Answer:
R = 28.125 ohms
Explanation:
Given that,
The voltage of a bulb, V = 4.5 V
Current, I = 0.16 A
We need to find the resistance of the filament. Using Ohm's law,
V = IR
Where
R is the resistance of the filament
So,
[tex]R=\dfrac{V}{I}\\\\R=\dfrac{4.5}{0.16}\\\\R=28.125\ \Omega[/tex]
So, the resistance of the filament is equal to 28.125 ohms.
Please helppppppp I need it todayyyyyt!!!!!
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s. Calculate his:
a. acceleration in his direction of motion
b. acceleration opposite to his direction of motion.
Answer: [tex]56.4\ m/s^2, 201.42\ m/s^2[/tex]
Explanation:
Given
Rocket attain a velocity of [tex]v=282\ m/s[/tex] in a time period of [tex]t=5\ s[/tex]
It was brought jarringly back to rest in only [tex]t'=1.4\ s[/tex]
Acceleration is the change in velocity of the object over a period of time
(a) Acceleration in his direction of motion
[tex]\Rightarrow a=\dfrac{v-0}{t}\\\\\Rightarrow a=\dfrac{282}{5}\\\\\Rightarrow a=56.4\ m/s^2[/tex]
(b) acceleration opposite to his direction of motion i.e. deceleration is
[tex]\Rightarrow a_d=\dfrac{0-v}{t'}\\\\\Rightarrow a_d=\dfrac{-282}{1.4}\\\\\Rightarrow a_d=-201.42 \ m/s^2\\\Rightarrow a_d=201.42\ \text{decelration}[/tex]
Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in the same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 2I0?
A. f/4
B. f/2
C. 3f/2
D.) 2f
Answer:
Option D
Explanation:
From the question we are told that:
The attractive magnetic force per unit length as
[tex]f = F/L[/tex]
Separation Distance [tex]x=2d[/tex]
Generally the equation for Magnetic force between two current carrying wire is mathematically given by
[tex]\frac{F}{\triangle l}=\frac{\mu_0I_1I_2}{\mu \pi x}[/tex]
[tex]\frac{F}{\triangle l }=\frac{I_1I_2}{ x}[/tex]
Where
[tex]x=2r[/tex]
And
[tex]I_1=I_2=>2I[/tex]
Then
[tex]\frac{F}{\triangle l}=>\frac{2*2}{2}*f[/tex]
[tex]\frac{F}{\triangle l}=>2f[/tex]
Therefore s the force per unit length between the wires if their separation is 2d
[tex]\frac{F}{\triangle l}=>2f[/tex]
Option D
Two workers are sliding 450 kg kg crate across the floor. One worker pushes forward on the crate with a force of 380 NN while the other pulls in the same direction with a force of 230 NN using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor
Answer:
The coefficient of kinetic friction on the floor is 0.138
Explanation:
Given;
mass of the crate, m = 450 kg
force applied by the first worker, F₁ = 380 N
force applied by the second worker in the same direction as the first worker, F₁ = 230 N
frictional force opposing the motion of the box = -[tex]F_k[/tex]
Apply Newton's second law of motion;
∑F = ma
[tex]F_1 + F_2 - F_k = ma[/tex]
If the crate slides with constant speed, acceleration is zero (0).
[tex]F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = \frac{F_1 + F_2}{mg} \\\\\mu _k = \frac{380 + 230}{450 \times 9.8} \\\\\mu _k = 0.138[/tex]
Therefore, the coefficient of kinetic friction on the floor is 0.138
Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object. For an object to change its state of motion, a force must be applied to it.
A. Velocity
B. Inertia
C. Distance
D. Area
Answer:
B) Inertia is the resistance of any physical object
А bus has started to move from
the rest with an acceleration of
0.25 m/s². find its final velocity
Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his car, mcartmcart, is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest.
Chuch then picks up a ball of mass mballmball and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is vc. The speed of the thrown ball relative to the ground is vb. Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie's speed relative to the ground after she catches the ball is vj.
