Answer:
The Dimension of mass is [F T v⁻¹] and Dimension of energy is [F T v]
Explanation:
(i) We know that,
Force = mass × acceleration
= mass × velocity/time
⇒ mass = force × time ÷ velocity
or. [mass] = [force] [time] ÷ [velocity]
= [F] [T] ÷ [v]
∴ [mass] = [F T v⁻¹]
(ii) Dimensions of energy are same as the dimensions of kinetic energy
∴ Energy = [ ½mv²] = [m] [v]²
= [F T v⁻¹] [v]²
Energy = [F T v]
Thus, The Dimension of mass is [F T v⁻¹] and Dimension of energy is [F T v]
-TheUnknownScientist
Explanation:
fvt-1 = mass
ftv = energy
How many inner shell electrons are there in Bohr model
Answer:
10 is the correct answer
Explanation:
It is ten because the electrons that encircled the two orbit before the third orbit if the bohr model is 10. In bohr model, the electrons encircled the nucleus of an atom called orbit.
If a bullet with a mass of 0.005kg is fired from a
gun at a speed of 1000 m/s, what is its
momentum?
what’s the v?
Answer:
5kgm/s
Explanation:
Given parameters:
Mass of bullet = 0.005kg
Speed = 1000m/s
Unknown:
Momentum = ?
Solution:
Momentum is the amount of motion a body possesses.
Mathematically;
Momentum = mass x velocity
Now insert the parameters and solve;
Momentum = 0.005 x 1000 = 5kgm/s
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/s.
a. Calculate the length of the incline
b. calculate the horizontal and vertical components of the velocity at the end of the incline.
A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal force of 98.0 N for 1.40 s.(a) Determine the magnitude of each force acting on the block of ice while you are pulling.Fpull = NFg = NFN = N(b) With what speed is the ice moving after you are finished pulling?m/s
Answer:
a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.
b) The final speed of the block of ice is 9.8 meters per second.
Explanation:
a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.
The weight of the block ([tex]W[/tex]), measured in newtons, is:
[tex]W = m\cdot g[/tex] (1)
Where:
[tex]m[/tex] - Mass of the block of ice, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]m = 14\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the magnitudes of the weight and normal force of the block of ice are, respectively:
[tex]N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]N = W = 137.298\,N[/tex]
And the pull force is:
[tex]F_{pull} = 98\,N[/tex]
The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.
b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:
[tex]m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}[/tex] (2)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the block, measured in meters per second.
[tex]\Sigma F[/tex] - Horizontal net force, measured in newtons.
[tex]\Delta t[/tex] - Impact time, measured in seconds.
Now we clear the final speed in (2):
[tex]v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}[/tex]
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]m = 14\,kg[/tex], [tex]\Sigma F = 98\,N[/tex] and [tex]\Delta t = 1.40\,s[/tex], then final speed of the ice block is:
[tex]v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}[/tex]
[tex]v_{f} = 9.8\,\frac{m}{s}[/tex]
The final speed of the block of ice is 9.8 meters per second.
4. You’re driving a car that can climb a maximum gradient of 500m/km. The hill in front of you starts at an elevation of 20m and reaches 100m. The total distance up the hill is 1.5km? What is the gradient of the hill and will your car make it?
Answer:
Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.
Explanation: Given that
Maximum gradient = 500 m/km
Total distance = 1.5 km
Starting elevation = 20 m
Final elevation = 100 m
Gradient = change in elevation/ total distance.
Now, substitute the values into the formula.
Gradient = (100m - 20m)/1.5km
= 80m/1.5km
= 53.33m/km
Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.
A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-direction with a speed 2v and the second particle is at rest. After the collision, the second particle is moving in the direction 45o below the x-axis and with a speed √2v.
a) Find the velocity of the first particle after the collision. (i.e. find the x-and y-components of the velocity.)
b) Find the total kinetic energy of the two particles before and after the collision.
c) Is the collision elastic or inelastic?
