Complete question is;
A cross-country skier moves 36 meters eastward, then 44 meters westward, and finally 22 meters eastward. What's the Magnitude and Direction?
Answer:
Magnitude = 14 m
Direction = eastward
Explanation:
The magnitude and direction of this cross-country skier will be the gotten from his displacement.
Now, from directions of 4 cardinal points, we can say that the eastward direction is on the positive x-axis and is positive while the westward direction is on the negative x-axis and is negative.
From those positive and negative signs above, we can represent the movement of the skier as;
moves 36 meters eastward = 36 m
moves 44meters westward = -44 m
Moves finally 22 meters eastward = 22
Thus, displacement will be:
36 - 44 + 22 = 14
It is positive, and thus, the magnitude is 14 and the direction is east ward.
Which structures are found in the middle ear? Check all that apply.
A. hammer
B. anvil
C. ear canal
D. cochlea
E. ear drum
F. stirrup
Answer:
E, A, B, F
Explanation:
The Middle Ear contains:
eardrum.
cavity (also called the tympanic cavity)
ossicles (3 tiny bones that are attached) malleus (or hammer) - long handle attached to the eardrum. incus (or anvil) - the bridge bone between the malleus and the stapes. stapes (or stirrup) - the footplate; the smallest bone in the body.
Answer: rawr
Explanation:
hammer
anvil
ear drum
stirrup
two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of 20.g and sphere B has a mass of 10.g.
If both spheres leave the edge of the table at the same instant, sphere A will land
a. at some time after sphere B.
b. at the same time as sphere B.
c. at some time before sphere B.
d. There is not enough information to decide.
A would land before since its heavier
What electromagnetic wave is a is a TV remtoe
Answer:
infared radiation
Explanation:
How would the mass and weight of an object on the Moon compare to the mass and weight of the same object on Earth? * Mass and weight would both be less on the Moon. Mass would be the same but its weight would be less on the Moon. Mass would be less on the Moon and its weight would be the same. Mass and weight would both be the same on the Moon.
Answer:
B. Mass would be the same but its weight would be less on the Moon.
Explanation:
The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
Thus, the mass of a body is constant either on the Earth or on the Moon. But the weight would be less on the Moon because the gravitational force on the Moon is far less than that on the Earth. Therefore the weight would be less on the Moon.
The appropriate option is B.
The mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
The prime focus to solve this problem is the mass and weight of an object. The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
So, the mass of a body is constant either on the Earth or on the Moon. But the weight of an object will depend on the mass and the gravitational acceleration.
W = mg
Here, W is weight, m is mass and g is gravitational acceleration.
Weight would be less on the Moon because the gravitational force on the Moon is far less (due to lower value of g) than that on the Earth. Therefore the weight would be less on the Moon.
Thus, we can conclude that the mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
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Define conductor and insulator, including how the resistance is different in the two, and give at least one example of each.
Answer:
Those substances which can conduct electricity are called conductors, while those substances which don't conduct electricity are called insulators.
Resistance is the obstruction provided by the material through which the current passes,so since conductors conduct electricity and insulators don't,so the obstruction i.e resistance provided by the conductor must be less,while insulators being unable to conduct electricity,has very high resistance.
Example of conductor is copper
Example of insulator is plastic
A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall?
a. 9.8 kN.
b. 8.4 kN.
c. 7.7 kN.
d. 9.1 kN.
e. 1.2 kN.
Given that,
Mass of the object, m = 1.2 kg
Initial speed of the object, u = 8 m/s
Final speed of the object, v = -6 m/s (in opposite direction)
Time, t = 2 ms
To find,
The average force on the object by the wall.
Solution,
Let F be the force. Using Newton's second law of motion,
F = ma, a is acceleration
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1.2\times ((-6)-8)}{2\times 10^{-3}}\\\\=8400\ N[/tex]
or
F = 8.4 N
So, the magnitude of average force in the object by the wall is 8.4 N.