When answering the questions in this problem, keep the following in mind:
1. The original mass mcartmcart of Chuck and his cart does not include the mass of the ball.
2. The speed of an object is the magnitude of its velocity. An object's speed will always be a non-negative quantity.
Required:
a. Find the relative speed u between Chuck and the ball after Chuck has thrown the ball.
b. What is the speed vb of the ball (relative to the ground) while it is in the air?
c. What is Chuck's speed vc (relative to the ground) after he throws the ball?
d. Find Jackie's speed vj (relative to the ground) after she catches the ball in terms of vb.
e. Find Jackie's speed vj (relative to the ground) after she catches the ball in terms of u.
Explanation:
a.)
We find the relative speed
u = vb + vc
b.)
chuck and the cart are at a rest position
mcartvc = mballvb
from part a above,
vc = u - vb
mcartu = vb(mcart + mball)
make vb the subject of the equation
[tex]vb=\frac{mcartu}{mcart +mball}[/tex]
c.)
from anser a,
vb = u - vc
then mcart vc = mball(u-vc)
[tex]vc= \frac{Mballu}{Mcart+Mball}[/tex]
d.
Mballvb = (Mcart + Mball)vj
we make vj the subject to get her relative speed
[tex]vj=\frac{MballVb}{Mcart+Mball}[/tex]
e.
given the solution in part d above,
we have
[tex]vj=\frac{MballVb}{Mcart+Mball}[/tex]
remember,
vb = u - vc
such that
[tex]Vj = \frac{Mball(u-v)}{Mball +Mcart}[/tex]
thank you!
different between pressure and force
Force is mass into acceleration
and pressure is force applied per unit area.
How do solar systems, galaxies, and the universe show different frames of reference about space?
Answer:
Many people are not clear about the difference between our Solar System, our Milky Way Galaxy, and the Universe.
Let’s look at the basics.
Our Solar System consists of our star, the Sun, and its orbiting planets (including Earth), along with numerous moons, asteroids, comet material, rocks, and dust. Our Sun is just one star among the hundreds of billions of stars in our Milky Way Galaxy. If we shrink the Sun down to smaller than a grain of sand, we can imagine our Solar System to be small enough to fit onto the palm of your hand. Pluto would orbit about an inch from the middle of your palm.
Artist diagram of Milky Way galaxy
On that scale with our Solar System in your hand, the Milky Way Galaxy, with its 200 – 400 billion stars, would span North America (see the illustration on the right). Galaxies come in many sizes. The Milky Way is big, but some galaxies, like our Andromeda Galaxy neighbor, are much larger.
The universe is all of the galaxies – billions of them! NASA’s telescopes allow us to study galaxies beyond our own in exquisite detail, and to explore the most distant reaches of the observable universe. The Hubble Space Telescope made one of the deepest images of the universe, called the Hubble Extreme Deep Field (image at the top of this article). Soon the James Webb Space Telescope will be exploring galaxies forming at the very beginning of the universe.
You are one of the billions of people on our Earth. Our Earth orbits the Sun in our Solar System. Our Sun is one star among the billions in the Milky Way Galaxy. Our Milky Way Galaxy is one among the billions of galaxies in our Universe. You are unique in the Universe!
You can observe objects in our solar system and even see other galaxies at a star party near you-and rest assured that everything you are seeing is a part of the same universe as you!
Explanation:
A solar system is the system of celestial bodies built around a central star, the Sun. All of the system bodies, be they dwarf planets, small bodies and large planets, are held in a gravitational bond around the central star. Our solar system has eight large planets:
Four inner planets which are terrestrial, made entirely of rock and metal: Mars, Mercury, Earth and Venus;
Four outer planets which are gas and ice giants: Jupiter and Saturn (composed entirely of helium and hydrogen), Uranus and Neptune (composed of ices such as water, ammonia and methane).
