Answer:
Explanation:
a) Maximum height id expressed as;
H = u²sin² theta/2g
H = 36.6²(sin42.2)²/2(9.8)
H = 1,339.56(0.6717)/19.6
H = 899.81/19.6
H = 45.91m
Hence the maximum height is 45.91m
b) The Time of flight is the total time in air expressed as;
T = 2usin theta/g
T = 2(36.6)sin42.2/9.8
T = 73.2sin42.2/9.8
T = 49.17/9.8
T = 5.017secs
Hence the total time in air is 5.017scs
c) Range = U²sin2(theta)/g
Range = 36.6²sin2(42.2)/9.8
Range = 1,339.5(0.9952)/9.8
Range = 1,333.11/9.8
Range = 136.03m
Hence the range is 136.03m
d) USing the rime of flight formula;
T =2Usintheta/g
1.5 = 2Usin42.2/9.8
2Usin42.2 =1.5*9.8
2Usin42.2 = 14.7
U = 14.7/2sin42.2
U = 14.7/1.3434
U = 10.94m/s
Hence the speed of the projectile is 10.94m/s
A block slides down an inclined plane from rest. Initially the block is at 4.5m above the ground. Find the speed of the block when it is 1.5m above the ground. 1) 7.7m/s 2) 9.4m/s 3) 5.4m/s 4) 3.2m/s
Since, no external force is acting , so the system is in equilibrium .
Initial total energy = Final total energy
[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )
Therefore, option 1) is correct.
Hence, this is the required solution.
What is the voltage drop across the 10.0 2 resistor?
10.00
120.0 V
20.00
30.00
Answer: 20.0V
Answer:
20.0V
SANA MAKATULONG
The voltage drop across the 10.0-ohm resistor would be 20.0 volt.
What is resistance?Resistance is the obstruction of electrons in an electrically conducting material.
The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.
V=IR
For calculating equivalent resistance in series combination.
Re = R1 + R2 + R3
For the given problem the total resistance of the circuit would be as all the three resistors are connected in the series combination.
Re= 10+20+30
Re=60 ohm
As given in the figure all the resistance are connected in the series combination therefore the current flowing through them would be the same.
For the given problem we have to design and construct a circuit that has two resistors connected in series.
By using Ohms law
V=IR
120 = 60×I
I = 2 ampere
Given that we have a 120 V battery, that will produce a current of 2 Ampere
By using Ohm's law we can calculate the voltage drop across a 10-ohm resistor
V=IR
=2×10
= 20 volt
Thus, the voltage drop across the 10.0-ohm resistor comes out to be 20.0 volt.
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what are the scales of measurement of temperature?
Answer:
There are three temperature scales in use today, Fahrenheit, Celsius and Kelvin. Fahrenheit temperature scale is a scale based on 32 for the freezing point of water and 212 for the boiling point of water, the interval between the two being divided into 180 parts
Answer: I have don't understand this question
Explanation:
If 10 calories of energy are added to 2 grams of ice at -30° C, calculate the final temperature of the ice. (Notice that the specific heat of ice is different from that of water.)
-30° C
40° C
-20° C
30° C
Answer:
-20°C
Explanation:
The specific heat capacity of ice using the cgs system is 0.5cal/g°C
The enthalpy change is calculated as follows
ΔH=MC∅ where M represents mass C represents specific heat and ∅ represents the temperature change.
10cal = 2g×0.5cal/g°C×∅
∅=10cal/(2g×0.5cal/g°C)
∅=10°C
Final temperature= -30°C+ 10°C= -20°C
The earliest mineral observed to showmagnetic properties is called
A leadstone
Blodestone
Cloadstone
Dnone of the above
E all of the above
Answer:
B: lodestone
Explanation:
Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.
Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.
Find the weight of an object with mass 80 kg on the moon ( g = 1.6 m/s^2)
Answer:
80kg = 133 Newtons I'm pretty sure this is right.
The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Weight of the object is force applied due to gravity, So force = mass.gravity .
on the moon gravity is 1/6 of that on the earth.
so weight or force = 80*(9.8/6)
= 130 Newtons
The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.
