Answer:
gravitational force
Explanation:
Gravitational force -an attractive force that exists between all objects with mass; an object with mass attracts another object with mass; the magnitude of the force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between the two objects
How much would a spring scale with k = 120 N/m stretch, if it had a 3.75 J of work done
on it?
Answer:
0.25m
Explanation:
Given parameters:
Spring constant , K = 120N/m
Work done = 3.75J
Unknown:
magnitude of extension = ?
Solution:
To solve this problem;
Work done = [tex]\frac{1}{2}[/tex]kx²
K is the spring constant
x is the extension
3.75 = [tex]\frac{1}{2}[/tex] x 120x²
3.75 = 60x²
x² = 0.06
x = √0.06 = 0.25m
A bullet is fired horizontally at a height of 2 meters at a velocity of 930 m/s. Assume no air resistance. How long until the bullet reaches the ground?
0.32 s
0.57 s
0.64 s
0.25 s
A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to rotate by pulling on the rope with a constant force of 195 N. If the disk starts from rest, what is its angular speed in rev/s at the end of 2.05 s?
Answer:
The angular speed is 0.13 rev/s
Explanation:
From the formula
[tex]\tau = I\alpha[/tex]
Where [tex]\tau[/tex] is the torque
[tex]I[/tex] is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
But, the angular acceleration is given by
[tex]\alpha = \frac{\omega}{t}[/tex]
Where [tex]\omega[/tex] is the angular speed
and [tex]t[/tex] is time
Then, we can write that
[tex]\tau = \frac{I\omega}{t}[/tex]
Hence,
[tex]\omega = \frac{\tau t}{I}[/tex]
Now, to determine the angular speed, we first determine the Torque [tex]\tau[/tex] and the moment of inertia [tex]I[/tex].
Here, The torque is given by,
[tex]\tau = rF[/tex]
Where r is the radius
and F is the force
From the question
r = 3.00 m
F = 195 N
∴ [tex]\tau = 3.00 \times 195[/tex]
[tex]\tau = 585[/tex] Nm
For the moment of inertia,
The moment of inertia of the solid disk is given by
[tex]I = \frac{1}{2}MR^{2}[/tex]
Where M is the mass and
R is the radius
∴[tex]I = \frac{1}{2} \times 325 \times (3.00)^{2}[/tex]
[tex]I = 1462.5[/tex] kgm²
From the question, time t = 2.05 s.
Putting the values into the equation,
[tex]\omega = \frac{\tau t}{I}[/tex]
[tex]\omega = \frac{585 \times 2.05}{1462.5}[/tex]
[tex]\omega = 0.82[/tex] rad/s
Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π
0.82 rad/s = 0.82/2π rev/s
= 0.13 rev/s
Hence, the angular speed is 0.13 rev/s,
Describe why you are doing the experimen?
Answer:
An experiment is a procedure carried out to support, refute, or validate a hypothesis. Experiments provide insight into cause-and-effect by demonstrating what outcome occurs when a particular factor is manipulated.
Explanation:
A guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beat/s when sounded with a 357 Hz tuning fork. What is the vibrational frequency (in Hz) of the string
Answer:
349 Hz
Explanation:
We are told that the guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beat/s when sounded with a 357 Hz tuning fork.
This means that for the 352 Hz tuning fork, the vibrational frequency is;
f = 352 ± 3
f = (352 + 3) or (352 - 3)
f = 355 Hz or 349Hz
For the 357 Hz tuning fork, the vibrational frequency is;
f = 357 ± 8
f = (357 + 8) or (357 - 8)
f = 365 Hz or 349 Hz
In both cases, 349 Hz is common;
Thus, the vibrational frequency of the string = 349 Hz
Suppose a certain object has a mass of 5.00 kilograms on the earth. On the
Moon, where g is 1.6 m/s/s what would its mass be?*
Answer:
it would be 49.03325 Newton.
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor?
Complete question:
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.
Answer:
The electric field inside this metal resistor is 3125 V/m
Explanation:
Given;
length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m
diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m
the potential difference due to electric field between the two ends of the resistor, V = 10 V
The electric field inside this metal resistor is given by;
ΔV = EL
where;
ΔV is change in electric potential
E = ΔV / L
E = 10 / (3.2 x 10⁻³ )
E = 3125 V/m
Therefore, the electric field inside this metal resistor is 3125 V/m
Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in rad/s) of simple harmonic oscillations of this system?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula to be used here is
ω = 2π/T
Where ω is the angular frequency (in rad/s)
T is the period - the time taken for Block A to complete one oscillation and return to it's original position.