Find the current if 55 C of charge pass a particular point in a circuit in 5 seconds.
Answer:
The current is 11 Amperes
Explanation:
Electric Current
The electric current is defined as a stream of charged particles that move through a conductive path.
The current intensity can be calculated as:
[tex]\displaystyle I=\frac{Q}{t}[/tex]
Where:
Q = Electric charge
t = Time taken by the charge to move through the conductor
The current intensity is often measured in Amperes.
The charge passing through a point in a circuit is Q= 55 c during t=5 seconds, thus the current intensity is:
[tex]\displaystyle I=\frac{55}{5}[/tex]
I = 11 Amp
The current is 11 Amperes
Two motorcycles are traveling due east with different velocities. However, 3.63 seconds later, they have the same velocity. During this 3.63-second interval, motorcycle A has an average acceleration of 4.55 m/s2 due east, while motorcycle B has an average acceleration of 18.9 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 3.63-second interval, and (b) which motorcycle was moving faster
Answer:
52.095 m/s
Motorcycle a was moving faster
Explanation:
We start by using one of the equations of motion
V = u + at
If the first motorcycle starts with an initial speed of u(a) and accelerates at a value of a(a) = 4.55 m/s², then the final speed after a time of 3.63 seconds is V(a). We then represent it as
V(a) = u(a) + a(a).t
If the second motorcycle starts with an initial speed of u(b) and accelerates at a value of a(b) = 18.9 m/s², then the final speed after a time of 3.63 seconds is V(b). We then represent it as
V(b) = u(b) + a(b).t
Assuming that the final speeds v(a) = v(b), and then subtract the equation of the second motorcycle from that of the first, we have
0 = u(a) - u(b) + a(a).t - a(b).t
-u(a) + u(b) = a(a).t - a(b).t, on rearranging, we have
u(b) - u(a) = [a(a) - a(b)]t
Since we have the values for acceleration and the time, we substitute so that
u(b) - u(a) = (4.55 - 18.9)3.63
u(b) - u(a) = -14.35 * 3.63
u(b) - u(a) = -52.095, or we rearrange to get
u(a) - u(b) = 52.095 m/s
A disk rotates at a constant angular velocity of 30 degrees per second. Consider a point on the edge of the disk. Through how many degrees has it rotated after 3 seconds?
Answer:
The disk covers a rotation of 90º after 3 seconds.
Explanation:
Since the disk rotates at constant angular speed, we can determine the change in angular position ([tex]\Delta \theta[/tex]), measured in sexagesimal degrees, by the following kinematic formula:
[tex]\Delta \theta = \omega\cdot \Delta t[/tex] (1)
Where:
[tex]\omega[/tex] - Angular velocity, measured in sexagesimal degrees per second.
[tex]\Delta t[/tex] - Time, measured in seconds.
If we know that [tex]\omega= 30\,\frac{\circ}{s}[/tex] and [tex]\Delta t = 3\,s[/tex], then the change in angular position is:
[tex]\Delta \theta = \left(30\,\frac{\circ}{s} \right)\cdot (3\,s)[/tex]
[tex]\Delta \theta = 90^{\circ}[/tex]
The disk covers a rotation of 90º after 3 seconds.
A heat pump has a coefficient of performance of 3.85 and operates with a power consumption of 7020 W. How much energy does it deliver into a home during 1 h of continuous operation?
Answer:
97.3 MJ
Explanation:
The formula for the coefficient of Perfomance is given as
COE = Q/W, where
COE is the coefficient of Perfomance
Q is the heat provided
W serves as the work input.
Dividing both sides of the equation by a factor of time t, we get the coefficient of Perfomance in terms of heating power and input power, so we say
COE = P / P(i),
making heating power, P the subject of formula, we have
P = COE * P(i)
P = 3.85 * 7020 * 1 * 3600
P = 97297200 J
P = 97.3 MJ
The speed of sound in air is 10 times faster than the speed of a wave on a certain string. The density of the string is 0.002kg/m. The tension in the string is __________.