The solar system also contains asteroid belts and the natural satellites of some of the planets. The trans-Neptunian region has the Kuiper belt, home to several dwarf planets, Pluto among them. Our solar system is located on the Orion Arm and is part of the Milky Way Galaxy. It was formed 4.6 billion years ago.
A galaxy is made out of billions of stars and their solar systems, held together by gravity, with a super- massive black hole at the center. Our Solar System is called the Milky Way; it is a spiral galaxy and the black hole in the center is called Sagittarius A*. Apart from the spiral shape, galaxies can also be elliptical or irregular in form. Galaxies gather in groups, clusters and super-clusters and there are billions of Galaxies in the Universe.
Some of these other galaxies are visible to the naked eye on a dark night and from places away from artificial light sources. The Andromeda Galaxy is the most recorded one throughout time and all over the world, its existence having been noted since the 10th century by Persian astronomer Al-Sufi, and having been the object of debate among other great thinkers up to the moment when the technology caught up to the discourse.
Solar System vs Galaxy
So what is the difference between a solar system and a galaxy?
A solar system represents the group of planets gravitationally bound to the central star. A galaxy has billions of stars and their solar systems. This difference in size is not only visible in the number of stars it is made out of, but also by how long it takes to cross it. It takes one light year to cross our solar system, and 100,000 light years to cross the galaxy.
While the biggest thing inside a solar system is the central sun, the biggest thing inside a galaxy is a massive black hole. The planets in a solar system orbit the sun, which is at the center, and the Sun, in turn, orbits the center of the Milky Way.
Comparison Chart
Solar system Galaxy
A group of planets orbiting the central sun A group of planetary systems whose central Suns are orbiting the center of the Galaxy
Gravitationally bound Gravitationally bound
Can be crossed in 1 light year Can be crossed in 100,000 light years
Most of the system mass is taken up by the central sun It hosts a super massive black hole, Sagittarius A*
More solar systems make up galaxies More galaxies make up the Universe
One of the earliest vertebrate animal groups that evolved in the early Paleozoic Era
are
Probability of a woman wearing green to walk into a restaurant
5. Charges added to a conductor quickly spread over the surface of the object.
a. True
b. False
Answer:
a. True
Explanation:
Conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.
A conductor can be defined as any material or physical object that allows the conduction (transfer) of electric charge or thermal energy. Some examples of conductors are metal, steel, aluminum, copper, frying pan, pot, metallic spoon, etc.
During heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles.
Furthermore, any charge that is added to a conductor would quickly spread over the surface of the conducting object due to the fact that it allows the movement of subatomic particles.
Answer: True
Explanation: Aced Test
An athlete training for an event does 6.53 104 J of work during a workout and gives off 5.97 105 J of heat. Consider the athlete to be like a heat engine. (a) Determine the magnitude of the change in internal energy of the athlete. J (b) What is the efficiency of the athlete
Answer:
(a) The magnitude of the change in internal energy is 6.623 x 10⁵ J
(b) the efficiency of the athlete is 10.94 %
Explanation:
Given;
work done by the athlete (system), W = 6.53 x 10⁴ J
the heat given off by the athlete (system), Q = 5.97 x 10⁵ J
The simple diagram below will be used to illustrate the direction of the energy flow assuming a heat engine.
Q← ⊕ →W
The work, W, points away from the system since the system does the work
The heat, Q, points away from the system since heat is given off
Apply first law of thermodynamic;
ΔU = Q + W
where;
q is the heat flowing into or out of the system
(+q if the heat is flowing into the system
(-q if the heat is leaving the system
w is the work done by or on the system
(+w if the work is done on the system by the surrounding
(-w if the work is done by the system to the surrounding
Thus, from the above explanation, the change in internal energy of the system is calculated as;
ΔU = -Q - W
ΔU = - 5.97 x 10⁵ J - 6.53 x 10⁴ J
ΔU = -6.623 x 10⁵ J
The magnitude of the change in internal energy = 6.623 x 10⁵ J
(b) the efficiency of the athlete;
[tex]Efficiency = \frac{W}{Q} \times 100\%\\\\Efficiency = \frac{6.53 \times 10^4}{5.97 \times 10^5} \times 100\%\\\\Efficiency = 10.94 \ \%[/tex]
A cart weighing 40 pounds is placed on a ramp incline 15 degrees to the horizon. The cart is held in place by a rope inclined 60 degrees to the horizontal. find the force that the rope must exert on the cart to keep it from rolling down the ramp.
