Answer:
There are three main sources of heat in the deep earth:
(1) heat from when the planet formed and accreted, which has not yet been lost;
(2) frictional heating, caused by denser core material sinking to the center of the planet; and (3) heat from the decay of radioactive elements.
Explanation:
hope it was helpful...
A 200N lamp is suspended from three cables as shown in the figure below. Find the tensions in each of the three cables.
Answer:
66.6N
Explanation:
Step one;
given data
the mass of the lamp = 200N
we are told that it is suspended by 3 cables.
Now we know that the weight will be distributed equally on the cables
Step two:
so, let the tension in each cable be T
T+T+T= 200
3T=200
T=200/3
T=66.6N
The tenion on each cable is 66.6N
what does the circulatory system consist CLASS 7
Answer:
the circulatory system consists of heart, blood, blood vessels.
hope this answer will help you
A hockey player strikes a puck that is initially at rest. The force exerted by the stick on the puck is 975 N, and the stick is in contact with the puck for 0.0049 s.
(a) What is the impulse imparted by the stick to the puck.
___________ kg m/s
(b) What is the speed of the puck (m= 1.67 kg)just after it leaves the hockey stick?
____________ m/s
Explanation:
Given that,
The force exerted by the stick on the puck is 975 N
The stick is in contact with the puck for 0.0049 s
Initial speed of the puck, u = 0 (at rest)
(a) We need to find the impulse imparted by the stick to the puck.
Impulse = Force × time
J = 4.7775 kg-m/s
(b) Mass of the puck, m = 1.76 kg
We need to find the speed of the puck just after it leaves the hockey stick.
Let the speed be v.
As impulse is equal to the change in momentum.
[tex]J=m(v-u)\\\\4.7775=1.67(v-0)\\\\v=\dfrac{4.7775}{1.67}\\\\v=2.86\ m/s[/tex]
So, when the puck leaves the hockey stick its speed is 2.86 m/s.
You hold your physics textbook in your hand. (Assume that no other objects are in contact with the book.)
1. Identify the forces acting on the book.
a. book on hand.
b. hand on book.
c. floor on book.
d. earth on book.
2. For each force you identified in part 1), indicate the direction.
a. book on hand
b. hand on book
c. floor on book
d. earth on book
3. Identify the forces acting on your hand.
a. book on hand
b. hand on book
c. floor on hand
d. earth on hand
4. For each force you identified in part 3), indicate the direction.
a. book on hand
b. hand on book
c. floor on hand
d. earth on hand
5. Identify the forces that form the action-reaction pair as defined by Newton's third law.
a. earth on hand
b. hand on book
Answer:
1) the correct answer is b and d
2) For force b its direction is vertical up
for the force d its direction is vertical down
3) the correct answers are: a, c and d
4) Force a is vertical down , force c is vertical up and force d is vertical down
5) the correct answer which is b
Explanation:
In this exercise it is asked to identify the forces, fundamentally on the free there are the forces of gravity and the support force of the hand, with these facts we answer the questions
1) the correct answer is b and d
the hand acts on the book with a contact force and the Earth acts on the book with the force of gravity.
2) For force b its direction is vertical up
for the force d its direction is vertical down
3) The forces on the hand are the weight of the book. The force of gravity due to the mass of the hand. As the hand is in balance, there must be a force applied by the arm to keep the hand in position; assuming the hand is in the air, if the hand is resting on the floor the force of the floor on the hand can perform this function
therefore the correct answers are: a, c and d
4) Force a is vertical down
force c is vertical up
force d is vertical down
5) The action and reaction forces are forces of equal magnitude, each applied to one of the bodies, we have
* the force of the hand on the free and its reaction the force of the book on the hand
* The force of the Earth on the book and the hand, giving the weight of each one and the relationship is the force of the book and the hand on the Earth
the correct answer which is b
A 10-KG mass is lifted upward, by a light cable . what is the tension in the cable if the acceleration is (A) zero, (B) 6m/s2 upward , and (C) 6m/s2 downward
Answer:
(a) 98 N
(b) 158 N
(c) 38 N
Explanation:
Part (a)When the acceleration is 0 m/s², the net force on the mass is 0 N. Therefore, the tension force is equal to the weight force due to Newton's Second Law:
∑F_y = T - w = ma_y ∑F_y = T - w = m(0 m/s²)∑F_y = T - w = 0 ∑F_y = T = wSince the tension in the cable and the weight of the mass are equal to each other, we can solve for the weight force of the mass by using:
w = mg w = (10 kg)(9.8 m/s²)w = 98 NSince T = w, we can say that T = 98 N.