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
A seagull flying horizontally over the ocean at a constant speed of 2.85 m/s carries a small fish in its mouth. It accidentally lets go of the fish, and 2.45 s after the bird lets go, the fish lands in the ocean.
a. Just before reaching the ocean, what is the horizontal component of the fish's velocity?
b. Just before reaching the ocean, what is the vertical component of the fish's velocity?
c. If the seagull's initial speed were decreased, which of the following regarding the fish's velocity upon reaching the ocean would be true?
1. The horizontal component of the fish's velocity would decrease.
2. The horizontal component of the fish's velocity would increase.
3. The horizontal component of the fish's velocity would stay the same.
4. The vertical component of the fish's velocity would decrease.
5. The vertical component of the fish's velocity would increase.
6. The vertical component of the fish's velocity would stay the same.
Answer:
a) 2.85 m/s
b) -24.01 m/s
c) 1
Explanation:
From the question, we can attest that the motion of the fish that was dropped by the seagul is that of a projectile motion. This motion is made up of two other motions which are, a horizontal uniform motion and a vertical motion, at constant acceleration. From the question, we are asked to find the horizontal motion. And then, the horizontal component of the fish's velocity does not change, therefore its the horizontal component of the fish's velocity is 2.85 m/s
To find the vertical component of the fish's velocity, we use the equation
v(y) = u(y) + gt
Where u(y) is the initial velocity which is zero, and g is the acceleration due to gravity which is -9.8 m/s. We are also told from the question that it took 2.45 s, and that's our time t. Applying this into the equation, we have
v(y) = 0 + -9.8 * 2.45
v(y) = -24.01 m/s
For the 3rd part, the horizontal component of the fish's velocity would decrease
One small speaker is placed 3m to the east of a second speaker, and a listener stands 4m directly south of one of the speakers. That listener finds that if they move in any direction, the sound gets louder. What is the longest possible wavelength of the sound from the speakers
Answer:
The value is [tex]\lambda = 2 \ m[/tex]
Explanation:
From the question we are told that
The distance of the speaker from the second speaker to the east is [tex]d = 3 \ m[/tex]
The distance of the speaker from the listener to the south is [tex]a = 4 \ m[/tex]
Generally given that if the speaker move in any direction, their sound become louder , it then mean that the position of the listener of minimum sound (i.e a position of minima ) ,
Generally the path difference of the sound produce by both speaker at a position of minima is mathematically represented as
[tex]y = \frac{\lambda}{2}[/tex]
Generally considering the orientation of the speakers and applying Pythagoras theorem we see that distance from the second speaker to the listener is mathematically represented as
[tex]b = \sqrt{d^ 2 + a^2 }[/tex]
=> [tex]b = \sqrt{3^ 2 + 4^2 }[/tex]
=> [tex]b = 5[/tex]
Generally the path difference between the two speaker with respect to the listener is
[tex]y = b - a[/tex]
=> [tex]y = 5 - 4[/tex]
=> [tex]y = 1[/tex]
So
[tex]1 = \frac{\lambda}{2}[/tex]
=> [tex]\lambda = 2 \ m[/tex]
What is the beat frequency heard when two organ pipes, each open at both ends, are sounded together at their fundamental frequencies if one pipe is 48 cm long and the other is 63 cm long?
Answer:
Beat frequency = 125.5 Hz
Explanation:
Given:
Length of first organ pipes = 43 cm = 0.43 m
Length of second organ pipes = 63 cm = 0.63 m
Computation:
Wavelength L = λ/2
length λ = 2L
Frequency f1 = v/λ = 340 / 2(0.43)
Frequency f1 = 395.34 Hz
Frequency f2 = v/λ2 = 340/2(0.63)
Frequency f2 = 269.84 Hz
Beat frequency = Frequency f1 - Frequency f2
Beat frequency = 395.34 - 269.84
Beat frequency = 125.5 Hz
A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg.
Required:
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 is not moving then its acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = 0
Fm - Ff = 0.