Answer: [tex]14.64\ N[/tex]
Explanation:
Given
Inclination of ramp is [tex]\theta=15^{\circ}[/tex]
Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal
Weight of cart [tex]W=40\ lb[/tex]
from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp
Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position
[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]
Question 11 of 22
A horse of mass 180 kg gallops at a speed of 8 m/s. What is the momentum
of the horse?
Answers
1440
22.5
845
1955
Momentum = (mass) x (speed)
If you work the problem in the same units as the given data, then you get the momentum in units of kilogram-meters per second, and your horse has 1,440 of them.
Answer:
A
Explanation:
1440 kg*m/s
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude
Answer:
Centripetal acceleration = 0.79 m/s²
Explanation:
Given the following data;
Radius, r = 2.6 km
Time = 360 seconds
Conversion:
2.6 km to meters = 2.6 * 1000 = 2600 meters
To find the magnitude of centripetal acceleration;
First of all, we would determine the circular speed of the car using the formula;
[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]
Where;
r represents the radius and t is the time.Substituting into the formula, we have;
[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]
[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]
Circular speed, V = 45.38 m/s
Next, we find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]
Where;
V is the circular speed (velocity) of an object.r is the radius of circular path.Substituting into the formula, we have;
[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]
[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]
Centripetal acceleration = 0.79 m/s²
If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force between them?
Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N
A train with mass 3.3 x 107 kg starts from rest and accelerates to a speed of 42
m/s. What is the initial kinetic energy of the train?
Answer:
kinetic energy of the train = 2,910.6 x 10⁷ joule
Explanation:
Given:
Mass of train = 3.3 x 10⁷ kg
Speed of train = 42 m/s
Find:
kinetic energy of the train
Computation:
kinetic energy = (1/2)(m)(v²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)
kinetic energy of the train = (3.3 x 10⁷)(882)
kinetic energy of the train = 2,910.6 x 10⁷ joule
Answer: The initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
Explanation:
Given: Mass = [tex]3.3 \times 10^{7} kg[/tex]
Speed = 42 m/s
Kinetic energy is the energy acquired by an object due to its motion.
Formula to calculate kinetic energy is as follows.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
m = mass of object
v = speed of object
Substitute the values into above formula as follows.
[tex]K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 3.3 \times 10^{7} kg \times (42 m/s)^{2}\\= 2910.6 \times 10^{7} kg m^{2}/s^{2} (1 J = 1 kg m^{2}/s^{2})\\= 2910.6 \times 10^{7} J[/tex]
Thus, we can conclude that the initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
what's the difference between mass and inertia in a tabular form
Answer:
to be honest i dont know
Explanation:
^^
If a second ball were dropped from rest from height ymax, how long would it take to reach the ground
Answer:
[tex](b)\ t_1 - t_0[/tex]
[tex](d)\ t_2 - t_1[/tex]
[tex](e)\ \frac{t_2 - t_0}{2}[/tex]
Explanation:
Given
See attachment for complete question
Required
How long to reach the ground from the maximum height
First, calculate the time of flight (T)
[tex]T =t_2 - t_0[/tex]
The time taken (t) from maximum height to the ground is:
[tex]t = \frac{1}{2}T[/tex]
So, we have:
[tex]t = \frac{t_2 - t_0}{2}[/tex]
Another representation is:
At ymax, the time is: t1
On the ground, the time is t2
The difference between these times is the time taken.