Part (b)Let's set the upwards direction to be positive and the downwards direction to be negative. We can use Newton's Second Law to solve for the tension in the cable if the acceleration is 6 m/s² upward:
∑F_y = T - w = ma_y∑F_y = T - mg = m(6 m/s²)∑F_y = T - mg = 6mPlug the known values into the equation and solve for T.
T - mg = 6mT - (10 kg)(9.8 m/s²) = 6(10 kg) T - 98 = 60 T = 158 NThe tension in the cable if the acceleration is +6 m/s² is 158 N.
Part (c)The process is the same, but this time acceleration is -6 m/s².
∑F_y = T - w = ma_y∑F_y = T - mg = m(-6 m/s²)∑F_y = T - mg = -6mPlug known values into the equation and solve for T.
T - mg = -6mT - (10 kg)(9.8 m/s²) = -6(10 kg) T - 98 = -60 T = 38 NThe tension in the cable if the acceleration is -6 m/s² is 38 N.
What magnitude point charge creates a 9,000 N/C electric field at a distance of 1.00 m? a. 1.00 C b. 1.00 mC c. 1.00 ?C d. 1.00 nC
Answer:
1 µC
Explanation:
From the question given above, the following data were obtained:
Electric field intensity (E) = 9000 N/C
Distance (r) = 1 m
Charge (Q) =?
NOTE: Electric force constant (K) = 9×10⁹ Nm²C¯²
The magnitude of the point charge can be obtained as follow:
E = KQ/r²
9000 = 9×10⁹ × Q / 1²
9000 = 9×10⁹ × Q
Divide both side by 9×10⁹
Q = 9000 / 9×10⁹
Q = 1×10¯⁶ C
Recall:
1 micro charge (µC) = 1×10¯⁶ C
Hence, the magnitude of the point charge is 1 µC
Answer:
1ηCExplanation:
The magnitude of the point charge is expressed as;
V = kQ/r
k is the coulombs constant = 9.0×10^9 N⋅m²/C²
V is the electric field = 9000N/C
Q is the charge
r is the distance = 1.00m
Get the required charge
Substitute the given parameters into the formula;
9000 = 9.0×10^9Q/1.0
9000 = 9.0×10^9Q
Q = 9000/9.0×10^9
Q = 9.0×10^3/9.0×10^9
Q = 9.0/9.0×10^(3-9)
Q = 1.0×10^6C
Q = 1ηC
hence the magnitude of the point charge is 1ηC
Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects that orbit the Sun –Not large enough to have enough gravitational pull to clear the path of their orbit –4 known in the Kuiper Belt (Eris, Haumea, Makemake, and Pluto) –1 in the asteroid belt (Ceres) Jasmine later copied the title of the notes from a friend. What title did Jasmine most likely copy? Comets The Kuiper Belt Asteroids Dwarf Planets
Answer:
D. Dwarf Planets
Explanation:
Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects that orbit the Sun –Not large enough to have enough gravitational pull to clear the path of their orbit, Jasmine is most likely to copy the Dwarf Planets, therefore the correct answer is option D.
What is a solar system?It is a system that collection of all the planets and spatial bodies revolving around the sun because of the gravitational pull of the sun.
