Answer:
The answer is 198.23 kg
Explanation:
To find the mass of an object given it's mass and acceleration we use the formula
where
m is the mass
F is the force
a is the acceleration
From the question
a = 92.54 m/s²
F = 18,344 N
We have
We have the final answer as
198.23 kg
Answer:
198.22
Explanation:
using m=f/a you divide the force by the acceleration to get the mass (18344 / 92.54) = 198.22
A block is attached to one end of a spring with the other end of the spring fixed to a wall. The block is vibrating horizontally on a frictionless surface. If the mass of the block is 4.0 kg, the spring constant is k
Complete Question
A block is attached to one end of a spring with the other end of the spring fixed to a wall. The block is vibrating horizontally on a frictionless surface. If the mass of the block is 4.0 kg, the spring constant is k = 100 N/m, and the maximum distance of the block from the equilibrium position is 20 cm, what is the speed of the block at an instant when it is a distance of 16 cm from the equilibrium position?
Answer:
The velocity is [tex]v = 0.6 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the block is m = 4.0 kg
The spring constant is k = 100 N/m
The maximum distance of the block from equilibrium position is d = 20 cm =0.20 m
The distance considered is [tex]d_k = 16 \ cm = 0.16 \ m[/tex]
Generally the maximum energy stored in the spring is mathematically represented as
[tex]E = \frac{1}{2} * k * d^2[/tex]
=> [tex]E = \frac{1}{2} *100 * 0.2^2[/tex]
=> [tex]E = 2.0 \ J[/tex]
Gnerally according to the law of energy conservation
The energy maximum energy of the spring = energy of the spring at [tex]d_k[/tex] + energy of the block at [tex]d_k[/tex]
Here energy of the block at [tex]d_k[/tex] is mathematically represented as
[tex]K_1 = \frac{1}{2} mv^2[/tex]
=> [tex]K_1 = \frac{1}{2} * 4* v^2[/tex]
=> [tex]K_1 = 2v^2[/tex]
Generally the energy of the spring at [tex]d_k[/tex] is mathematically represented as
[tex]E_2 =\frac{1}{2} * k * d_k^2[/tex]
=> [tex]E_2 =\frac{1}{2} * 100 * (0.16)^2[/tex]
=> [tex]E_2 =1.28 \ J[/tex]
So
[tex]2.0 = 1.28 + 2v^2[/tex]
=> [tex]v = 0.6 \ m/s[/tex]
Type of tissue that helps with movement.Immersive Reader
a. Epithelial
b.Muscle
c.Connective
d.Nervous
Help please!!!!!!!!!!!!!
Answer:
The second choice
Explanation:
I think the answer is the second choice because if the surface is smooth, there is less friction. With a boat, it is easier to pull it on water than on the sand, because water has less friction, and thus, the answer is the second choice because rough surfaces have more friction.
Hopefully the explanation and answer helps!
A cheetah can run at a maximum speed
91 km/h and a gazelle can run at a maximum speed of 72.7 km/h.
If both animals are running at full speed,
with the gazelle 87.5 m ahead, how long before
the cheetah catches its prey?
Answer in units of s.
Answer:
Approximately 17.21 seconds
Explanation:
With subtraction, we have the gazelle 18.3 km/h slower than the cheetah, which is about 5.08333 m/s. As the gazelle is 87.5 meters ahead of the cheetah, 87.5 divided by 5.083333333 is about 17.21 seconds.
can you describe your own perspective whats Physical Science all about? PLS GUYS HELP
Answer:
Physical science is the study of the inorganic world. That is, it does not study living things. The four main branches of physical science are astronomy, physics, chemistry, and the Earth sciences, which include meteorology and geology.
An archer shoots an arrow with vertical velocity of 10 m/s and horizontal velocity of
30m/s. What is the maximum height the arrow reaches?
Answer:
5.1 m
Explanation:
Given in the y direction:
v₀ = 10 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (10 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 5.1 m
A rectangular loop with an area of 2 m2 is placed perpendicular to a uniform magnetic field of 1 Tesla. The field’s magnitude is increased to 6 Tesla in 4 seconds. The magnitude of the induced emf is equal to:
Answer:
Induced emf = 0
Explanation:
An emf can be induced due to the change in magnetic field. It can be given by :
[tex]\epsilon=\dfrac{d\phi}{dt}\\\\\because \phi=BA\cos\theta\\\\\epsilon=\dfrac{d(BA\cos\theta)}{dt}\\\\\epsilon=A\cos\theta\dfrac{dB}{dt}[/tex]
As the loop is placed perpendicular to a uniform magnetic field of 1 Tesla. It means that [tex]\theta=90^{\circ}[/tex] and cos(90) = 0. Hence, the induced emf is equal to 0.