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Bob rides his bike with a constant speed of 10 miles per hour. How long will he take to travel a distance of 15 miles?
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➡ 150hrs.
[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]
➡ Time = distance × speed
➡ Time = 15*10
➡ Time = 150hrs ans.
A gene carries the blank for the trait
A doctor examines a mole with a 15.0 cm focal length magnifying glass held 12.4 cm from the mole.
a. What is its magnification?
b. Where is the image?
c. How big is the image of a 5.00 mm diameter mole?
Answer:
a. Magnification = 6.1
b. The image formed is virtual, and on the same side of the lens as the object.
c. Image size = 119.8 squared millimetres
Explanation:
Magnification = [tex]\frac{Image distance}{Object distance}[/tex]
But, focal length, f = 15.0 cm, and object distance, u = 12.4 cm. Let the image distance be represented by v.
a. Applying the lens formula, we have;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{15}[/tex] = [tex]\frac{1}{12.4}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{15}[/tex] - [tex]\frac{1}{12.4}[/tex]
= -[tex]\frac{13}{930}[/tex]
v = -75.1538
The image distance, v = -75.2 cm
Magnification = [tex]\frac{75.2}{12.4}[/tex]
= 6.0645
Magnification = 6.1
b. The image formed is virtual, and on the same side of the lens as the object.
c. Given that diameter of mole = 5.00 mm.
Its radius = [tex]\frac{diameter}{2}[/tex] = [tex]\frac{5.0}{2}[/tex]
= 2.5 mm
Thus, the area of the mole would be;
A = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x [tex](2.5)^{2}[/tex]
= 19.643
A = 19.64 square millimetres.
Thus, the size of the image can be determined by;
Magnification = [tex]\frac{Image size}{Object size}[/tex]
Image size = Magnification x object size
= 6.1 x 19.64
= 119.804
The size of the image is 119.8 squared millimetres.
You accidentally drop a book out of the window of a tall building. Assuming
no air resistance, how fast will the book be moving after 2.8 seconds?
Answer:
depends on how talll the building is but lets say its 100 ft tall 12MPH
Explanation:
Answer:
around 9.81m/s i think
Explanation:
PRACTICE
z
1.
What happens to the force of gravity between two objects if the distance between them is doubled?
Answer:
If the mass of one of the objects is doubled, then the force of gravity between them is doubled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
Explanation:
which three statements about gravity in the formation of the solar system are true
The answer is A, B, D.
Answer:
The answer is A, B, D.
Explanation:
ap3x verified
A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3. B) 0.00337 m3. C) 0.00457 m3. D) 0.00766 m3 E) 0.00980 m3
Answer: B) 0.00337 m3.
Explanation:
Given data:
Mass of the ball = 10kg
Weight of the ball in air = 98N
Weight of the ball in water = 65N
Solution:
To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.
= 98 - 65 / 9.8
= 33 / 9.8
= 3.37kg
The volume of the ball is 3.37kg
The density of water is 1kg per Liter.
So 3.37 kg of water would have a volume of 3.37 Liters.
Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).
A household refrigerator consumes electrical energy at the rate of 200 W. lf electricity costs 5 k per kWh, calculate the cost of operating the appliance for 30 days
Answer:
= 720000 [k]
Explanation:
The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.
[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]
= 720000 [k]
If the knee has coordinates of (1.2,2.7) and the ankle has coordinates of (1.5,2.1), with the shank weighing 3.2kg, what is the moment of inertia of the shank about the proximal end point? (%k=0.528)
Answer:
The value is [tex]I = 0.48 \ kg m^2[/tex]
Explanation:
From the question we are told that
The coordinate of the knee is [tex]knee = (1.2,2.7)[/tex]
The coordinate of the ankle is [tex]ankle = (1.5,2.1)[/tex]
The mass of the shank is [tex]m = 3.2 \ kg[/tex]
Generally the length between the knee and the ankle is mathematically represented as
[tex]D = \sqrt{( 1.5 - 1.2)^2+ [ 2.1- 2.7]^2 }[/tex]
=> [tex]D =0.67 \ m[/tex]
Generally the moment of inertia of the shank about the proximal end point is mathematically represented as
[tex]I= m * \frac{D^2}{3}[/tex]
=> [tex]I = 3.2 * \frac{0.67^2}{3}[/tex]
=> [tex]I = 0.48 \ kg m^2[/tex]
If I travel 300 m east, then 400 m west, what is my distance &
displacement?