Our Solar System is based on a heliocentric model in which the Sun is assumed to reside at the central point of the planetary system.
As given in the problem Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects that orbit the Sun –Not large enough to have enough gravitational pull to clear the path of their orbit –4 known in the Kuiper Belt (Eris, Haumea, Makemake, and Pluto) –1 in the asteroid belt (Ceres) Jasmine later copied the title of the notes from a friend.
Thus, Jasmine is most likely to copy the Dwarf Planets, therefore the correct answer is option D.
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Jupiter is the largest planet in our solar system, having a mass and radius that are, respectively, 318 and 11.2 times that of earth. Suppose that an object falls from rest near the surface of each planet, and that each object falls the same distance before striking the ground. Determine the ratio of the time of fall on Jupiter to that on earth.
Answer:
0.62
Explanation:
we know that [tex]g=\frac{G M}{R^{2}}[/tex]
[tex]\frac{g_{g}}{g_{J}}=\frac{M_{g}}{M_{J}} \times \frac{R_{j}^{2}}{R_{z}^{2}}=\frac{M}{318 M} \times \frac{(11.2)^{2} R_{g}^{2}}{R_{g}^{2}}[/tex]
[tex]\frac{g_{g}}{g_{j}}=\frac{125.44}{318}=0.394[/tex]
We know that [tex]=\sqrt{\frac{2 h}{g}}[/tex]
here given that each object falls the same distance
[tex]\therefore t \alpha \sqrt{\frac{1}{g}}[/tex]
[tex]\therefore \frac{t_{J}}{t_{B}}=\sqrt{\frac{g_{g}}{g_{J}}}=\sqrt{\frac{g}{0.394}}[/tex]
[tex]\therefore \frac{t_{j}}{t_{g}}=\sqrt{0.394}=0.62[/tex]
15. A vessel containing a liquid 'L' is balanced by a solid 'S' as shown below. Now, two identical pieces of cork (which float on the liquid) are placed gently, one on the solid and one in the liquid. What will happen to the balance?
Answer:
C) There will be no change in the balance
Explanation:
Before and after the cork is placed, the masses in each pan are the same. There will be no change in the balance.
How many words can be made with five letters
Answer:
agree eager eagre ragee
Explanation:
hope this help
Two spheres of equal mass, A and B, are projected off the edge of a 1.0 m bench. Sphere A has a horizontal velocity of 10 m/s and sphere B has a horizontal velocity of 5 m/s.
__ 6. If both spheres leave the edge of the table at the same instant, sphere A hits the floor at the spot marked X. Sphere B will hit the floor
a. at some point between the edge of the table and X.
b. at some point past X.
c. at the same distance from the table as X.
d. there is not enough information to decide.
Sphere A travels a horizontal distance of (10 m/s) t after time t, while sphere B travels a distance of (5 m/s) t. So sphere B lands on the floor some point between the table's edge and the point X (A).
Consider two cars that are on course for a head-on collision. If they have masses m1 = 1,300 kg and m2 = 1,800 kg and are both traveling at 34 m/s, what is the magnitude of the total momentum?
Answer:
105400kgm/s
Explanation:
Momentum is the product of the mass and velocity of an object.
Momentum = mass × velocity
Given
Masses m1 = 1,300 kg and m2 = 1,800
velocity = 34m/s
Magnitude of the total momentum = (m1+m2)v
Magnitude of the total momentum = (1300+1800)((34)
Magnitude of the total momentum = 3100(34)
Magnitude of the total momentum = 105400kgm/s
A current is induced in a wire by moving the wire through a magnetic field. Which is one factor that affects the direction of the current? the direction in which the wire moves the thickness of the wire the direction of the needle on an ammeter the type of magnet Mark this and return
Answer:
I belive it is the thickness of the wire
Explanation:
There are no other answers and I searched it up and it seems that it is the thickness of the wire.