Fm is the moving force
Ff is the frictional force
Fm = Ff
This means that the moving 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
m is the mass of the crate = 31.2kg
g is the acceleration due to gravity = 9.8m/s²
R = 31.2 × 9.8
R = 305.76N
Recall that;
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
what is the role of heat in the formation
of new rocks
Answer: Igneous rocks are created by heat. They start off as magma, which is hot, melted rock deep within a volcano. When magma cools and hardens, igneous rock forms.
Explanation:
Heat or increase in temperature plays a vital role in formation of igneous, metamorphic rocks.
What is the role of heat in the formation of new rocks?On Earth, igneous sedimentary or metamorphic rocks are created. When rocks are heated to their melting point and magma is produced, igneous rocks are created. Rocks that undergo metamorphosis are created when heat and pressure transform the original or parent rock into a whole new rock.
The heat causes a physical weathering process where the rock splits apart into fragments as it expands and contracts. When oxygen or moisture in the air change the chemical makeup of rock minerals, this also contributes to chemical weathering.
Heat and pressure cause an existing rock to change into a new rock, creating metamorphic rocks. When hot magma hits rock, contact metamorphism takes place. Large sections of existing rocks are altered by regional metamorphism as a result of the intense heat and pressure caused by tectonic forces.
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A car is traveling at a constant speed along the road ABCDE shown in the drawing. Section Ab and DE are straight. Rank the acceleration in each of the four sections according to magnitude (smallest first).
-AB & DE tie, CD, BC
-all tie
-CD, BC, AB & DE tie
-AB & DE tie, BC , CD
-BC, CD, AB & DE tie
Answer:
AB = DE <CD <BC
Explanation:
This is an exercise in kinetics, the accelerations defined as the change in velocity over the time interval, therefore the accelerations of a vector.
Because the acceleration is a vector, it has two parts, the modulus that the numerical value of the magnitude and the direction, a change in any of them implies the existence of a relationship.
Let's apply these reasoning to our problem.
AB Path
this path is straight and as they indicate that the constant speed the acceleration is zero
DE path
This path is straight and since the velocity is constant the zero steps
BC path
This path is a curve and the velocity modulus is constant, but its directional changes therefore there is an acceleration called centripetal, given by the expression
[tex]a_{c}[/tex] = v² / r
where r is the radius of the curve and the direction of acceleration is towards the center of the curve
CD path
This path is a curve and it also has centripetal acceleration, as can be seen in the drawing, the radius of the curve is greater than in section BC, therefore the acceleration is less
[tex]a_{BC}[/tex] > [tex]a_{CD}[/tex]
In summary lower accelerations are
AB = DE <CD <BC
An object has mass 4 kg. What is its weight (in newton) on earth?
Answer:
Should be -39.2 N
Explanation:
w=mg
w=4 x -9.8 m/s2
= -39.2 N
Calculate the escape velocity
the moon's surface given that a man on the moon has 1/6 his weight on earth
Answer:
v = 2.38 × 10³ m/s
Explanation:
Escape velocity, v = √(2gR) where g = acceleration due to gravity on planet and R = radius of planet.
Since it is given that the weight of the man on the moon is 1/6 his weight on earth, and g' = acceleration due to gravity on moon and g = acceleration due to gravity on earth and m = mass of man,
mg' = mg/6
g' = g/6
Since g = 9.8 m/s²,
g'= 9.8 m/s² ÷ 6
g' = 1.63 m/s²
The escape velocity of the moon is thus v = √(2g'R) where R = radius of moon = 1.737 × 10⁶ m.
Substituting these into v, we have
v = √(2g'R)
v = √(2 × 1.63 m/s² × 1.737 × 10⁶ m)
v = √[5.663 × 10⁶ (m/s)²]
v = 2.38 × 10³ m/s
Sugar crystals enter a dryer at the rate of 1000 kg h-1 and at 20% w.b. moisture content. They leave the dryer at 3% w.b. moisture content. If the drying process requires 3000 kJ kg-1 of water removed, estimate the amount of heat required per hour and the rate of dry crystals out of the dryer.