To solve for this period T, the formula below should be used
T = 2π√m/k
where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)
Any conclusion reached by analogy is worth accepting
True
False
During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in the cannon,determine the average net force exerted on him in thebarrel of the cannon.
Answer:
2872.8 N
Explanation:
We have the following information
m =n72kg
Δy = 18m
t = 0.95s.
From here we use the equation
Δy=1/2at2 in order to solve for the acceleration.
So a
=( 2x 18m)/(0.95s²)
= 36/0.9025
= 39.9m/s2.
From there we use the equation
F = ma
F=(72kg) x (39.9)
= 2872.8N.
2872.8N is the average net force exerted on him in the barrel of the cannon.
Thank you!
branches of sicence
Answer: Natural science can be divided into two main branches
Explanation:
life science and physical science. Life science is alternatively known as biology, and physical science is subdivided into branches: physics, chemistry, astronomy and Earth science.
A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?
Answer:
The number is [tex]N = 300[/tex]
Explanation:
From the question we are told that
The net charge is [tex]Q = -4.8 *10^{-17 } \ C[/tex]
Generally the charge on a electron is [tex]e = - 1.60 *10^{-19 } \ C[/tex]
Generally the number of excess electrons is mathematically represented as
[tex]N = \frac{Q}{e}[/tex]
=> [tex]N = \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}[/tex]
=> [tex]N = 300[/tex]
Zinc has a work function of 4.3 eV. a. What is the longest wavelength of light that will release an electron from a zinc surface? b. A 4.7 eV photon strikes the surface and an electron is emitted. What is the maximum possible speed of the electron?
Answer:
a
[tex]\lambda_{long} = 288.5 \ nm[/tex]
b
The velocity is [tex]v = 3.7 *0^{5} \ m/s[/tex]
Explanation:
From the question we are told that
The work function of Zinc is [tex]W = 4.3 eV[/tex]
Generally the work function can be mathematically represented as
[tex]E_o = \frac{hc}{\lambda_{long}}[/tex]
=> [tex]\lambda_{long} = \frac{hc}{E_o}[/tex]
Here h is the Planck constant with the value [tex]h = 4.1357 * 10^{-15} eV s[/tex]
and c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda_{long} = \frac{4.1357 * 10^{-15} * 3.0 *10^{8}}{4.3}[/tex]
=> [tex]\lambda_{long} = 2.885 *10^{-7} \ m[/tex]
=> [tex]\lambda_{long} = 288.5 \ nm[/tex]
Generally the kinetic energy of the emitted electron is mathematically represented as
[tex]K = E -E_o[/tex]
Here E is the energy of the photon that strikes the surface
So
[tex]E- E_o = \frac{1}{2} m * v^2[/tex]
Here m is the mass of electron with value [tex]m = 9.11*10^{-31 } \ kg[/tex]
Generally [tex]1 ev = 1.60 *10^{-19} \ J[/tex]
=> [tex]v = \sqrt{ \frac{2 (E - E_o ) }{ m } }[/tex]
=> [tex]v = \sqrt{ \frac{2 (4.7 - 4.3 )* 1.60 *10^{-19} }{ 9.11 *10^{-31} } }[/tex]
=> [tex]v = 3.7 *0^{5} \ m/s[/tex]
Two particles are separated by 0.38 m and have charges of -6.25 x 10-9C
and 2.91 x 10-9 C. Use Coulomb's law to predict the force between the
particles if the distance is doubled. The equation for Coulomb's law is
Fe = kq92, and the constant, k, equals 9.00 x 10°N-m/c2.
A. -2.83 x 10-7N
B. 2.83 x 10-7N
C. -1.13 x 10-6N
D. 1.13 x 10-6N
Answer:A
Explanation:
Answer:
A. -2.83 x 10-7N
Explanation:
Answered: A 4 kg mass is attached to a horizontal spring with the spring constant of 600 N/m and rests on a frictionless surface on the ground. The spring is compressed 0.5 m past its equilibrium. What is the initial energy of the system.
Answer: 75 joules
In the video your blood is compared to a __________________ that delivers oxygen to your body and picks up CO2 to be released out when you breath.
Answer:
delivery truck
Explanation:
because i got it right
how does tom and jerry movie character influence your attitude
Answer:
it makes me wish I was a cartoon
Answer:
goofy and stupid and act like a kid
Explanation:
7. What does the changing colour perceived by the person as the filter changes indicate to you
about white light?
Answer:
lo
Explanation:
What becomes V if we use 2 resistors of 4W in parallel?
