What
is the the improvetanse of water in our body and how do
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Answer:
Or open Meet and enter this code to the source of definition and how to use it to make
An electric lamp consumes 60W at 220 volts. How many dry cells of 1.5 V and internal resistance 1 Ohm are required to glow the lamp?
Answer:
1. Number of dry cells of 1.5 V required is 40.
2. Number of internal resistance of 1 ohm required is 807
Explanation:
We'll begin by calculating the resistance. This can be obtained as follow:
Power (P) = 60 W
Voltage (V) = 220 V
Resistance (R) =?
P = V²/R
60 = 220² / R
Cross multiply
60 × R = 220²
60 × R = 48400
Divide both side by 60
R = 48400 / 60
R ≈ 807 Ohm
1. Determination of the number of dry cells of 1.5 V required.
Voltage (V) = 220
Dry Cells = 1.5 V
Number of dry cells (n) =?
n = Voltage / Dry cells
n = 60 / 1.5
n = 40
2. Determination of the number of internal resistance of 1 ohm required.
Resistance (R) = 807 Ohm
Internal resistance (r) = 1 ohm
Number of internal resistance (n) =?
n = R/r
n = 807 / 1
n = 807
SUMMARY:
1. Number of dry cells of 1.5 V required is 40.
2. Number of internal resistance of 1 ohm required is 807
The following contribute to accidents when a teen driver has other teens as passengers
Answer:
When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
What are the factors contribute to accidents when a teen driver has other teens as passengers?When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.
Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.
Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
To learn more about factors contribute to accidents refer to:
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Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story. If a 1500 kg car is driven in sand at 4.9 m/s , the coefficient of rolling friction is 0.060. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
Required:
a. What propulsion force is needed to keep the car moving forward at a constant speed?
b. What power is required for propulsion at 5.0 m/s?
c. If the car gets 15 mpg when driving on sand, what is the car's efficiency? One gasoline contains 1.4×10 ^8 J of chemical energy.
Answer:
a) [tex]F_p=882N[/tex]
b) [tex]P=4410W[/tex]
c) [tex]V_p'=24135[/tex] ,[tex]n=15.2\%[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=1500kg[/tex]
Velocity [tex]v=4.9m/s[/tex]
Coefficient of Rolling Friction [tex]\mu=0.06[/tex]
a)
Generally the equation for The Propulsion Force is mathematically given by
[tex]F_p=\mu*mg[/tex]
[tex]F_p=0.06*1500*9.81[/tex]
[tex]F_p=882N[/tex]
b)
Therefore Power Required at
[tex]V_p=5.0m/s[/tex]
[tex]P=F_p*V_p[/tex]
[tex]P=882*5[/tex]
[tex]P=4410W[/tex]
c)
[tex]V_p' =15mpg[/tex]
[tex]V_p'=15*\frac{1609}[/tex]
[tex]V_p'=24135[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=F_p*V_p'[/tex]
[tex]W=882*15*1609[/tex]
[tex]W=2.13*10^7[/tex]
Therefore
Efficiency
[tex]n=\frac{W}{E}*100\%[/tex]
Since
Energy in one gallon of gas is
[tex]E=1.4*10^8J[/tex]
Therefore
[tex]n=\frac{2.1*10^7}{1.4*10^8}*100\%[/tex]
[tex]n=15.2\%[/tex]
The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Complete question:
A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Answer:
The tension on the rope is 1.95 N
Explanation:
The general equation of a progressive wave is given as;
[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]
Compare the given equation with the general equation of wave, the following parameters will be deduced.
A = 0.75 cm
k = 0.400π cm⁻¹
ω = 250π s⁻¹
The frequency of the wave is calculated as;
ω = 2πf
2πf = 250π
2f = 250
f = 250/2
f = 125 Hz
The wavelength of the wave is calculated as;
[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]
The velocity of the wave is calculated as;
v = fλ
v = 125 x 0.05
v = 6.25 m/s
The tension on the rope is calculated as;
[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]
Therefore, the tension on the rope is 1.95 N
A 300 kg block of dimensions 1.5 m × 1.0 m × 0.5 m lays on the table with its largest face.
Calculate:
Area of the largest face
Answer:
1.5
x 1.0
1.50
x 0.5
075.00
answer: 75.00m
Explanation:
I hope this help
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of with the horizontal
Complete Question
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of
40° 16' with the horizontal.
Answer:
[tex]d=8.01m[/tex]
Explanation:
From the question we are told that:
Length of ladder [tex]l=12.5m[/tex]
Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]
Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by
[tex]d=l sin \theta[/tex]
[tex]d=12.5 sin 40.26[/tex]
[tex]d=8.01m[/tex]
3. Calculate the force it would take to accelerate a 50 ka bike at a rate of 3 m/s2 (6 points)
Answer:
150 N
Explanation:
Given that,
Acceleration (a) = 3 m/s²Mass of the bike (m) = 50 kgWe are asked to calculate force required.
