. An object will undergo constant acceleration if it is in equilibrium. B. Subject to the same net force, a larger mass will accelerate at a greater rate than a smaller mass. C. If the vector sum of the forces on an object is not zero, the object is in equilibrium. D. A net force of 125 N acts horizontally on an object that weighs 125 N. Therefore, the object does not accelerate. E. If an object is at rest, then there are no forces acting on the object. F. None of these statements are false.
Answer:
A, B, C, D and E
Explanation:
A. An object will undergo constant acceleration if it is in equilibrium.
This is false because an object in equilibrium has no acceleration.
B. Subject to the same net force, a larger mass will accelerate at a greater rate than a smaller mass.
This is false because, since F = ma and a = F/m where a is acceleration and m = mass. Since F is constant, a ∝ 1/m. So acceleration is inversely proportional to mass. So, the smaller mass would accelerate at a greater rate than the larger mass.
C. If the vector sum of the forces on an object is not zero, the object is in equilibrium.
This is false because an object can only be in equilibrium if the vector sum of the forces acting on it is zero.
D. A net force of 125 N acts horizontally on an object that weighs 125 N. Therefore, the object does not accelerate.
This is false because, the weight acts in the vertical direction and has no effect in the horizontal direction. The only forces in the horizontal direction are the frictional force and the 125 N horizontally applied force.
The objects accelerates if the 125 N applied force is greater than the frictional force and since frictional force = μN = μW where μ = coefficient of friction, N = normal force and W = weight of the object. Since μ < 1, f < W. So f < 125 N. So, the object accelerates.
E. If an object is at rest, then there are no forces acting on the object.
This is false because, an object might be at rest when the net force acting on it is zero not only when no force acts on it.
A 100-n object and a 50-n object are placed on scales a and b respectively inside an elevator ascending with constant velocity 3.0m/s which statement below correctly describes the readings on the scales inside the elevator
Answer: b
Explanation:
The reading of the scale of the elevator ascending with constant velocity is 150 N.
Reading of the scale
The reading of the scale on the elevaor is calculated by applying Newton's second law of motion;
R = m(a + g)
R = ma + mg
R = F + W
where;
a is the acceleration of the objectsAt constant velocity, the acceleration of the object is zero (0).
R = 0 + 100 + 50
R = 150 N
Thus, the reading of the scale of the elevator ascending with constant velocity is 150 N.
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When a wave of light strikes a diffraction screen (or grating) it will cause the wave to pass through each slit, causing the emerging light to radiate outward as if from a new source. These newly formed waves create alternating patterns of constructive and destructive interference, which can be observed when projected onto a screen. Select the variables listed below that determine the angle at which these patterns occur (you might need to select more than one answer):
Answer:
The answer is "the angle depends on wavelength and distance of slits".
Explanation:
[tex]d \sin \theta = m\lambda[/tex] as well as angular, of mth order darkened border is indicated only by angular position, the [tex]\theta[/tex] mth order bright border generated by a light wavelength [tex]\lambda[/tex] , and use the grating of the segregation slit.
[tex]d \sin \theta = ( m + \frac{1}{2}) \lambda[/tex]
Therefore, [tex]\theta[/tex] depends on d and [tex]\lambda[/tex] for a specific order.
An electric motor operates a pump that irri-
gates a farmer's crop by pumping 10 000 L of
water a vertical distance of 8.0 m into a field
each hour. The motor has an operating resis-
tance of 22.0 2 and is connected across a
110-V source.
a. What current does it draw?
b. How efficient is the motor?
Answer:
a. 5A
b. 39.60%
Explanation:
The computation is shown below:
a. The current does it draw is
= v ÷ R
= 110v ÷ 22
= 5A
b. Now the efficiency of the motor is
n = mgh ÷ vlt
= (10,000 × 9.8 × 8) ÷ (5 × 3600 × 110)
= 784000 J ÷ 1,980,000
= 39.60%
hence, the above formulas are applied & the same is relevant
A classic demonstration illustrating eddy currents is performed by dropping a permanent magnet inside a conducting cylinder. The magnet does not go into free fall. Instead it reaches terminal velocity and can take a few seconds to drop a length of about a meter. Suppose the mass of the magnet is 70 g and width of 1.0 cm. It falls with a terminal velocity of 10 cm/s and the length of the pipe is 80 cm. The magnitude of the Joule heating from the eddy currents is approximately:________
Answer:
The correct solution is "0.69 N".