A current is induced in a wire by moving the wire through a magnetic field. A factor that affects the direction of the current is the direction in which the wire moves.
What is magnetic field?Magnetic field is the region of space where an object experiences the magnetic force towards a magnetic material.
The direction of flow of electric current in a current carrying conductor wire depends on many factors. The direction of the wire moves in or out and current will flow towards the direction of a given wire.
Hence, a factor that affects the direction of the current is in the direction in which the wire moves.
Learn more about Magnetic field.
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Which statements explain the special theory of relativity? Check all that apply.
- Time and space are relative.
- The speed of light in a vacuum is constant for all observers.
- Physical laws change based on an observer's motion.
- Physical laws remain constant regardless of an observer's motion.
- The special theory of relativity applies to objects with constant velocity.
- The special theory of relativity applies to accelerating objects.
Answer:
A. Time and space are relative.
B. The speed of light in a vacuum is constant for all observers.
D. Physical laws change based on an observer's motion.
E. Physical laws remain constant regardless of an observer's motion.
Explanation:
The statements that explain the special theory of relativity are a),b) and e) respectively.
The special theory of relativity, developed by Albert Einstein, is based on the idea that the laws of physics are the same for all observers who are moving uniformly relative to each other. This means that there is no "absolute" or "preferred" frame of reference in the universe. Instead, all frames of reference are equally valid, and any observer can use their own frame of reference to describe physical phenomena.
One consequence of this idea is that time and space are relative. This means that the measurements of time and distance depend on the observer's frame of reference. For example, if two observers are moving relative to each other, they will measure different lengths and times for the same event. This effect is known as time dilation and length contraction.
Finally, the special theory of relativity applies only to objects with constant velocity. It does not apply to accelerating objects, which require the more general theory of relativity. The special theory of relativity is a fundamental theory in modern physics, and it has important implications for our understanding of space, time, and the nature of reality.
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The correct question is :
Which statements explain the special theory of relativity? Check all that apply.
a) Time and space are relative.
b) The speed of light in a vacuum is constant for all observers.
c) Physical laws change based on an observer's motion.
d) Physical laws remain constant regardless of an observer's motion.
e) The special theory of relativity applies to objects with constant velocity.
f) The special theory of relativity applies to accelerating objects.
What is the strength of the electric field in a region where the electric potential is constant?
Answer:
Where the electric potential is constant, the strength of the electric field is zero.
Explanation:
As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e
Eₓ = - dV / dx ----------(i)
From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.
Therefore, a constant electric potential means that electric field is zero.
30 points? I have no clue
Answer:
The second graph, B
Explanation
At a distance of 10 km from a radio transmitter, the amplitude of the E-field is 0.20 volts/meter. What is the total power emitted by the radio transmitter
Answer:
The total power is [tex]P = 6.665 *10^{4} \ W[/tex]
Explanation:
From the question we are told that
The distance is [tex]r = 10 \ km = 1000 \ m[/tex]
The amplitude of the electric field is [tex]E = 0.20 \ volt/meter[/tex]
Generally the average intensity of the electromagnetic field from the radio transmitter is mathematically represented as
[tex]I = \frac{E^2}{ 2 \mu_o * c }[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
So
[tex]I = \frac{0.2^2}{ 2 * 4\pi *10^{-7} * 3.0*10^{8} }[/tex]
=> [tex]I = 5.307 *10^{-5} \ W/m^2[/tex]
Generally this intensity can also be mathematically represented as
[tex]I = \frac{P }{ 4 \pi r^2 }[/tex]
=> [tex]P = I ( 4 \pi r^2 )[/tex]
=> [tex]P = 5.307 *10^{-5} ( 4 * 3.142 * 1000^2 )[/tex]
=> [tex]P = 6.665 *10^{4} \ W[/tex]
The total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].