The number of
• in the atom of an element determines its chemical properties.
Answer:
Yes, the number of electrons determines the chemical properties of the atom.
Explanation:
(DUE IN FIVE MINUTES, QUICK)
Explain why your weight would change if you went to the moon, but your mass wouldn’t.
The moon's gravitation force is determined by the mass and the size of the moon. Since the moon has significantly less mass than the Earth, it will not pull objects toward itself at the strength that Earth will.
The two main types of weathering are (4 points)
A. mechanical and physical
B. physical and kinetic
C. chemical and physical
D. chemical and acidic
Answer:
b
Explanation:
Answer:
its acually c
Explanation:
PLZZZ HURRY!!! I will freind and make brainliest
What are the basic building blocks of matter?
electrons
atoms
molecules
compounds
Answer:
an atom
Explanation: atom then matter
Answer:
the basic building blocks that make up matter are called atoms.
The Newton unit obtained from
A)Kg.m/s2
B) Kg. m
C) kg/m
D) kg ml
Answer:
Explanation:
The Newtons unit is kg. m/s2
Option A is the correct answer
What is the acceleration of the object if a 300-N force acting on a 25 kg object
Given parameters:
Force on the object = 300N
Mass of object = 25kg
Unknown:
Acceleration = ?
Solution:
According to Newton's second law of motion, force is a product of mass and acceleration.
Force = mass x acceleration
Input the parameters and solve for the acceleration;
300 = 25 x acceleration
Acceleration = 12m/s²
The acceleration is 12m/s²
PLEASE HELP
Discussion: If you put something like a piece of cardboard between a magnet and an iron nail, the magnet still holds the nail in place, even though the magnet is not touching the nail Explain how that happens. Use the words induce, magnetic field, permanent magnet and temporary magnet in your response.
Part B
You should find that your Interpolated and extrapolated values are not even close to the actual recorded values for these
displacement and velocity readings. Describe the basic assumption behind Interpolation and extrapolation, then for at least
one of these values explain why the calculated value was significantly larger or smaller than the recorded value.
BI
x
Font Sizes
A- A-EIE 3
I
Characters used: 0 / 15000
Answer:
When ever we use interpolation and extrapolation in our case we use linear approximation but the displacement verses time graph as well as velocity verses time grph are not linear so that whenever we use interpolatio and extrapolation we did not get close readings to the actual recorded values.
Explanation:
is newton's first law true on earth?
Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
THIS LAW IS TRUE AS IT ALSO HAVE A REAL LIFE EXAMPLE.
Examples of Newton's 1st Law : If you slide a hockey puck on ice, eventually it will stop, because of friction on the ice. It will also stop if it hits something, like a player's stick or a goalpost.
3. An object with a mass of 3.2 kg has a force of 6.2 N applied to it. What is the resulting acceleration
of the object?
Answer:
The answer is 1.94 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the acceleration
From the question we have
[tex]a = \frac{6.2}{3.2} \\ = 1.9375[/tex]
We have the final answer as
1.94 m/s²Hope this helps you
Engineers are using computer models to study train collisions to design safer
train cars. They start by modeling an elastic collision between two train cars
traveling toward each other. Car 1 is traveling east at 6 m/s and has a mass
of 3,154 kg, Car 2 is traveling west at 23 m/s and has a mass of 8,296 kg.
After the collision, car 1 has a final velocity of 7 m/s west. What is the final
velocity of car 2?
A. 18 m/s east
B. 23 m/s west
C. 23 m/s east
D. 18 m/s west
SUN
Answer:
18 m/s west
Explanation:
The correct option is C. 23 m/s east
What is law of conservation of momentum ?Conservation of momentum states that For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.
using conservation of momentum
m1v1 + m2v2 = m1 v1' + m2 v2'
Given
m1 = 3154 kg
v1= 6m/s
m2 = 8296 kg
v2 = 23 m/s
v1' = 7m/s
v2' = ?
3154 * 6 + 8296 * 23 = 3154 * 7 + 8296 v2'
209732 - 22078 = 8296 v2'
187654 = 8296 v2'
v2' = 22.62 ≈ 23 m/s east
correct option is C. 23 m/s east
learn more about conservation of momentum
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Electricity & Magnetism
4
Electricity can be used to produce powerful forces.
What type of energy is electricity converted to in an electromagnet?
A. sound energy
B.
heat energy
C. light energy
D. magnetic energy
The resultant force is equel to the.......of all the force
A) sum
B) product
C) subtraction
D) Division
Answer:
A) sum
Explanation:
That's the answer bro
Answer:
the answer is C subtraction
1) Which of the following is considered an effective treatment for someone with hearing loss based on nerve damage?