Answer:
The tension on the string is 2.353 N.
Explanation:
Given;
the speed of sound in air, v₀ = 343 m/s
then, the speed of sound on the string, v = 343 / 10 = 34.3 m/s
mass per unit length, m/l = μ = 0.002 kg/m
The speed of sound on the string is given as;
[tex]v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu[/tex]
where;
T is the tension on the string
T = (34.3)²(0.002)
T = 2.353 N
Therefore, the tension on the string is 2.353 N.
A popular ride at an amusment park lifts
customers up to a height of 50 m and then
drops them threw a displacement of 50 m
before slowing them to a stop. How fast
are the customers going at the 50 m
mark?
Answer:
[tex]31.32\ m/s[/tex]
Explanation:
[tex]We\ are\ given\ that:\\Height\ to\ which\ there're\ lifted=50m\\Displacement\ during\ the\ descent=50m\\Now,\\In\ order\ to\ find\ the\ velocity\ of\ the\ customers\ at\ 50\ m,\\We\ can\ use\ the\ Third\ Equation\ Of\ Motion,\ which is:\\2as=v^2-u^2\\As\ we\ know\ that,\\Acceleration\ due\ to\ gravity=9.81\ m/s^2\ or\ roughly\ 10\ m/s^2\\Displacement=50\ m\\Initial\ velocity=0\ m/s^2\\ [As\ they\ stop\ when\ they\ reach\ the\ maximum\ height\ of\ 50\ m\\ and\ begin\ their\ descent][/tex]
[tex]By\ reconstructing\ the\ Third\ Equation\ Of\ Motion,\ we\ have:\\2gs=v^2\\Hence,\\v^2=2*9.81*50 \\v^2=981\ m^2/s^2 \\v=\sqrt{981\ m^2/s^2} \\v=31.32\ m/s[/tex]
Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on?
Answer:
None of the mass or the radius of the sphere
Explanation:
When a uniform solid sphere of any given mass, say M and any given radius, say R, rolls without slipping downwards an inclined plane that starts from rest. The linear velocity of the sphere at about the bottom of the inclined happens not to depend on either of its mass or that of the radius of its sphere.
PHYSICS! HELP ME PLEASE ASAP!
1.) If an ambulance passes you while you're walking your dog, a change in pitch is ____?
A.) not perceived by either the pedestrian or you
B.) perceived by the you
C.) perceived by the driver
2.) In the star wars movies, the soundtrack is very important. musicians and sound technicians spend weeks perfecting the noise of the light sabers, lasers, tie fighters and explosions. in real life, there would be no sound at all in space. Why?
A.) mechanical waves require a medium to travel
B.) sound is an electromagnetic wave
C.) mechanical waves do not require a medium to travel
D.) sound is a product of vibration
3.) what is a compression?
A.) region of zero pressure in a medium caused by a wave passing
B.)region of a high altitude in a medium caused by a passing wave
C.) region of a high pressure in a medium cause by a passing wave
D.) region of low pressure in a medium caused by a passing wave
Answer:
1B, 2B and 3C hope this helps
Explanation:
A 2150 kg car, moving east at 10.0 m/s, collides and joins with a 3250 kg car. The cars move east together at 5.22 m/s. What is the 3250 kg car’s initial velocity calculated to the nearest tenth? Record your answer in the boxes below. Be sure to use the correct place value.