Answer:
The dryer evaporate 200 - 24.74 kg of water per hour
To remove 1kg of water it need 3000 K J
So to remove, 175.26 Kg
it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.
Explanation:
In one hour, the amount of sugar entering = 1000 kg
w.b moisture content is defined as,
weight of water / weight of water + weight of dry
[tex]W_{w}[/tex]/[tex]W_{w}[/tex] + [tex]W_{d}[/tex] x 100
[tex]W_{w}[/tex] + [tex]W_{d}[/tex] = 1000 kg when entering
it has 20% moisture content when entering
[tex]W_{w}[/tex] = 0.2 x 1000 = 200 kg
when leaving it has 3% moisture content then weight of dry material
[tex]W_{d}[/tex] = 1000 - 200 = 800 Kg
[tex]\frac{W_{w}^{'} }{W_{w}^{'} + W_{d}^{'} }[/tex] = 0.03
[tex]\frac{W_{w}^{'} }{W_{w}^{'} + 800 }[/tex] = 0.03
[tex]W_{w} ^{'}[/tex] = 0.03 x [tex]W_{w} ^{'}[/tex] + 0.03 x 800
[tex]W_{w} ^{'}[/tex] = 24.74 kg
When leaving the dryer, the crystals has total weight of the water = 24.74 kg per hour.
The dryer evaporate 200 - 24.74 kg of water per hour
To remove 1kg of water it need 3000 K J
So to remove, 175.26 Kg
it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.
A(n) ____ is just an abstract description of what an object will be like if any objects are ever actually instantiated.
Answer:
Class
Explanation:
A class is just an abstract description of what an object will be like if any objects are ever actually instantiated.
Its purpose is to provide an appropriate superclass out of which other classes can confirm take from and even go ahead to share a mutual and particular design. In many other languages, the class name is often used as the name for the template itself, it can also be used as the name for the default constructor of the class too.
A crane raises a 12,000 N marble sculpture at a constant velocity onto a pedestal 1.5 m above the ground outside an art museum. How much work is done by the crane
For the work, applicate formula:
[tex]\boxed{\boxed{\green{\bf{W = F\times d}}}}[/tex]
According our data:
Replacing:W = 12000 N * 1,5 m
Resolving:W = 18000 J
The work done is 18000 Joules.
You heat up a 3.0 kg aluminum pot with 1.5 kg of water from 5 to 90o Celsius. The specific heat capacity of Aluminum is 900 J/kgK. (a) How much energy did it take to heat the pot of water
Answer:
765,000Joules or 765kJ
Explanation:
The Quantity of heat required is expressed as;
Q = (mcΔt)al + (mcΔt)water
m is the mass
c is specific heat capacity
Δt is the change in temperature
Q = (3(900)(90-5)) + (1.5(4200)(90-5))
Q = 2700*85 + 6300*85
Q = (2700+6300)85
Q = 9000*85
Q = 765,000
Hence the amount of energy needed is 765,000Joules or 765kJ
A pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the roof of a building, as shown. The rope pulls horizontally on the lower crate, and the coefficient of static friction between the lower crate and the roof is 0.666. What is the weight of the heaviest pallet of bricks that can be supported this way?
A. 400 lb
B. 350 lb
C. 250 lb
D. 266 lb
Answer:
C.250 lb is the answer...A person holds a 4.0 kg block at position A shown above on the left, in contact with an uncompressed vertical spring with a spring constant of 500 N/m. The person gently lowers the block from rest at position A to rest at position B. Which of the following describes the change in the energy of the block-spring-Earth system as a result of the block being lowered?
Answer:
Energy decrease by approximately 1.5J
The change in the energy of the block-spring-Earth system as a result of the block being lowered will be 1.424 J.
