Explanation:
m = 19 oz × (28.3 g/1 oz) = 537.7 g
V = 92.8 mL
[tex]\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}[/tex]
The image shows the right-hand rule being used for a current-carrying wire.
An illustration with a right hand with fingers curled and thumb pointed up.
Which statement describes what the hand shows?
When the current flows down the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows down the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Answer:
The answer is (D): When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Explanation:
A ship is flying away from Earth at 0.9c (where c is the speed of light). A missile is fired that moves toward the Earth at a speed of 0.5c relative to the ship. How fast does the missile move relative to the Earth
Answer:
the required speed with which the missile move relative to the Earth is -0.727c
Explanation:
Given the data in the question;
relative velocity relation;
u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]
so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]
and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]
hence, u' is the speed of B as seen by A
so
u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]
now, given that; V[tex]_A[/tex] = 0.9c and V[tex]_B[/tex] = 0.5c
we substitute
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)
u' = ( -0.4c ) / 1 - 0.45
u' = -0.4c / 0.55
u' = -0.727c
Therefore, the required speed with which the missile move relative to the Earth is -0.727c
A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replaced with a 2.8kg mass.What is the stretch of the spring?
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
Explanation:
Given: Mass = 5 kg
Spring constant = 6 N/m
Formula used to calculate period is as follows.
[tex]T = 2 \pi \sqrt\frac{m}{k}[/tex]
where,
T = period
m = mass
k = spring constant
Substitute the values into above formula as follows.
[tex]T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s[/tex]
Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
A car moving in a straight line uniformly accelerated speed increased from 3 m / s to 9 m / s in 6 seconds. With what acceleration did the car move?
a.
2 m/s2
b.
1 m/s2
c.
0 m/s2
d.
3 m/s2
Answer:
b) 1 m/s
I am sure...........
a. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.
1. the force of the horse pulling on the cart
2. the force of the cart pulling on the horse
3. the force of the horse pushing on the road
4. the force of the road pushing on the horse
b. Suppose that the horse and cart have started from rest; and as time goes on, their speed increases in the same direction. Which one of the following conclusions is correct concerning the magnitudes of the forces mentioned above?
1. Force 1 exceeds Force 2.
2. Force 2 is less than Force 3.
3. Force 2 exceeds Force 4.
4. Force 3 exceeds Force 4.
5. Forces 1 and 2 cannot have equal magnitudes.
Answer:
a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3
Explanation:
a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies
Forces 1 and 2 are action and reaction forces F₁ = F₂
Forces 3 and 4 are action and reaction forces F₃ = F₄
as it indicates that the
b) how the car increases if speed implies that force 1> force3
F₁ > F₃
therefore the correct answer is 3
A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string
Answer:
9.42 m/s
Explanation:
Applying,
V' = Aω.............. Equation 1
Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.
But,
ω = 2πf................. Equation 2
Where f = frequency, π = pie
And,
f = v/λ................ Equation 3
Where, λ = wave length, v = velocity
Also,
v = √(T/μ)................. Equation 4
Where T = Tension, μ = linear density.
From the question,
Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m
Substitute into equation 4
v = √(50/0.005)
v = √(10000)
v = 100 m/s
Also Given: λ = 2.0 m
Substitute into equation 3
f = 100/2
f = 50 Hz.
Substitute the value of f into equation 2
Where π = constant = 3.14
ω = 2(3.14)(50)
ω = 314 rad/s
Finally,
Given: A = 3.0 cm = 0.03 m
Substitute into equation 1
V' = 0.03(314)
V' = 9.42 m/s
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying pan is only 0.150 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)
Answer:
[tex]f_n=3.75N[/tex]
Explanation:
From the question we are told that:
Frictional force [tex]F=0.150N[/tex]
Coefficient of kinetic friction [tex]\mu=0.04[/tex]
Generally the equation for Normal for is mathematically given by
[tex]f_n=\frac{F}{\mu}[/tex]
Therefore
[tex]f_n=\frac{0.150}{0.04}[/tex]
[tex]f_n=3.75N[/tex]
A digital signal differs from an analog signal because it a.consists of a current that changes smoothly. b. consists of a current that changes in pulses. c.carries information. d. is used in electronic devices.