Answer:100m west
Explanation:
2.5kg of water (C = 4189 J/(Kg x K)) is heated from T1 = 13.5 C to T2 = 22.5 C
To the heated water 0.5 kg of water at T3= 20 C is added. What is the final temperature of the water in Kelvin?
Answer:
338 K
Explanation:
[tex]m_{1}[/tex] = 2.5 kg, c = 4189 J/(kg K), [tex]T_{1}[/tex] = 13.5 [tex]^{o} C[/tex], [tex]T_{2}[/tex] = 22.5 [tex]^{o} C[/tex]
[tex]m_{2}[/tex] = 0.5 kg, [tex]T_{3}[/tex] = 20 [tex]^{o} C[/tex]
Heat loss by hotter water = heat gained by cooler water
[tex]m_{1}[/tex]cΔT = [tex]m_{2}[/tex]cΔT
[tex]m_{1}[/tex]c([tex]T_{2}[/tex] - [tex]T_{1}[/tex]) = [tex]m_{2}[/tex]c([tex]T_{4}[/tex] - [tex]T_{3}[/tex])
2.5 x 4189 x (22.5 - 13.5) = 0.5 x 4189 x ([tex]T_{4}[/tex] - 20)
2.5 x 4189 x 9 = 2094.5 ([tex]T_{4}[/tex] - 20)
94252.5 = 2094.5[tex]T_{4}[/tex] - 41890
94252.5 + 41890 = 2094.5[tex]T_{4}[/tex]
136142.5 = 2094.5[tex]T_{4}[/tex]
[tex]T_{4}[/tex] = [tex]\frac{136142.5}{2094.5}[/tex]
= 65
[tex]T_{4}[/tex] = 65 [tex]^{o} C[/tex]
But,
θ K = 273 + θ[tex]^{o} C[/tex]
= 273 + 65
= 338
The final temperature of the water is 338 K.
The value of the temperature in kelvin is 338 K.
The given parameters;
mass of the initial water, m = 2.5 kgspecific heat capacity of water, C = 4189 J/kg.Kinitial temperature of water, T₁ = 13.5 ⁰Cfinal temperature of the water, T₂ = 22.5 ⁰Cmass of the final water, m = 0.5 kgtemperature of the final water, = 20 ⁰CThe final temperature of the water is calculated as follows as shown below;
[tex]m_1C\Delta \theta_1 = m_2C\Delta \theta_2\\\\m_1\Delta \theta_1 = m_2\Delta \theta_2\\\\2.5(22.5 - 13.5) = 0.5(t_4- 20)\\\\22.5 = 0.5t_4 - 10\\\\0.5t_4 = 32.5\\\\t_4 = \frac{32.5}{0.5} \\\\t_4 = 65 \ ^0C[/tex]
The value of the temperature in kelvin is calculated as;
[tex]t_4 = 273 + 65\\\\t_4 = 338 \ K[/tex]
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It is more difficult to climb a vertical staircasethan a slanted one give reason
Answer:
The reason is similar to the reason why it is difficult to roll an object on a surface with a positive incline than rolling it on the ground
The more the path becomes vertical, the more force we have to apply to oppose the force of gravity
but when we are moving horizontally, we don't have to move against the gravity and hence, it is less difficult than going vertically upwards
A 235-kg merry-go-round at the Great Escape in Lake George is in the shape of a uniform, solid, horizontal disk of radius 1.50 m. It is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s?