[tex]\longrightarrow[/tex] F = ma
[tex]\longrightarrow[/tex] F = (50 × 3) N
[tex]\longrightarrow[/tex] F = 150 N
a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of
Answer:
Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?
This problem has been solved!
This problem has been solved!See the answer
This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of 4.9 m/s. What is the magnitude of the net force acing on the student?
Explanation:
use this R= m(g-a), where R = reaction = weight, m= mass, a= acceleration and g= acceleration due to gravity
The density of blood is 1055 kg/m3 . If the blood at the very top of your head exerts a minimum gauge pressure of 45 mm Hg (6000 Pa), estimate the gauge pressure at your heart in pascals.
Answer:
P = 10135.6 Pa
Explanation:
For this exercise we use that the pressure varies with the height
P = P₀ + ρ g h
where h is the height from the head to the heart, which is approximately
h = 40 cm = 0.40m and P₀ is the head pressure P₀ = 6000 Pa
P = 6000 + 1055 9.8 0.40
P = 6000 + 4135.6
P = 10135.6 Pa
what is the dimensional formula of force and torque
Answer:
Units. Torque has the dimension of force times distance, symbolically T−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule. The unit newton metre is properly denoted N⋅m.
Dimension: M L2T−2
In SI base units: kg⋅m2⋅s−2
Other units: pound-force-feet, lbf⋅inch, ozf⋅in
Answer:
hope it is helpful to you
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state the story of archimedes
Answer:
Archimedes was born about 287 BCE in Syracuse on the island of Sicily. He died in that same city when the Romans captured it following a siege that ended in either 212 or 211 BCE. One story told about Archimedes' death is that he was killed by a Roman soldier after he refused to leave his mathematical work.
Is the actual height the puck reached greater or less than your prediction? Offer a possible reason why this might be.
Answer:
Answer to the following question is as follows;
Explanation:
The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.
A 5.70 g lead bullet traveling at 490 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet
Answer:
461.73 K
Explanation:
Given that,
The mass of a bullet, m = 5.7 g
Speed of the bullet, v = 490 m/s
Half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree.
Using the conservation of energy,
[tex]\dfrac{1}{2}(\dfrac{1}{2}mv^2)=mc\Delta T\\\\\Delta T=\dfrac{v^2}{4c}[/tex]
Where
x is the specific heat of lead, c = 130 J/kg K
So,
[tex]\Delta T=\dfrac{(490)^2}{4\times 130}\\\\=461.73\ K[/tex]
So, the increase in temperature of the bullet is 461.73 K.
Preocupada com o aumento da tarifa na conta de luz, uma pessoa resolve economizar diminuindo o tempo de banho de 20 para 15 minutos. Seu chuveiro possui as seguintes especificações: 4200 W e 220V. Sabendo que o kWh custa R$0,30, a economia feita em 10 dias foi de aproximadamente
Answer:
The mount saved is $ 0.105.
Explanation:
Concerned about the increase in the electricity bill, a person decides to save by reducing bathing time from 20 to 15 minutes. Your shower has the following specifications: 4200 W and 220V. Knowing that the kWh costs R$0.30, the savings made in 10 days were approximately.
The electrical energy is given by
E = P x t
where, P is the electrical power and t is the time.
When he is using the shower for 20 minutes, the energy consumed is
E = 4200 x 20 x 60 = 5040,000 J = 1.4 kWh
When he is using the shower for 15 minutes, the energy consumed is
E' = 4200 x 15 x 60 = 3780000 J = 1.05 kWh
The difference in energy is
E'' = E - E' = 1.4 - 1.05 = 0.35 kWh
The money saved is
= 4 0.3 x 0.35 = $ 0.105
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.
What is the magnitude of the velocity after it hits the ground?
Answer:
9.25 m/s
Explanation:
How many loops are in this circuit?
I see six (6) loops.
I attached a drawing to show where I get six loops from.
Upon completing an interview, it is important that you send a follow-up thank you
note/letter/e-mail because it will show that you are a person who appreciates an opportunity.
A True
B
False
If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
spring? Show MATH, answer and unit.
Answer:
17.8cm
Explanation:
1.3kg --> 4cm
1kg --> 3, 1/13cm
5.8kg --> 18.8cm
A 49.5-turn circular coil of radius 5.10 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.535 T. If the coil carries a current of 26.5 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm
Explanation:
Given;
number of turns of the circular coil, N = 49.5 turns
radius of the coil, r = 5.10 cm = 0.051 m
magnitude of the magnetic field, B = 0.535 T
current in the coil, I = 26.5 mA = 0.0265 A
The magnitude of the maximum possible torque exerted on the coil is calculated as;
τ = NIAB
where;
A is the area of the coil
A = πr² = π(0.051)² = 0.00817 m²
Substitute the given values and solve for the maximum torque
τ = (49.5) x (0.0265) x (0.00817) x (0.535)
τ = 0.00573 Nm
τ = 5.73 x 10⁻³ Nm
Stars have different colors. What causes stars to have colors?