Answer:
d.it is used in electronic devices
The efficiency of a machine can be increased by
Explanation:
the efficiency of a machine can be increased by reducing the friction
please mark the brainliest
In young Goodman’s Brown hawthornes reveals his feelings about his Puritan ancestors when
Answer:
In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others.
Explanation:
hey mate i know it is right so don't worry this is the correct
Answer:
In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others. so that means the answer is D.
Explanation:
A. brown strives to resist his dark mission
B. faith expresses her anxieties about young brown's departure
C. brown discovers his catechism teacher is on speaking terms with the devil
D. the dark man reveals that he helped brown's forebears persecute others(correct answer)
the unit of area is called a derived unit.why?
Explanation:
the unit of area is called a derived unit because it is made of two fundamental unit metre and metre.
A light year is a unit that measures time
Answer:
Light-year is the distance light travels in one year. Light zips through interstellar space at 186,000 miles (300,000 kilometers) per second and 5.88 trillion miles (9.46 trillion kilometers) per year.
Explanation:
A scenario where reaction time is important is when driving on the highway. During the delay between seeing an obstacle and reacting to avoid it (or to slam on the brakes!) you are still moving at full highway speed. Calculate how much distance you cover in meters before you start to put your foot on the brakes if you are travelling 65 miles per hour.
Answer:
66.83 meters
Explanation:
After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.
1 mile = 1609.34 meters (multiply these meters by 65)
65 miles = 104,607 meters
1 hour = 3600 seconds
Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.
[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]
Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...
29.0575m * 2.3s = 66.83 meters
According to the model, when was the universe at its most dense?
A) During the Dark Ages where matter increased in mass.
B) Just before the Big Bang where all matter existed in a singularity.
C) During the nuclear fusion events, as the atoms become more massive.
D) Current day, as the number of galaxies, solar systems, and planets have increased.
Answer:
The Answer is D
Explanation:
Hope this helps!!!!
A hoop rolls with constant velocity and without sliding along level ground. Its rotational kinetic energy is:______a- half its translational kinetic energyb- the same as its translational kinetic energyc- twice its translational kinetic energyd- four times its translational kinetic energy
Answer:
The same as its translational KE.
The easy way to do this is to make up numbers and use them.
So, I'll say m=2 and r=3. I will also say v=3 .
Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)
note: (1/2*I*w^2)
Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)
Now, lets plug our made up values in:
Rot Ke : 1/2 (9*2)(3/3) *note w = v/r
Tran Ke: 1/2(2)(9)
Rot Ke: 9
Tran Ke: 9
9=9, same.
What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?
Answer:
1.2 x 10^-12
Explanation:
3/2.5 x 10^-6/10^6
1.2 x 10^-6 x 10^-6
1.2 x 10^-12
A man weighing 720 N and a woman weighing 500 N have the same momentum. What is the ratio of the man's kinetic energy Km to that of the woman Kw?
Answer:
Because weight W = M g, the ratio of weights equals the ratio of masses.
(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)
but p's are equal, so
K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662Explanation:
A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initially at rest. Which of the following correctly identifies the change in total kinetic energy and the resulting speed of the objects after the collision? Kinetic Energy Speed
(A) Increases 2 m/s 3.2 m/s
(B) Increases Soold 2 m/s
(C) Decreases 3.2 m/s
(D) Decreases
Answer: (d)
Explanation:
Given
Mass of object [tex]m=2\ kg[/tex]
Speed of object [tex]u=5\ m/s[/tex]
Mass of object at rest [tex]M=3\ kg[/tex]
Suppose after collision, speed is v
conserving momentum
[tex]\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s[/tex]
Initial kinetic energy
[tex]k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J[/tex]
Final kinetic energy
[tex]k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J[/tex]
So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.
why the walls of tyres becomes warm as the car moves
Answer:
the particles vibrate inside the tyre
Explanation:
as the car moves kinetic energy is transfered in the tyres which causes the particles to vibrate inside the tyre so the kinetic store is. transferred into thermal
Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of the earth. The second satellite travels at aspeed that is less than v.At what distance from the center of the earth does the secondsatellite orbit?At a distance that is less than r.At a distance equal to r.At a distance greater than r.Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed v_e. An identical satellite is orbiting the moon at thesame distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make onerevolution compare to the time T_e it takes the satellite orbiting the earth to make onerevolution?T_m is less than T_e.T_m is equal to T_e.T_m is greater than T_e.