Answer:
The constant force to be exerted on the rope is 221.55 N
Explanation:
Given;
mass of the merry, m = 235 kg
radius, r = 1.5 m
number of revolution per second, = 0.4 rev/s
time of motion, t= 2.00 s
The angular acceleration is given by;
[tex]\alpha = \frac{0.4 \ rev}{s} *\frac{2\pi \ rad}{rev} *\frac{1}{2.0\ s} = 1.257 \ rad/s^2[/tex]
Torque is given by;
τ = F.r
Also torque in uniform solid disk is given by;
τ = ¹/₂mr²α
Thus, equating the two;
F.r = ¹/₂mr²α
F = ¹/₂mrα
F = ¹/₂(235)(1.5)(1.257)
F = 221.55 N
Therefore, the constant force to be exerted on the rope is 221.55 N
Two speeding lead bullets, one of mass 15.0 g moving to the right at 295 m/s and one of mass 7.75 g moving to the left at 375 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0°C. Assume the change in kinetic energy of the system appears entirely as increased internal energy. We would like to determine the temperature and phase of the bullets after the collision. (Lead has a specific heat of 128 J/(kg K), a melting point of 327.3°C, and a latent heat of fusion of 2.45 104 J/kg.)
Answer:
The final temperature of the bullets is 327.3 ºC.
Explanation:
Let suppose that a phase change does not occur during collision and collided bullets stop at the end. We represent the phenomenon by the First Law of Thermodynamics:
[tex]K_{A, o} + K_{B, o}-K_{A}-K_{B}+U_{A,o} + U_{B,o}-U_{A}-U_{B} = 0[/tex] (1)
Where:
[tex]K_{A,o}[/tex], [tex]K_{A}[/tex] - Initial and final translational kinetic energies of the 15-g bullet, measured in joules.
[tex]K_{B,o}[/tex], [tex]K_{B}[/tex] - Initial and final translational kinetic energies of the 7.75-g bullet, measured in joules.
[tex]U_{A,o}[/tex], [tex]U_{A}[/tex] - Initial and final internal energies of the 15-g bullet, measured in joules.
[tex]U_{B,o}[/tex], [tex]U_{B}[/tex] - Initial and final internal energies of the 7.75-g bullet, measured in joules.
By definitions of translational kinetic energy and sensible heat we expand and simplify the equation above:
[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T) = 0[/tex] (2)
Where:
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the 15-g and 7.75-g bullets, measured in kilograms.
[tex]v_{A,o}[/tex], [tex]v_{A}[/tex] - Initial and final speeds of the 15-g bullet, measured in meters per second.
[tex]v_{B,o}[/tex], [tex]v_{B}[/tex] - Initial and final speeds of the 7.75-g bullet, measured in meters per second.
[tex]c[/tex] - Specific heat of lead, measured in joules per kilogram-Celsius degree.
[tex]T_{o}[/tex], [tex]T[/tex] - Initial and final temperatures of the bullets, measured in Celsius degree.
Now we clear the final temperature of the bullets:
[tex](m_{A}+m_{B})\cdot c \cdot (T-T_{o}) = \frac{1}{2}\cdot [m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})][/tex]
[tex]T-T_{o} = \frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex]
[tex]T= T_{o}+\frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex] (3)
If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the final temperature of the collided bullets is:
[tex]T = 30\,^{\circ}C+\frac{(15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right)}[/tex]
[tex]T = 852.534\,^{\circ}C[/tex]
Given that found temperature is greater than melting point, then we conclude that supposition was false. If we add the component of latent heat of fussion, then the resulting equation is:
[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)-U = 0[/tex] (4)
[tex]U=\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)[/tex]
If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]T = 327.3\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then latent heat received by the bullets during impact is:
[tex]U =\frac{1}{2}\cdot (15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right) \cdot (30\,^{\circ}C-327.3\,^{\circ}C)[/tex][tex]U = 331.872\,J[/tex]
The maximum possible latent heat ([tex]U_{max}[/tex]), measured in joules, that both bullets can receive during collision is:
[tex]U_{max} = (m_{A}+m_{B})\cdot L_{f}[/tex] (5)
Where [tex]L_{f}[/tex] is the latent heat of fusion of lead, measured in joules per kilogram.