Answer:
2.1 m/s
Explanation:
According to law of conservation of momentum;
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the common velocity
Given
m1 = 2150kg
m2 = 3250kg
u1 = 10.0m/s
u2 = ?
v = 5.22m/s
Substitute and get u2
2150(10) + 3250u2 = (2150+3250)5.22
21,500 + 3250u2 = 5400(5.22)
3250u2 = 28,188 - 21500
3250u2 = 6688
u2 = 6688/3250
u2 = 2.1 m/s
Hence the 3250 kg car’s initial velocity has an initial velocity of 2.1 m/s
Two spheres, 1.00 kg each, whose centers are 2.00 m apart, would have what gravitational force between them? A. 3.14 X 10-17 N
B. 1.67 X 10-11 N
C. 8.17 X 10-6N
D. 5.78 X 10-6 N
Answer: B
Explanation: the teacher just told us the answer
The gravitational force between the two spheres is [tex]1.67 \times 10^{-11} \ N[/tex].
The given parameters;
mass of each sphere, m = 1.00 kgdistance between their center mass, r = 2 mThe gravitational force between the two spheres is determined by applying Newton's law of universal gravitation as shown below;
[tex]F = \frac{Gm_1 m_2 }{r^2} \\\\[/tex]
where;
G is universal gravitation constant = 6.67 x 10⁻¹¹ N/m[tex]F = \frac{(6.67\times 10^{-11})\times (1\times 1)}{2^2} \\\\F = 1.67 \times 10^{-11} \ N[/tex]
Thus, the gravitational force between the two spheres is [tex]1.67 \times 10^{-11} \ N[/tex].
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A bicycle has a momentum of 36 kg* m/s and a very!I city of 4 m/s.What is the mass of the bicycle?
p = 36 kgm/s
v = 4m/s
we know that,
p = mv
so,
[tex]m = \frac{p}{v} [/tex]
[tex]m = \frac{36}{4} [/tex]
[tex]m = 9kg[/tex]
A carousel at the local carnival rotates once every 45 seconds.
(a) What is the linear speed of an outer horse on the carousel, which is 2.75 m from the axis of rotation?
(b) What is the linear speed of an inner horse that is 1.75 m from the axis of rotation?
Answer:
We know that the carousel does a complete rotation in 45 seconds.
Then the frequency of this carousel will be f = 1/45 seconds.
And the angular frequency will be 2*pi times the frequency, then we have:
angular frequency = w = 2*3.14*(1/45s) = 0.1396 s^-1
Now, the linear speed of an object that rotates with a radius R, and an angular frequency W is:
S = R*W
then:
a) in this case the radius is 2.75m, then the linear speed is:
S = 2.75m*0.1396 s^-1 = 0.3839 m/s
b) in this case the radius is 1.75m, then the linear speed here is:
S = 1.75m*0.1396 s^-1 = 0.2443 m/s
(a) The linear speed of an outer horse on the carousel is 0.384 m/s.
(b) The linear speed of an inner horse on the carousel is 0.244 m/s.
Given data:
The time interval for the rotation of carousel is, t = 45 s.
The distance of the outer horse from the axis of rotation is, r = 2.75 m.
The distance of an inner horse from the axis of rotation is, r' = 1.75 m.
(a)
The linear speed in this problem can be obtained from the concept of rotational mechanic, in which the ratio of the circumference and the time gives required linear speed. So,
v = 2 π r/t
Solving as,
v = 2 π (2.75) / 45
v = 0.384 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.384 m/s.
(b)
Now similarly the linear speed of an inner horse is calculated as,
v' = 2 π r' / t
Solving as,
v' = 2 π (1.75) / 45
v' = 0.244 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.244 m/s.
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20 pts.
Which of the following statements is true?
O Electromagnets use electrlcity and magnets.
O Magnetic fields are strongest around the poles of a magnet.
O The south pole of a magnet will repel the south pole of another magnet.
O all of the above
Answer:
all are true so d is right
Explanation:
Electromagnets use electrlcity and magnets is true.
Magnetic fields are strongest around the poles of a magnet is true.
The south pole of a magnet will repel the south pole of another magnet is true
and since all of them is true the answer is d all of the above
A gas cylinder holds 0.10 mol of O2 at 150 C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halved
Part A
What is the final volume?