Answer:
a. At a distance greater than r
b. T_m is greater than T_e.
Explanation:
a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?
Since the centripetal force on any satellite, F equals the gravitational force F' at r,
and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.
Now, F = F'
mv²/r = GMm/r²
v² = GM/r
v = √GM/r
Since G and M are constant,
v ∝ 1/√r
So, if the speed decreases, the radius of the orbit increases.
Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.
So, the distance the second satellite orbits is at a distance greater than r
b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?
Since the speed of the satellite, v = √GM/r where M = mass of planet
Since the satellite is orbiting at the same distance, r is constant
So, v ∝ √M
Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r
Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite
Since r is constant for both orbits, T ∝ 1/v
Now, since the speed of the speed of the satellite on earth orbit v is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e
So, T_m is greater than T_e.
The tendon from the flexor hallucis longus muscle in the hindlimb of a turkey (Meleagris gallopavo) has a resilience of 0.93. If 9.00 J of work are done on the tendon to stretch it out, how many Joules of work does the tendon do as it is relaxing?
Answer:
8.37 Joules
Explanation:
The amount of energy that a substance, such as animal tissue or rubber can take and yet return to its original condition is known as resilience. When we stress such a substance and then let it restore to its previous state, we are referring to the substance to maintain its elastic area of the stress-strain curve.
From the given information:
the resilence is 0.93
The amount of work done during stretching of the tendon = 9.00 J
Thus,
the work done when relaxing = 0.93 × 9.00 J
= 8.37 Joules
why are cows is important?
Answer:
cause they give u milk
Explanation:
Answer:
Cows are important as they provide humans many things for survival. They provide milk, meat, and leather, all of these are important resources.
If 1.02 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved through it?
Answer:
Explanation:
one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]
If the pressure of a gas is really due to the random collisions of molecules with the walls of the container, why do pressure gauges – even very sensitive ones – give perfectly steady readings? Shouldn't the gauge be continually jiggling and fluctuating? Explain.
Answer:
there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.
Explanation:
The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²
To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.
In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface
# _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1
#_molecular = 6.02 10¹⁷ molecules per second
therefore the variation of the number of molecules is not very important
Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.
The latent heat of vaporization of water is roughly 10 times the latent heat of fusion of water. The amount of heat required to boil away 1 kg of water is __________ the amount of heat required to melt 1 kg of ice.
Answer:
The amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice
Explanation:
let the latent heat of fusion of ice = L
then, the latent heat of vaporization of water = 10L
The heat of fusion of 1 kg of ice = 1 x L = L
The heat of vaporization 1 kg of water = 1 x 10L = 10L
Therefore, the amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice
A strong trough in a Rossby wave occurs when the jet stream A. bends towards the Equator. B. bends toward the poles. C. does not bend but maintains an east to west flow. D. does not bend but maintains a west to east flow.
Answer:
A. bends towards the Equator.
Explanation:
Rossby waves are inertial waves that are naturally occurring in a rotating fluids. These waves are also called as the planetary waves.
The Rossby waves are undulated that occur in the polar front jet stream when there is a significant differences in the temperatures between the polar and the tropical air masses.
It occurs when the polar air masses moves towards the equator and when the tropical air masses moves towards the pole. It is formed when the air bends away from the poles and bends towards the equator.
Hence the correct option is (A).
1. Convert the following length into meters
a. 123.50mm
b. 560cm
c. 100dm
d. 125.89km
What are the messing forces that would make the object be in equilibrium?
Answer:
A) 20 N, B) 20 N, & C) 8 N
Explanation:
For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.
1. Determination of A and B.
Forward forces = Backward forces
A + 10 + B = 25 + 25
A + 10 + B = 50
Collect like terms
A + B = 50 – 10
A + B = 40
Assume A and B to be equal. Thus, A is 20 N and B is 20 N.
2. Determination of C
Upward forces = Downward forces
C + 112 = 20 + 100
C + 112 = 120
Collect like terms
C = 120 – 112
C = 8 N
Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.
A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energy will the ball have at the top of its flight? (Assume there is no air resistance.) A. 43.9 J B. 37.5 J C. 48.5 J D. 41.2 J
Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
[tex]v_f=v_0+at[/tex] where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
Δx = [tex]21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2[/tex] which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