The given parameters;
amplitude of the electric field, E = 0.2 V/mdistance of the transmitter, = 10 km = 10,000 mThe intensity of the radio wave from the transmitter is calculated as follows;
[tex]I = \frac{E^2}{2\mu_0 c} \\\\I = \frac{0.2^2 }{2\times 4\pi \times 10^{-7} \times 3\times 10^8} \\\\I = 5.305 \times 10^{-5} \ W/m^2[/tex]
The total power emitted by the radio transmitter is calculated as follows;
[tex]I = \frac{P}{A} \\\\P = IA\\\\P = I \times 4\pi r^2\\\\P = (5.305\times 10^{-5} )\times 4\pi \times (10,000)^2\\\\P = 6.67\times 10^{4} \ W[/tex]
Thus, the total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].
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A plane is flying 700 km/hr to the east into a head wind that is moving at 20 km/hr west.
Calculate the planes velocity.
O 720 km/hr
680 km/hr
0-680 km/hr
0 -720 km/hr
Answer:
getc+d
Explanation:
gger
A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg.
a. If the crate doesn't move, what's the magnitude of the static friction force (in N)?
b. What is the minimum possible value of the coefficient of static friction between the crate and the floor?
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate doesn't move (static), acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = Fm - Ff = 0.
Fm is the applied force
Ff is the frictional force
Since Fm - Ff = 0
Fm = Ff
This means that the applied force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
R = 31.2 × 9.8
R = 305.76N
From the formula
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
Vector A with arrow lies in the xy plane. Both of its components will be negative if it points from the origin into which quadrant? A. the first quadrant B. the second quadrant C. the third quadrant D. the fourth quadrant E. the second or fourth quadrants
Answer:
C
Explanation:
From the question we are told that a vector on the x and y plane face their negative axis
Generally in the x and y plane thr negative y axis is made to face down opposite the positive y axis
Whilst the negative x axis faces the left which is also alternate to the positive x axis
Generally A vector pointing towards the x and y negative axis fro the origin (0) will definitely be in the third quadrant
Is Chocolate Milk a compound,mixture or a element
Answer:
mixture
Explanation:
because of many Ingredients it is called a mixture
Which option is an element? A. water B.Sodum cloride C.oxygen.D air
Answer:
C. Oxygen
Explanation:
Oxygen is the 8th element.
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 1.60 kg and 16.0 g whose centers are separated by about 3.30 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
Answer:
The value is [tex]F = 1.568 *10^{-9} \ N[/tex]
Explanation:
From the question we are told that
The mass of the first lead sphere is [tex]m = 1.60 \ kg[/tex]
The mass of the second lead sphere is [tex]M = 16 \ g = 0.016 \ kg[/tex]
The separation between masses is [tex]r = 3.30 \ cm = 0.033 \ m[/tex]
Generally the gravitational force between each sphere is mathematically represented as
[tex]F = \frac{G * m * M }{r^2 }[/tex]
Here G is the gravitational constant with value [tex]G = 6.67 *10^{-11 } \ m^3 \cdot kg^{-1} \cdot s^{-2}[/tex]
[tex]F = \frac{6.67 *10^{-11 } * 1.60 * 0.016 }{0.033^2 }[/tex]
=> [tex]F = 1.568 *10^{-9} \ N[/tex]
This table shows statistics about the US population in 2010. Which demographic trend does this table best support? People are marrying later than they used to. There are more single-parent families. The American population is more diverse. There are fewer younger adults than older adults.
Answer:
D. There are fewer younger adults than older adults.
Explanation:
Failed the assignment but got this answer correct.
Answer:
D is correct
Explanation:
A 15.0kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 degrees above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3.
a. Draw a free body diagram. Draw your coordinate system and label the axes.
b. Calculate the work done on the block by the 70 N force.
c. Calculate the work done on the block by the normal force.
d. Calculate the work done on the block by the gravitational force.
e. Calculate the work done on the block by the force of friction.