So;
[tex]t = t_2 - t_1[/tex]
Since air resistance is to be ignored, then
[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height
If a fisherman applies a horizontal force with magnitude 47.0 NN to the box and produces an acceleration of magnitude 3.20 m/s2m/s2, what is the mass of the box
Answer:
The correct solution is "14.6875 kg".
Explanation:
Given values:
Force,
F = 47.0 N
Acceleration,
a = 3.20 m/s²
Now,
⇒ [tex]Force=Mass\times Acceleration[/tex]
or,
⇒ [tex]F=ma[/tex]
⇒ [tex]47.0=m\times 3.20[/tex]
⇒ [tex]m=\frac{47.0}{3.20}[/tex]
⇒ [tex]=14.6875 \ kg[/tex]
pls solve this:
if u solve this, I will surely mark u as brilliant
Answer: The frequency of a pendulum is [tex]0.2 s^{-1}[/tex].
Explanation:
Time period is defined as the time required to produce complete wave.
As we know that the frequency and time are inversely proportional to each other.
That means,
[tex]\nu =\frac{1}{T}[/tex]
Where,
[tex]\nu[/tex] is frequency of pendulum
T is time period
Given:
Time period = 5 seconds
Now putting all the given values in the above formula, we get the frequency of the pendulum.
[tex]\nu =\frac{1}{T}\\\\\nu =\frac{1}{5s}\\\\\nu =0.2s^{-1}[/tex]
Therefore, the frequency of a pendulum is [tex]0.2 s^{-1}[/tex].
is anyone online??just asking
Answer:
me...:(
Explanation:
Answer:
hello I'm online here thanks for the points (◔‿◔)
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W
Complete question:
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.
Answer:
The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Explanation:
Given;
power of the flux produced, P = 1 W = 1 J/s
energy per photon, E = 78 keV
Convert the energy per photon to J
E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon
let the number of photons = n
n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s
[tex]n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s[/tex]
Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.
Answer:
4.408 [tex]\mathsf{M_{sun}}[/tex]
Explanation:
According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:
[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]
combined mass (m₁+m₂) = 4.408 [tex]\mathsf{M_{sun}}[/tex]
A red apple reflects ______
light and absorbs all other
colours.
A red apple reflects Red light and absorbs all other colours.
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Where does the extra energy go?
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Then the extra energy goes in overcoming the gravitational acceleration and heating up of body etc
The pair of forces described by Newton third law must be
Answer:
The answer is Newton's third law of motion states that every action has an equal and opposite reaction. This means that force always act in pairs
The pair of forces described by Newton third law must be in opposite direction.
What is Newton's third law of motion ?Every action have equal and opposite reaction. for example when we fire bullet from a gun, the gun will recoil back and bullet moves forward. In case of rocket, rocket is fired, thrust is reaction of force applied by the gas on the floor.
The motion of lift from an airfoil in which the air is diverted downward by the airfoil's action and the wing is pushed upward in response.
When a spinning ball moves, the air is deflected to one side, and the ball responds by travelling in the other direction.
A jet engine's motion generates thrust, and hot exhaust gases rush out the back of the engine, producing thrust in the opposite direction.
To know more about motion :
https://brainly.com/question/26603017
#SPJ3.
g You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest
You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest
A box attached to a spring is being pulled across a flat frictionless surface. The spring constant is 45 N/m, the box is accelerating at
1.3 m/s2, and the spring is stretched by 0.88 m. What is the mass of the box?
26 kg
O 38 kg
30 kg
O 33 kg
Plzzzzz help
Answer:
Mass = 30.46 kg
Explanation:
Given the following data;
Spring constant = 45 N/m
Acceleration = 1.3 m/s²
Extension = 0.88 m
To find the mass of the box;
First of all, we would determine the force acting on the spring.
Force = spring constant * extension
Force = 45 * 0.88
Force = 39.6 N
Next, we find the mass using Newton's second equation of motion.
Force = mass * acceleration
39.6 = mass * 1.3
Mass = 39.6/1.3
Mass = 30.46 kg