TAD
Cochlear implant
Hearing aid
OBI
No treatment available
2) Sylvester is dealing with hearing loss. The doctor informs him that his basilar membrane is damaged. What type of hearing loss is Sylvester experiencing?
Nerve deafness
Conduction hearing loss
Cochlear hearing loss
Conduction deafness
Sensory hearing loss
Answer:
For number 1 no treatment available , number 2 cochlear hearing loss
Explanation:
nerve damage is permanent
A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).
The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.
In figure a, the block is to the left of the spring, and an arrow above the block points to the right.
In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.
In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.
In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.
In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.
(a)
Find the distance of compression d (in m).
m
(b)
Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
m/s
(c)
Find the distance D (in m) where the object comes to rest.
m
(d)
What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
m
Answer:
(a) Approximately [tex]0.335\; \rm m[/tex].
(b) Approximately [tex]1.86\; \rm m\cdot s^{-1}[/tex].
(c) Approximately [tex]0.707\; \rm m[/tex].
(d) Approximately [tex]0.228\; \rm m[/tex].
Explanation:
[tex]v_i[/tex] denotes the velocity of the object in the first diagram right before it came into contact with the spring. Let [tex]m[/tex] denote the mass of the block. Let [tex]\mu[/tex] denote the constant of kinetic friction between the object and the surface. Let [tex]g[/tex] denote the constant of gravitational acceleration.Let [tex]k[/tex] denote the spring constant of this spring.(a)Consider the conversion of energy in this object-spring system.
First diagram: Right before the object came into contact with the spring, the object carries kinetic energy [tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2[/tex].
Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.
Third diagram: After the velocity of the object becomes zero, it has moved a distance of [tex]D[/tex] and compressed the spring by the same distance.
Energy lost to friction: [tex]\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D[/tex]. Elastic potential energy that the spring has gained: [tex]\displaystyle \frac{1}{2}\,k\, D^2[/tex].The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:
[tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
Assume that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex]. In the equation above, all symbols other than [tex]D[/tex] have known values:
[tex]m =1.10\; \rm kg[/tex].[tex]v_i = 2.60\; \rm m \cdot s^{-1}[/tex].[tex]\mu = 0.250[/tex].[tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].[tex]k = 50.0\; \rm N \cdot m^{-1}[/tex].Substitute in the known values to obtain an equation for [tex]D[/tex] (where the unit of [tex]D\![/tex] is [tex]m[/tex].)
[tex]3.178 = 2.69775\, D + 25\, D^2[/tex].
[tex]2.69775\, D + 25\, D^2 + 3.178 = 0[/tex].
Simplify and solve for [tex]D[/tex]. Note that [tex]D > 0[/tex] because the energy lost to friction should be greater than zero.
[tex]D \approx 0.335\; \rm m[/tex].
(b)The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:
[tex]\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J[/tex].
As the object moves to the left, part of that energy will be lost to friction:
[tex](\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J[/tex].
The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:
[tex]2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J[/tex].
Calculate the velocity corresponding to that kinetic energy:
[tex]\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}[/tex].
(c)As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ([tex]1.91\; \rm J[/tex]) would be lost to friction.
How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is [tex]\mu \cdot m \cdot g[/tex].
[tex]\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m[/tex].
(d)Similar to (a), solving (d) involves another quadratic equation about [tex]D[/tex].
Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) [tex]1.91\; \rm J[/tex].
Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.
[tex]\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
[tex]25\, D^2 + 2.69775\, D - 1.90811\approx 0[/tex].
Again, [tex]D > 0[/tex] because the energy lost to friction is greater than zero.
[tex]D \approx 0.228\; \rm m[/tex].