Part B
What is the final temperature?
Answer:
V2 = 1.899*10^-3 m^3
T2 = 347.125 K
Explanation:
Using gas law, we know that
PV = nRT,
Where
V1 = 0.00115743 m^3.
gamma = 1.4
Now, when we solve for final volume, V2 we get
V2 = V1/((P2/P1)^(1/gamma))
V2 = 1.899*10^-3 m^3
Using the same law and method, when we try to solve for the temperature, we find that the final temperature, T2 is
T2 = T1*((V1/V2)^(gamma-1))
T2 = 347.125 K
The final volume is 1.899*10^-3 m^3
And, the final temperature is 347.125 K
Gas law:here we used gas law,
we know that
PV = nRT,
Here
V1 = 0.00115743 m^3.
gamma = 1.4
Now final volume is
V2 = V1/((P2/P1)^(1/gamma))
V2 = 1.899*10^-3 m^3
Now the final temperature is
T2 = T1*((V1/V2)^(gamma-1))
T2 = 347.125 K
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What is the condition required of the phase difference (in radians) between two waves with the same wavelength if these waves interfere constructively?
a. (2m +1)π where m= 0, +1, +2, etc.
b. mπ where m = 0, +1, +2, etc.
c. 2mπ where m = 0, +1, +2, etc.
d. (m+1)π where m = 0, +1, +2, etc.
Answer:
c.
Explanation:
In order to two waves with the same wavelength can interfere constructively, their crests and valleys must coincide in space, so the phase difference must be equal to an integer number of wavelengths, i.e. m *(2 π rad), where m= 0, +1, +2, etc.This is equal to the stated by the answer c) , so c) it's the right answer.what is mean by combination reaction ?
[tex] \underline{\purple{\large \sf Combination \: reaction :-}} [/tex]
Those reaction in which two or more substances combine to form a one new substance are called Combination reaction
In this reaction, We can add :
Two or more elements can combine to form a compound.Two or more compounds can combine to from a one new compound.An element and a compound can combine to form a new compound.[tex] \underline{\green{\large \sf For\: example :}} [/tex]
[tex] \sf 2H_{2} + O_{2} \: \underrightarrow{Combination} \: 2H_{2}o[/tex]
In this, Hydrogen is an element and Oxygen is another element. Both are combined to form compound 'Hydrogen oxide'. Hydrogen oxide is commonly known as water.
Through what angle in degrees does a 33 rpm record turn in 0.32 s?
63°
35°
46°
74°
Answer:
1 rev = 2(pi) rad pi(rad) = 180 degrees
so 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s = 63.36 degrees
Explanation:
63.36 estimated to 63 so 63
The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.
Calculation of the angle:Since we know that1 rev = 2(pi) rad
So here pi(rad) = 180 degrees
Now for 33 rpm it should be like
= 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s
= 63.36 degrees
= 63 degrees
hence, The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.
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A tsunami, an ocean wave generated by an earthquake, propagates along the open ocean at 700 km/hr and has a wavelength of 750 km. What is the frequency of the waves in such a tsunami?A. 6.8 HzB. 0.93 HzC. 0.00026 HzD. 1.1 HzE. 0.15 Hz
Answer:
C) 0.00026 Hz
Explanation:
In any wave, there exists a fixed relationship between the speed v, the frequency f and the wavelength λ, as follows:[tex]v = \lambda * f (1)[/tex]
Replacing by the givens in (1), and solving for f, we get:[tex]f = \frac{v}{\lambda} = \frac{700km/hr}{750 km} = 0.93 1/hr (2)[/tex]
Converting the units to Hz (1/sec), we get:[tex]f = 0.93 \frac{1}{hr} *\frac{1 hr}{3600sec} = 2.6e-4 = 0.00026 Hz (2)[/tex]
The answer C. is the right one.Help me please,
A ball is thrown straight up in the air. What is the velocity and acceleration at the top of the path?