Explanation:
The given values are:
Mass of magnet,
m = 70 g
or,
= 0.07 kg
Width,
= 1.0 cm
Velocity,
= 10 cm/s
Length of the pipe,
= 80 cm
Whenever the velocity is constant, then the net force which is acting on the magnet will be "0".
On the magnet,
The up-ward force will be:
⇒ [tex]F=mg[/tex]
On substituting the values, we get
⇒ [tex]=0.07\times 9.8[/tex]
⇒ [tex]=0.69 \ N[/tex]
a problem solving method that involves trying all possible solutions until one works is using_____.
Answer:
trial and error.
Explanation:
a problem solving method that involves trying all possible solutions until one works is using trail and error.
These two questions are connected to the figure.
Answer:
A, and E
Explanation:
Tarnish is produced by a redox reaction that occurs when a metal reacts with
a nonmetallic compound. The green tarnish on a copper penny might be
produced by a reaction between copper and hydrogen sulfide. What occurs
during this reaction?
A. Hydrogen sulfide acts as an acid, and copper acts as a base.
B. Copper atoms gain electrons from sulfur atoms.
C. A double-replacement reaction takes place.
D. Copper atoms lose electrons to sulfur atoms.
Answer:
Copper atoms lose electrons to sulfur atoms
Explanation:
a p e x (:
The 10-lb block A attains a velocity of 2ft/s in 5 seconds, starting from rest. Determine the tension in the cord and the coefficient of kinetic friction between block A and the horizontal plane. Neglect the weight of the pulley. Block B has a weight of 8 lb. Please work on this by using the principle of linear impulse-momentum
Answer:
Explanation:
Let T be the tension in the cord.
Impulse by cord = change in momentum of block A .
T x 5s = 10 ( 2 -0) = 20
T = 4 poundal .
acceleration of block B = 2 / 5 = 0.4 m /s²
Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .
= 8 ( 32 + .4 ) = 259.2 poundal
Frictional force on block A = 259.2 - 4 = 255.2 poundal
μ x 10 x 32 = 255.2
320μ = 255.2
μ =0 .8 .
solve this with figure.help me ......
[tex] \huge\mathfrak\purple{Hope \: it \: helps}\ [/tex]
do u ever think that how are u living cause we could not even be here and God but made us but had did it all started I believe in god btw I'm just saying how
Answer:
What is the question
Explanation:
lol
why do we preserve food?
Answer:
The primary objective of food preservation is to prevent food spoilage until it can be consumed. Gardens often produce too much food at one time—more than can be eaten before spoilage sets in. Preserving food also offers the opportunity to have a wide variety of foods year-round. It's economic.
Explanation:
A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is trans- formed into elastic potential energy of the bending board, if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the fore. Ultimately the board breaks at an applied force of 800 N and a deflection of 1.2 cm.
a. To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand.
b. If the hand is moving this fast and comes to rest in a dis- tance of 1.2 cm, what is the average force on the hand?
(a) The velocity is "6.2 m/s".
(b) The average force is "800.83 N".
According to the question,
Force,
F = 800 NDeflection,
x = 1.2 cm= [tex]1.2\times 10^{-2} \ m[/tex]
As we know,
The work done,
→ [tex]W = F\times d[/tex]
[tex]= 800\times 1.2\times 10^{-2}[/tex]
[tex]= 9.6 \ J[/tex]
(a)
Given:
Mass of hand,
m = 0.50 kgNow,
→ [tex]\frac{1}{2} mv^2 = 9.6 \ J[/tex]
[tex]v = \sqrt{\frac{2\times 9.6}{0.50} }[/tex]
[tex]= 6.2 \ m/s[/tex]
(b)
→ [tex]v^2 = u^2 +2ax[/tex]
→ [tex]a= \frac{v^2}{2x}[/tex]
[tex]= \frac{(6.2)^2}{2\times 1.2\times 10^{-2}}[/tex]
[tex]= 1601.67 \ m/s^2[/tex]
hence,
The average force will be:
→ [tex]F_{avg} = m\times a[/tex]
[tex]= 0.50\times 1601.67[/tex]
[tex]= 800.83 \ N[/tex]
Thus the above answers are correct.
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