Answer:
[tex]W=70 * 5cos20 = 328.89 J[/tex]
[tex]W_n = 0[/tex]
[tex]W_g=0[/tex]
[tex]W_f= -184.59J[/tex]
Work done is 0
Explanation:
From the question we are told that
Weight of block =15.0kg
Force acting on the block = 70.0N
At an angle of 20 degree
Displacement of block is 5m
Coefficient of kinetic friction 0.3
b) Generally work done by force is give by [tex]W=fdcos \theta[/tex]
therefore
[tex]W=70 * 5cos20 = 328.89 J[/tex]
c) there is no work done by the normal force in this scenario because
normal force in this case is perpendicular to the displacement of the motion
[tex]W_n = 0[/tex]
d) The displacement in the vertical direction is 0
Therefore the gravitational work done is 0 [tex]W_g=0[/tex]
e)Generally in finding work done by friction we first find frictional force
Mathematically the equation for frictional force is given [tex]f = \alpha N[/tex]
Given that
[tex]N=mg-Fsin20[/tex]
[tex]N= 15.0*9.8 - 70 sin20[/tex]
[tex]N=123 N[/tex]
[tex]f=0.3* 123.06 = 36.92N[/tex]
Mathematically solving to get work done by frictional force [tex]W_f[/tex]
[tex]W_f= -fd\\W_f = -36.92 * 5[/tex]
[tex]W_f= -184.59J[/tex]
the frictional force work done is [tex]W_f= -184.59J[/tex]
A skier is traveling at a speed of 34.7 m/s when she reaches the base of a frictionless ski hill. This hill makes an angle of 10^o with the horizontal. She then coasts up the hill as far as possible. What height (measured vertically above the base of the hill) does she reach?
Answer:
h = 61.5m
Explanation:
Initial speed = 34.7m/s
Net force =
F= -mg Sin 10
We are finding distance moved by this skier upon the hill.
ma = - mgSin10
a = -gSin10
g = 9.8
Sin 10 = 0.1564
a = -9.8(0.1564)
a = -1.53
This is the acceleration
(-34.7)² = 2as
= -34.7² = (-2x1.53)s
-1204.09 = -3.06s
We find the value of s
S = -1204.09/-3.06
S = 393.5m
Sin10 = h/393.5m
0.1564 = h/393.5m
h = 0.1564x393.5m
h = 61.5
A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at this instant?
KEi = _________________J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
KEf/KEi=______________
Answer:
(a) the runner's kinetic energy at the given instant is 308 J
(b) the kinetic energy increased by a factor of 4.
Explanation:
Given;
mass of the runner, m = 64.1 kg
speed of the runner, u = 3.10 m/s
(a) the kinetic energy of the runner at this instant is calculated as;
[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J[/tex]
(b) when the runner doubles his speed, his final kinetic energy is calculated as;
[tex]K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J[/tex]
the change in the kinetic energy is calculated as;
[tex]\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4[/tex]
Thus, the kinetic energy increased by a factor of 4.
What is the acceleration down the ramp of an object on a 17° ramp?
A
2.9 m/s2
B
4.2 m/s2
C
5.5 m/s2
d
9.4 m/s2
a body of mass 1kg is made to oscillate on a spring of force constant 16 n/m calculate 1 the angular frequency 2 the frequency of oscillation
Answer :
[tex]\omega=r\ rad/s[/tex] and [tex]f=\dfrac{2}{\pi}\ Hz[/tex]
Explanation:
Given that,
Mass of a body, m = 1 kg
Force constant, k = 16 N/m
We need to find the angular frequency and the frequency of oscillation.
(a) The angular frequency of a body is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s[/tex]
(b) The frequency of oscillation is given by :
[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz[/tex]
Hence, this is the required solution.
PLEASE HELP ME!!!!!!!!
Answer:
I believe C
Explanation:
i never learned this sorry