Given information;
A person holds a 4.0 kg block at position A in contact with an uncompressed vertical spring with a spring constant, k, of 500 N/m.
The person gently lowers the block from rest at position A to rest at position B. The decrease in height, h, of the object or the compression in spring, x, will be 10 cm.
Due to the decrease in height of the block, the decrease in potential energy will be,
[tex]PE_b=-mgh\\=-4\times9.81\times0.1\\=-3.924\rm\;N[/tex]
Due to compression, the increase in potential energy of the spring will be,
[tex]PE_s=\dfrac{1}{2}kx^2\\=\dfrac{1}{2}500\times(0.1)^2\\=2.5\rm\;J[/tex]
So, the net change in the energy of the system will be,
[tex]E=-3.924+2.5\\=1.424\rm\;J[/tex]
Therefore, the change in the energy of the block-spring-Earth system as a result of the block being lowered will be 1.424 J.
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A powered winch is used to pull a sailboat to shore. The winch uses a 900 W motor.
If the motor is used for 30 s, how much work does it do? (Power: P = W/t)
0.03 J
30 J
960 J
27,000 J
Answer: 27,000 J :)
Explanation:
If the winch uses a 900 W motor and the motor is used for 30 s to pull a sailboat then work needed to do this work is 27000 J. hence option D is correct
What is Power?Power is the rate of doing work. Power is also defined as work divided by time. i.e. Power = Work ÷ Time. Its SI unit is Watt denoted by letter W. Watt(W) means J/s or J.s-1. Something makes work in less time, it means it has more power. Work is Force times Displacement. Dimension of Power is [M¹ L² T⁻³]
Given,
Power P = 900W
Time = 30s
Work in joule =?
By using formula,
P = Work ÷ Time
Work = Power × time
W= 900 × 30
W= 27000 J
Hence Work of 27000J is needed to pull a sailboat to shore in 30s.
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If Nell was blocking a defensive lineman for the South Grand Prairie Warriors with a force of 126 Newtons and had a mass of 56.2 kg…what would his acceleration be equal to?
Answer:
2.24 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]
m is the mass
f is the force
From the question we have
[tex]a = \frac{126}{56.2} \\ = 2.241992...[/tex]
We have the final answer as
2.24 m/s²Hope this helps you
An object traveling a circular path of radius 5 m at constant speed experiences an acceleration of 3 m/s2. If the radius of its path is increased to 10 m, but its speed remains the same, what is its acceleration?
Answer:
New acceleration = 1.5 m/s²
Explanation:
The acceleration of an object when it moves in a circular path is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
Where
v is speed of an object
r is radius of path
If v remains constant,
[tex]a\propto \dfrac{1}{r}[/tex]
or
[tex]\dfrac{a_1}{a_2}=\dfrac{r_2}{r_1}[/tex]
Put r₁ = 5 m, a₁ = 3m/s², r₂ = 10 m
[tex]a_2=\dfrac{a_1r_1}{r_2}\\\\a_2=\dfrac{3\times 5}{10}\\\\a_2=1.5\ m/s^2[/tex]
Hence, if the radius of its path is increased to 10 m, its acceleration will be 1.5 m/s².
PLEASE HELP!!!!
Suppose a tractor was moving forward at 6.0 m/s. It runs into a pile of hay and is
brought to a stop in 1.1 s. If the tractor had a mass of 2550 kg, with what force did
the pile of hay stop the tractor?
A. -5.5N
B. -14000N
C. 15000N
D. -15000N
E. -0.0021N
F. -0.000072
Answer:
B )-14000N
Explanation:
F=mv-mu
t
F =2550(0)-2550(6)
1.1
F = -13909.09
approximately -14000N
Suppose a tractor was moving forward at 6.0 m/s. It runs into a pile of hay and is brought to a stop in 1.1 s. If the tractor had a mass of 2550 kg, the force did the pile of hay stop the tractor 14000N
What are the types of force ?Force can be a unit of pushing or pulling of any object which result from the object’s interaction or movement, without applying force the objects can not be moved, can be stopped or change the direction.