If we know that [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex] and [tex]L_{f} = 2.45\times 10^{4}\,\frac{J}{kg}[/tex], then the maximum possible latent heat is:
[tex]U_{max} = (15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(2.45\times 10^{4}\,\frac{J}{kg} \right)[/tex]
[tex]U_{max} = 557.375\,J[/tex]
Given that [tex]U < U_{max}[/tex], the final temperature of the bullets is 327.3 ºC.
What is the energy contained in a 0.950 m3 volume near the Earth's surface due to radiant energy from the Sun?
An example of a short-term goal is
a.) becoming fluent in Japanese
b.) finishing the annual report in two weeks
c.) earning a college degree
d.) advancing to vice-president of the company
Answer:
a
Explanation:
all the other ones take years and hard work but learning japanese is pretty easy and you dont need to work to hard
Aluminum wire with a diameter of 0.8650 mm is wound onto a spool. The wire is insulated, but you have access to both ends. The resistivity of aluminum at 20.0 °C is 2.65 x 10^-8 Ω-m. You measure the resistance of the wire at that temperature, and it is 2.48 Ω. What is the length of the wire?
a. 8.10 x 10^4 m
b. 22.0 m
c. 5.68 m
d. 0.111 m
e. 55.0 m
Answer:
e. 55.0 m
Explanation:
Given;
diameter of the aluminum wire, d = 0.865 mm
radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m
resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m
resistance of the wire, R = 2.48 Ω
The resistance of a wire is given by;
[tex]R = \frac{\rho \ L}{A} \\\\[/tex]
where;
L is length of the wire
A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²
Substitute the givens and solve for L,
[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]
Therefore, the length of the wire is 55.0 m
hi! hina can you help me with those 2 questions
ONLY ANSWER IF YOU PLAY ADOPT ME ON ROB.LOX!!! tell me your best pet and user name in your answer plz.
READ!!!!! p.s. i have a fly ride T rex and more! also, STAY ON THE PAGE SO I COULD RESPOND AND WE COULD PLAY! yay!!
Answer:
bruh
Explanation:
this is to help people not play games lolllll
a 1000kg car uses a breaking force of 10,000N to stop in two second. What impulse acts on the car?
Answer:
5,000
Explanation:
Vf = Vi + a * t
(iii) Calculate the distance travelled by the car in part Q.
Use the equation
distance travelled = average speed x time
(2)
distance travelled = ....... m
Answer:
distance travelled = 3000m
Explanation:
distance travelled = average speed x time
=30m/s*100s
=3000m
Convert 451 milliliters to fluid
ounces. Round your answer to 2
decimal places. **There are 29.57
milliliters in 1 fluid ounce***
Answer:
451 milliliters equals 15.25 fluid ounces
Explanation:
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.
To solve a direct rule of three, the following formula must be followed:
a ⇒ b
c ⇒ x
So: [tex]x=\frac{c*b}{a}[/tex]
The direct rule of three is the rule applied in this case where there is a change of units.
In this case, the rule of three can be applied in the following way: if there are 29.57 milliliters in 1 fluid ounce, in 451 milliliters how many fluid ounces are there?
[tex]fluid ounces=\frac{451 mL*1 fluid ounce}{29.57 mL}[/tex]
fluid ounces= 15.25
451 milliliters equals 15.25 fluid ounces
2 Magnetism comes from the word
A Magnesia
B Magnentia
C Magnesium
D Magenta
E None of the above
Answer:
A Magnesia
Explanation:
The word magnetism comes from the word Magnesia, which is the name for a region in Asia Minor, where fragments of Fe3O4 ore (magnetite) were found, which attracts other metal objects.
Magnetism is a physical phenomenon by which we describe the attractive or repulsive force between materials.
This phenomenon has been known for thousands of years.
a current of 200 mA through a conductor converts 40 joules of electrical energy into heat in 30 seconds determine the p
otential drop across the conductor
Answer:
ou have I=200mA, E=40J, t=30s, and you want to find the voltage drop.
First, you should know that P=V⋅I , so V=PI
Second, you have the amount of energy converted in a certain amount of time, so E=P⋅t
So, find the power and use it to find the voltage drop.
this works , but i thought energy was defined by W = P * t whitch would then be P = W/t