A) v 0m/s, = 0m/s/s
B) v = 0m/s, a 10m/s/s
C) v = 10m/s, a 10m/s/s
D) v = 10m/s, a = 0m/s/s
E) None of the above
Option B
Explanation:
no distance was given only the acceleration due to the fact that it went up (10m/s/s)
s0 it is
0 m/s and 10m/s/s (option B)
PLEASE HELP ME ASAP!! GUYSSS!! I AM IN CLASS AND DYING! LITERALLY
Billy and Ashley live in the same time zone. Billy lives in Brazil (blue smiley face on the image below). Ashley lives in Eastern Canada (yellow smiley face on the image below).
One day, Billy and Ashley are both outside at 1:32 pm. They are talking on the phone to each other. As they talk, Billy notices the sky in Brazil getting progressively darker. Eventually, it feels like it is nighttime, because it is so dark. Billy thinks the world is coming to an end. He asks Ashley if she is experiencing the same thing in Canada. Ashley has no idea what he is talking about. “It’s perfectly bright and sunny where I am,” she says.
Ashley and Billy conclude that the world is not coming to an end. They reach out to some 7th graders to figure out what is happening. The 7th graders tell Billy that he is experiencing a solar eclipse. To help Billy and Ashley understand, create a model to show why Billy is experiencing an eclipse in Brazil, but Ashley is not experiencing the eclipse in Canada.
Your model must include:
The sun, the earth, the moon, and solar energy (clearly labeled).
How accurate your scale is.
How solar energy interacts with both the moon and with Earth.
The tilt of the moon’s orbit relative to the Earth’s orbit
Why Billy is experiencing the solar eclipse and why Ashley is not.
Answer:
sounds like a you problem
Explanation:
yeah
1+1=?
FIRST ONE TO ANSWER GETS BRAINLIEST!
Answer:
Thx for points merry Christmas answer 2
Explanation:
Answer:2
Explanation:
An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with the surface. The astronaut falls back down to the surface after a short time interval. Which of the following forces CANNOT be neglected when analyzing the motion of the astronaut?
Asteroids are known through the help of artificial gravity, to have small gravity. The forces that cannot be neglected when analyzing the motion of the astronaut is that the gravitational force between the astronaut and the asteroid.
The gravitational force between two objects is said to be inversely proportional to the distance between them when squared. Therefore, when an individual halve the distance then the force increases by four times.
Unbalanced forces are simply known to be brought about due to a change in motion, speed, and/or direction. If two forces act in the same direction on an object, the net force is said to be equal to the sum of the two forces.
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*
If a rock falls for 3 seconds off of a bridge, how far will the rock fall?
-30 m
-45m
-60m
-75m
A police officer standing at the side of the road uses a radar emitting frequency of 24.15 GHz. A car is going away from the officer at a speed of 50mph. What will be the difference in frequency of the beam reflected by the car received back by the radar?
a. 4.0 kHz (lower frequency)
b. 1.8 kHz (higher frequency)
c. 4.0 kHz (higher frequency)
d. 1.8 kHz (lower frequency)
Answer:
The correct answer is C
Explanation:
From the question we are told that
The frequency of the radar is [tex]f = 24.15 \ GHz = 24.15 *10^{9} \ Hz[/tex]
The speed of the car is [tex]v = 50 mph = \frac{50}{2.237} = 22.35 \ m/s[/tex]
Generally the difference frequency reflected by the car and the frequency which the radar receives back is mathematically represented as
[tex]\Delta f = \frac{f * 2 * v }{ c }[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
=> [tex]\Delta f= \frac{2 * 24.15 *10^{9} * 22.35}{ 3.0*10^{8}}[/tex]
=> [tex]\Delta f = 4000 \ Hz[/tex]
=> [tex]\Delta f = 4.0 \ kHz[/tex]
Given that the value is positive then it a higher frequency