A. 2.66 V
B. 6 V
C. 12 V
D. 24 V
Answer:
This question is incomplete.
Explanation:
This question is incomplete. However, it should be noted that the voltage, V, across resistors in parallel is the same (although there currents are not the same). Thus, if a voltage has been provided, it remains the same but if not provided, you can solve for it using the formulas below
V = IR
where V is the voltage. I is the current and R is the resistance
R in parallel can be calculated as R = 1/R₁ + 1/R₂ + 1/R₃ + ......
38. You are fishing and catch a fish with a mass of
6kg. If the fishing line can withstand a maximum
tension of 30 N, what is the maximum acceleration
you can give the fish as you reel it in?..*
(10 Points)
Enter your answer
Answer:
1.7333333m/s²
Explanation:
Tension of the line = the weight + force from pulling up the fish
30N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6a
∴ a = 1.7333333m/s²
You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is 1.7333333 m/s².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.
Tension of the line = the weight + force from pulling up the fish
30 N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6 a
a = 1.7333333 m/s²
You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is 1.7333333 m/s².
To learn more about acceleration refer to the link:
brainly.com/question/12550364
#SPJ2
Please help!!! I will give brainliest,
Answer:
C. a liter of salt water.
Explanation:
Defination of Solution =>
a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).
what can i yeet baby or toddler
Answer:
both
Explanation:
baby for fun, toddler for vengence
What is the result of increasing the speed at which a magnet moves in and
out of a wire coil?
A. The current in the wire increases.
B. The magnetic field around the magnet decreases.
C. The current in the wire decreases.
D. The magnetic field around the magnet increases.
Answer:
A. The current in the wire increases.
Explanation:
Increasing the speed at which a magnet moves in and out of a wire coil increases the current in the wire.
This phenomenon shows the inter-relationship between electricity and magnetic fields.
Magnetic fields are induced by passage of electric current. Also, electric current can be produce by magnetic fields. When the speed at which a magnet moves in and out of a wire coil increases, the current also increases.A box of mass 7.0 kg is accelerated from rest across a floor at a rate of 2.0 m/s2 for 9.0 s .Find the net work done on the box. Express your answer to two significant figures and include the appropriate units.
Answer:
Explanation:
Step one:
given data
mass = 7kg
acceleration =2m/s^2
time= 9seconds
acceleration = velocity/time
velocity= acceleration *time
velocity=2*9
velocity= 18m/s
distance moved= velocity* time
distance= 18*9
distance=162m
we also know that the force on impulse is given as
Ft=mv
F=mv/t
F=7*18/9
F=126/9
F=14N
work done = Force* distance
work done=14*162
work=2268Joules
work= 2.27kJ
A train is traveling at 55m/s begins to slow down as it approaches a bend in the tracks. If it travels around the bend at a speed of9 m/s and it takes 49 s to properly slow down, what distance does the train travel while slowing down?
Answer:
x = 1127 [m]
Explanation:
In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.
[tex]v_{f} =v_{o} -a*t[/tex]
where:
Vf = final velocity = 9 [m/s]
Vo = initial velocity = 55 [m/s]
a = acceleration o desacceleration [m/s²]
t = time = 49 [s]
Now replacing:
9 = 55 - a*49
a*49 = 55 + 9
a = 1.306 [m/s²]
Note: The negative sign in the above equation means that the speed decreases.
Now using the second equation.
[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]
(9)² = (55)² - 2*(1.306)*x
2944 = 2.612*x
x = 1127 [m]
There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center) and repel each other with an electrostatic force of F1=0.0720 N . A thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed, the spheres still repel, but with a force of F2=0.115 N . The Coulomb force constant is k=1/(4π????0)=8.99×109 N⋅m2/C2 . Using this information, find the initial charge on each sphere, q1 and q2 , if q1 is initially less than q2 .
Answer:
[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex]
Explanation:
[tex]F_1=0.072\ \text{N}[/tex]
[tex]F_2=0.115\ \text{N}[/tex]
r = Distance between shells = 40.4 cm
[tex]q_1[/tex] and [tex]q_2[/tex] are the charges
[tex]k[/tex] = Coulomb constant = [tex]8.99\times10^{9}\ \text{Nm}^2/\text{C}^2[/tex]
Force is given by
[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}[/tex]
[tex]F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}[/tex]
[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}[/tex]
Substituting the above value of [tex]q_1[/tex] we get
[tex]\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}[/tex]
[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}[/tex]
[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}[/tex]
Since we know [tex]q_1<q_2[/tex]
[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex].