The energy transferred between the object and the spring as a closed system, therefore, conserved are;
(a) The distance of compression, d ≈ 0.3354 meters
(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s
(c) The distance where the object comes to rest, D ≈ 0.7071 m
(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m
The reason the above values are correct are as follows;
The known parameters are;
Mass of the object, m₁ = 1.10 kg
Coefficient of friction, μ = 0.250
The initial speed of the object, [tex]v_i[/tex] = 2.60 m/s
Force constant of the spring, K = 50.0 N/m
Distance the spring is compressed by the object = d
(a) Conservation of energy principle
[tex]Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2[/tex]
Work done = Force × Distance
Friction force, [tex]F_f[/tex] = W × μ
Weight, W = m·g
Weight = Mass × Acceleration
Energy transferred by object = Work done by spring + Work done by friction
[tex]Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718[/tex]
Energy transferred by object = 3.718 J
[tex]Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2[/tex]
[tex]Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2[/tex]
[tex]W_{spring}[/tex] = 25·d²
Work done by friction, [tex]W_{friction}[/tex] = 1.10×9.81×0.250×d = 2.69775·d
Therefore;
3.718 = 25·d² + 2.69775·d
25·d² + 2.69775·d - 3.718 = 0
Solving gives
The distance of the compression d ≈ 0.3354 m
(b) The energy given by the spring = 25·d²
The work done by friction, [tex]W_{friction}[/tex] = 2.69775·d
Kinetic energy given to object = 0.55·v²
0.55·v² = 25·d² - 2.69775·d
0.55·v² = 25×0.3354² - 2.69775×0.3354
∴ v = √(3.4682) = 1.8623
The velocity of the object at the un stretched position, v ≈ 1.8623 m/s
(c) The kinetic energy, K.E. of the object on the way left is given as follows;
K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J
The work done by friction before object comes to rest = 2.69775·D
[tex]D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m[/tex]
The distance where the object comes to rest, D ≈ 0.7071 m
(d) The work done on spring, [tex]W_{spring}[/tex] = 25·D'²
Work done on friction, [tex]W_{friction}[/tex] = 2.69775·D'
Kinetic energy of object, K.E. ≈ 1.90751 J
K.E. = [tex]W_{spring}[/tex] + [tex]W_{friction}[/tex]
1.90751 ≈ 25·D'² + 2.6775·D'
25·D'² + 2.6775·D' - 1.90751 = 0
Solving with a graphing calculator gives;
D' ≈ 0.2278 m
The new value of the distance D = 0.2278 m
Learn more about the energy conservation principle here:
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What causes tides to occur in the ocean?
Waves
Wind
Gravitational pull
Coriolis effect
Answer:
Gravitational pull
Explanation:
The moon pulls the tides
Answer: gravitational pull
Tides are caused by a gravitational pull from the Moon. Ocean/bay tides rise because of this pull for the gravity under the water. This can happen every day up to 6 times.
How are gas giants similar to one another?
Answer:
they are all made of gass and they are all giants.
Explanation:
Answer:
How are the gas giants similar to one another? dont have solid surfaces and are much larger than earth. Why do all of the gas giants have thick atmospheres? Because they are so massive, the gas giants exert a much stronger gravitational force than the terrestrial planets
Explanation:
Which of the following does not discribe a mineral
Answer:
give us some further context to answer your question as well
Explanation:
why a dam is thicker at the bottom than it's top
Answer: Due to water pressure.
Explanation:
As depth increases so does the pressure.
A boy pushes a box with a force of 150 N at an angle of 40 with a flat floor. What component of his force is directed downward , or into the floor . PLEASE ANSWER!!!!!
Answer:
[tex]F_y=96.4N[/tex]
Explanation:
Hello.
In this case, considering the force diagram shown on the attached picture, we can see that the component of his force is directed downwards is:
[tex]F_y=F\times sin (\theta)[/tex]
Because the other component is the horizontal one:
[tex]F_x=F\times cos(\theta)[/tex]
In this case, the y-component force turns out:
[tex]F_y=150N\times sin (40\°)\\\\F_y=96.4N[/tex]
Moreover, the x-component force is also computed if required:
[tex]F_x=150N\times cos(40\°)\\\\F_x=114.9N[/tex]
Best regards.
A helpful association method like remembering the Allies during World War II as BAR
(Britain, America, and Russia) is called
O an acronym
O the DAP flashcard method
O a visual image
O a mind map
Answer:
an acronym because it is shorted to remember like mvemjsun it's the planet
A boy and a girl are pulling a heavy crate at the same time with 7 units of firce each. What is the net force acts on the ibject? Is the object balanced or unbalanced?
Answer:
Net force= 14 units
The object is unbalanced
Explanation:
The net force refers to the sum of all forces applied to an object. However, the direction of force applied determine the net force. In this question, a boy and girl is pulling a heavy crate at the same time.
This means that the force is in the same direction, hence, the net force will be:
F(N) = 7 + 7 = 14 unit
However, since the pull is occuring at the same direction. This means that the object has a net force, therefore, will move in a particular direction. This means that the OBJECT IS UNBALANCED
The nervous system has two distinct branches. They are the:
Answer:
central nervous system
peripheral nervous system
Explanation:
The nervous system has two main parts: The central nervous system is made up of the brain and spinal cord. The peripheral nervous system is made up of nerves that branch off from the spinal cord and extend to all parts of the body.Oct 1, 2018
Answer:
The central nervous system
The peripheral nervous system
Explanation:
The central nervous system (CNS) is the brain and spinal cord, and the peripheral nervous system (PNS) is everything else