A. location
B. temperature
C. oxygen
D. carbon dioxide
Answer:
temperature
Explanation:
temperature change forms different elements and different element sustain different colour
The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resistor that makes up part of the parallel circuit?
Answer:
The current across the resistance is 0.011 A.
Explanation:
Total resistance, R = 25 ohms
Total current, I = 100 mA = 0.1 A
Let the voltage is V.
By the Ohm's law
V = I R
V = 0.1 x 25 = 2.5 V
Now the resistance is R' = 220 ohm
As they are in parallel so the voltage is same. Let the current is I'.
V = I' x R'
2.5 = I' x 220
I' = 0.011 A
When an automobile moves with constant velocity the power developed is used to overcome the frictional forces exerted by the air and the road. If the power developed in an engine is 50.0 hp, what total frictional force acts on the car at 55 mph (24.6 m/s)
P = F v
where P is power, F is the magnitude of force, and v is speed. So
50.0 hp = 37,280 W = F (24.6 m/s)
==> F = (37,280 W) / (24.6 m/s) ≈ 1520 N
When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]
Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by 2.76 m/s, they then have the same kinetic energy. Calculate the original speeds of the two cars.
Let m be the mass of the second car, so the first car's mass is 2m.
Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.
Let u and v be the speeds of the first car and the second car, respectively. At the start,
• the first car has kinetic energy
K/2 = 1/2 (2m) u ² = mu ² ==> K = 2mu ²
• the second car starts with kinetic energy
K = 1/2 mv ²
It follows that
2mu ² = 1/2 mv ²
==> 4u ² = v ²
When their speeds are both increased by 2.76 m/s,
• the first car now has kinetic energy
1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²
• the second car now has kinetic energy
1/2 m (v + 2.76 m/s)²
These two kinetic energies are equal, so
m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²
==> 2 (u + 2.76 m/s)² = (v + 2.76 m/s)²
Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.
A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18 m/s in the x-direction, 24 m/s in the y-direction, and 72 m/s in the z-direction. The segment has length 0.50 m and is parallel to the z-axis as it moves.
Required:
a. Find the motional emf induced between the ends of the segment.
b. What would the motional emf be if the wire segment was parallel to the y-axis?
Answer:
Explanation:
From the information given:
The motional emf can be computed by using the formula:
[tex]E = L^{\to}*(V^\to*\beta^{\to})[/tex]
[tex]E = L^{\to}*((x+y+z)*\beta^{\to})[/tex]
[tex]E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)[/tex]
[tex]E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))[/tex]
[tex]E = 0.50*((18*0.800)[/tex]
E = 0.72 volts
According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.
As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .
Then the motional emf will be:
[tex]E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))[/tex]
E = 0 (zero)
I need help on this physics problem.
Answer:
the speed of the nerve impulse in miles per hour is 201.59 mi/hr
Explanation:
Given;
the speed of the nerve impulse, v = 90.1 m/s
To convert this speed in meters per second to miles per hour, we use the following method;
1,609 meter = 1 mile
3,600 s = 1 hour
[tex]v(mi/h) = 90.1 \ \frac{m}{s} \times \frac{1 \ mile}{1,609 \ m} \times \frac{3,600 \ s}{1 \ hour} = (\frac{90.1 \times 3,600}{1,609} )\frac{mi}{hr} = 201.59 \ mi/hr[/tex]
Therefore, the speed of the nerve impulse in miles per hour is 201.59 mi/hr
A cat's displacement is 15 meters to the right in 7.0 seconds. If, at the start of the 7.0 seconds, the cat was moving at a velocity of 2.0 m/s left what was its final velocity?
Answer:
6.3 m/s
Explanation:
From the given information:
The displacement (x) = 15 m
time (t) = 7.0 s
initial velocity = -2.0 m/s (since it is moving in the opposite direction)
We need to determine the acceleration then find the final velocity.
By applying the kinematics equation:
[tex]x = ut + \dfrac{1}{2}at^2[/tex]
[tex]15 = (-2.0)(7.0) + \dfrac{1}{2}a(7.0)^2\\ \\ 15 = -14.0 + \dfrac{49}{2}a \\ \\ 29= 24.5a \\ \\ a= \dfrac{29}{24.5} \\ \\ a = 1.184 \ m/s^2[/tex]
Now, to determine the final velocity by using the equation:
v = u + at
v = -2 + 1.184(7.0)
v = 6.288 m/s
v ≅ 6.3 m/s
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.
Answer:
[tex]\lambda=6.83\times 10^{-5}\ m[/tex]
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]
So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].
In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horse-
power motor that can be used to save the ship?
Answer:
P = 0.14 hp
Explanation:
The power required by the ship is given as:
[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]
where,
P = Power = ?
m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg
g = acceleration due to gravity = 9.81 m/s²
h = height = 2 m
t = time = 1 s
Therefore,
[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]
Converting to horsepower (hp):
[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]
P = 0.14 hp
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