Can someone help me with my physics
A pendulum is a body that is suspended from a fixed point so that it can swing back and forth through an exchange of kinetic energy and gravitational potential energy. Using 1–2 sentences, explain what happens to the kinetic energy and gravitational potential energy of the pendulum at the highest point and at the lowest point of its swing.
Answer:
Because mechanical energy (the sum of potential and kinetic energy) is conserved, as the kinetic energy increases, the potential energy decreases. The maximum kinetic energy is achieved when the pendulum passes through the lowest point, and the maximum potential energy is achieved at the highest point.
A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The other end of the spring is fixed to a wall. The spring is compressed by 10.0 cm from its equilibrium position and released from rest.
1) What is the speed of the object when it is 8.00 cm from equilibrium? (Express your answer to three significant figures.)
2) What is the speed when the object is 5.00 cm from equilibrium? (Express your answer to three significant figures.)
3) What is the speed when the object is at the equilibrium position? (Express your answer to three significant figures.)
Answer:
1) v = 0.45 m/s
2) v = 0.65 m/s
3) v = 0.75 m/s
Explanation:
1) We can find the speed of the object by conservation of energy:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
Where:
k: is the spring constant = 280 N/m
v: is the speed of the object =?
m: is the mass of the object = 5.00 kg
x: is the displacement of the spring
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]
2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]
3) When the object is at the equilibrium position, the speed of the object is:
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]
I hope it helps you!
(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s
(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s
(3) the speed of the object when the spring is at equilibrium is 0.748 m/s
Compression of spring and conservation of energy:
Given that the mass of the object, m = 5 kg
spring constant, k = 280 N/m
compression of the spring , x = 10 cm = 0.1m
(i) the spring compression is at d = 8cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]
v = 0.449 m/s
(ii) the spring compression is at d = 5cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]
v = 0.648 m/s
(iii) the spring is at equilibrium so compression is at d = 0cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]
v = 0.748 m/s
Learn more about conservation of energy:
https://brainly.com/question/14245799?referrer=searchResults
A particle with charge q1 C is moving in the positive z-direction at 5 m/s. The magnetic field at its position is B-3 4j1T What is the magnetic force on the particle? A. (20i+15j) N B. (207-15j) N C. (-20i+15j) N D. (-20/-15) N E. none of these
Answer:
D. [tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]
Explanation:
The statement is not correctly written, the correct form is now described:
A particle with charge [tex]q = -1\,C[/tex] is moving in the positive z-direction at 5 meters per second. The magnetic field at its position is [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex]. What is the magnetic force on the particle?
From classic theory on Magnetism, we remember that the magnetic force exerted on a particle ([tex]\vec F_{B}[/tex]), measured in newtons, is determined by the following vectorial formula:
[tex]\vec F_{B} = q\cdot \vec v \,\times \,\vec B[/tex] (1)
Where:
[tex]q[/tex] - Electric charge, measured in coulombs.
[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.
[tex]\vec B[/tex] - Magnetic field, measured in teslas.
If we know that [tex]q = -1\,C[/tex], [tex]\vec v = 5\,\hat{k}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex], then the magnetic force on the particle is:
[tex]\vec F_{B} = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\0\,\frac{C\cdot m}{s}&0\,\frac{C\cdot m}{s} &(-1\,C)\cdot (5\,\frac{m}{s} ) \\3\,T&-4\,T&0\,T\end{array}\right|[/tex]
[tex]\vec F_{B} = -(-4\,T)\cdot (-1\,C)\cdot \left(5\,\frac{m}{s} \right)\,\hat{i}+(-1\,C)\cdot\left(5\,\frac{m}{s} \right)\cdot (3\,T)\,\hat{j}[/tex]
[tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]
Which corresponds to option D.
A 37.0-kg child swings in a swing supported by two chains, each 3.06 m long. The tension in each chain at the lowest point is 410 N. (a) Find the child's speed at the lowest point.______m/s (b) Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.)_______ N(upword)
Answer:
1. 6.15m/s
2. 820N
Explanation:
The total upward force
= 410x2
= 820
g = 9.81
a = v²/r
= 2xT - msg = m x v²/r
= 820-37*9.81 = 37v²/3.06
= 820-362.97 = 37v²/3.06
= 457.03 = 12.09v²
To get v²
V² = 457.03/12.09
V² = 37.8
V = √37.8
V = 6.15m/s
B. We already have the answer to this question
The force exerted is simply gotten by this calculation
2x410
= 820N