Force is a quantitative parameter between two physical bodies, means an object and its environment, there are various types of forces in nature.
If an object present in its moving state will be either static or motion, the position of the object will be changed if it is pushed or pulled and The external push or pull upon the object is mainly called as Force.
The contact force that occurs when we apply some effort on an object such as Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force
Non-Contact forces are the type of forces that occur from a distance such as Electromagnetic Force, Gravitational Force, Nuclear Force
F=mv-mu
t
F =2550(0)-2550(6)
1.1
F = -13909.09
approximately -14000N
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This is science btw please help me
Answer:
c
Explanation:
An athlete swings a 6.90-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.830 m at an angular speed of 0.680 rev/s.
(a) What is the tangential speed of the ball?
m/s
(b) What is its centripetal acceleration?
m/s2
(c) If the maximum tension the rope can withstand before breaking is 110 N, what is the maximum tangential speed the ball can have?
m/s
Answer:
a) 0.5644 m/s
b) ~0.384 m/s^2
c) ~3.638 m/s
Explanation:
a) Tangential speed is found be the radius*rotational speed, so it is 0.83*0.68 = 0.5644 m/s
b) Centripetal acceleration is found by v^2/r, so it is (0.5644^2)/0.83 = ~0.384 m/s^2
c) Let the tangential speed be v. The maximum centripetal force 110 N (as given). Centripetal force = mass*centripetal acceleration = mass*v^2/r (because centripetal acceleration is found by v^2/r). Inputting the values from the problem and solving for v, we get:
110 = 6.9*v^2/0.83
v = sqrt(110*0.83/6.9) = ~3.638 m/s
I hope this helps! :)
Do field forces exist in nature?
Answer:
Yes, field forces exist in nature.
Explanation:
A field force is a force experience due to an interaction with fields. In this case, a contact is not required before the force can be felt.
The three major field forces are: magnetic field, electric field and gravitational field. The magnetic fields are produced due to the interaction between the north and south poles of a magnet, the electric field is one from charges, while gravitational field is a force of attraction due to gravity on the earth. All these fields occur in nature, therefore making field forces to exist in nature.
A 1.40 kg block is attached to a spring with spring constant 16.5 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 45.0 cm/s . What are
Answer:
A = 0.13 m
Explanation:
Given that,
Mass of a block, m = 1.4 kg
Spring constant of the spring, k = 16.5 N/m
While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 45.0 cm/s or 0.45 m/s
We need to find the amplitude of the subsequent oscillations.
We can use the conservation of energy here. Initially, the kinetic energy of the block is maximum and then it gets converted to the potential energy of the spring.
Mathematically,
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2[/tex]
A is the amplitude of subsequent oscillations.
[tex]A=\sqrt{\dfrac{mv^2}{k}} \\\\A=\sqrt{\dfrac{1.4\times (0.45)^2}{16.5}} \\\\A=0.13\ m[/tex]
So, the amplitude of subsequent oscillations is 0.13 m.
baseball player hits a line drive estimated to have traveled 145 meters the ball leaves the bat with a horizontal velocity of 40.0 m/s . how much time was the ball in the air
Answer:
5800
Explanation:
caluculatior
The ball was in the air for 3.625 second.
What is velocity?Velocity is the direction at which an object is moving and serves as a measure of the rate at which its position is changing as seen from a specific point of view and as measured by a specific unit of time (for example, 60 km/h northbound).
In kinematics, the area of classical mechanics that deals with the motion of bodies, velocity is a fundamental idea.
A physical vector quantity called velocity must have both a magnitude and a direction in order to be defined.
Given that: the a horizontal velocity of the ball: v = 40.0 m/s .
The ball traveled 145 meters.
Hence, the time during which the ball is in the air:
= distance travelled/ horizontal velocity of the ball
= 145 meter / 40.0 m/s .
= 3.625 second.
Learn more about velocity here:
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