Answer:
The equation to calculate the final velocity of the object.---
Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.
Explanation:
Also just so you know this isn't my answer I got it from here https://brainly.com/question/20016151 from im2slowwwww
A sinusoidal wave with wavelength 0.500 m travels along a string. The maximum transverse speed of a point on the string is. 4.00 m/s and the maximum transverse acceleration is 1.00 x 105 m/s2. What is the propagation speed of the wave
Answer:
The velocity [tex]v = 1989.2 \ m/s[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 0.500 \ m[/tex]
The maximum transverse speed is [tex]v = 4.0 \ m/s[/tex]
The maximum transverse acceleration is [tex]a = 1.00 *10^{5} \ m/s^2[/tex]
Generally the frequency of the wave is mathematically represented as
[tex]f = \frac{w}{2 \pi }[/tex]
Here w is the angular speed which is mathematically evaluated as
[tex]w = \frac{a}{v}[/tex]
=> [tex]w = \frac{1.00 *10^{5}}{4}[/tex]
=> [tex]w = 25000 \ rad/sec[/tex]
So
[tex]f = \frac{ 25000 }{2 * 3.142 }[/tex]
=> [tex]f = \frac{ 25000 }{2 * 3.142 }[/tex]
=> [tex]f = 3978.4 \ Hz[/tex]
Gnerally the propagation speed of the wave is mathematically represented as
[tex]v = f * \lambda[/tex]
=> [tex]v = 3978.4 * 0.500[/tex]
=> [tex]v = 1989.2 \ m/s[/tex]
An object has an average acceleration of + 6.24 m/s ^ 2 for 0.300 s . At the end of this time the object's velocity is + 9.31 m/s . What was the object's initial velocity?
Answer:
initial velocity = 7.44 m/s (3 s.f.)
Explanation:
a = 6.24 m/s² t = 0.300 s v = 9.31 m/s u = ?
v = u + at
9.31 = u + (6.24 x 0.300)
9.31 = u + 1.872
u = 7.438
= 7.44 m/s (3 s.f.)
Hope this helps!
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.
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I’m confused help me
Answer:
you select what ever feels right to you if you think what it tells you is true then you would say strongly agree and if you don't agree then it would be any of the other options
Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline. Which has the larger speed at the bottom?
The question incomplete, the complete question is;
Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline.
(A)the one falling vertically
(B)the one on the incline
(C)Both have the same speed.
(D)cannot be determined
Answer:
(C)Both have the same speed.
Explanation:
When we consider the question closely, we will discover that an object falling down a frictionless incline is comparable to an object falling freely under gravity.
In both instances, the acceleration of objects is just the same irrespective of mass.
Hence, the object falling vertically and the object sliding down a frictionless plane will have the same speed at the bottom.
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest and rolls without slipping.
Height is 7.05 m and not 7.05 mm
Answer:
9.603 m/s
Explanation:
We are dealing with rotation, so velocity of centre of mass is given by;
v_cm = Rω
Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²
Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).
Using conservation of energy, we have:
P.E = K.E_t + K.E_r
Formula for rotational and kinetic energy here are;
K.E_t = ½mv²
K.E_r = ½Iω²
mgh = ½mv² + ½Iω²
Since we want to find translational speed(v), let's get rid of ω.
Earlier, we saw that v_cm = Rω
Thus; ω = v/R
Also, we know that I = ½mR².
Thus;
mgh = ½mv² + ½(½mR²)(v/R)²
This gives;
mgh = ½mv² + ¼mv²
Divide through by m to get;
gh = v²(½ + ¼)
gh = ¾v²
Making v the subject gives;
v = √(4gh/3)
v = √((4 × 9.81 × 7.05)/3)
v = 9.603 m/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 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
A stone is dropped into a deep well and is heard to hit the water 5 s after being dropped. Determine the depth of the well.
Answer:
122.5 m
Explanation:
From the question given above, the following data were obtained:
Time (t) = 5 s
Depth (h) =?
The depth of the well can be obtained as:
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
h = ½gt²
h = ½ × 9.8 × 5²
h = 4.9 × 25
h = 122.5 m
Thus, the depth of the well is 122.5 m
A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release. What is the magnitude of its velocity just before it strikes the ground?
Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately
Based on the three-cell model of global circulation, surface winds diverge in the vicinity of:________.
a. the polar front
b. subpolar lows
c. the equator
d. the north and south poles
Answer:
a. the polar front
Explanation:
When the surface winds move towards the pole at angles that range from 30 to 60 degrees, they usually collide with cold air moving towards the equator. Since these two winds do not mix with ease, the polar front helps to separate them.
The three-cell model assumes that the earth is entirely filled with water thus removing any interactions with the land. It also assumes that there are no seasons nor Coriolis force.
What is the potential energy of a spring that is compressed 0.65 m by a 25 kg block if the spring constant is 95 N/m?
A. 1.6J
B. 7.9J
C. 15J
D. 20J
Answer:
D. 20J
Explanation:
Answer:
20 JExplanation:
yes
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
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
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! :)
Fon of the assignment will be graded by your teacher Question: In a lab experiment, A square of black paper and a square of white paper are place directly under 100-watt lamps for 30 minutes. Predict what happens to the temperature of each paper during the experiment.
Answer: OVERVIEW
Students will use black and white construction paper and a light source to learn that dark objects
absorb more light and reflect less light than bright objects. The activity also demonstrates the conversion
of radiant light energy into heat energy.
CONCEPTS
• Dark surfaces absorb more visible light energy than bright surfaces
• Dark surfaces reflect less visible light energy than bright surfaces
• Energy can change forms, in this case from radiant light energy to heat
• Clouds, being bright, reflect significant amounts of sunlight and help to regulate Earth’s temperature
MATERIALS
• 2 thermometers
• Flood lamp, desk lamp, or area in direct sunlight
• Ruler
• Construction paper, 1 piece white, 1 piece black, or 2 sheets photocopy paper
• Scissors
• Cellophane tape or rubber bands
• 2 empty metal food cans, same size (be sure rims are not jagged)
PREPARATION
The paper and the cans can be prepared beforehand or prepared as part of the activity (see Procedure).
Although two cans with their tops completely removed can be used, the experiment will be more effective
(have fewer external effects), if only holes are placed in the cans’ lids, e.g., two holes from a bottle opener
to empty material out of the can, and one center hole created with an awl for the thermometer. Only one
hole is actually needed for the experiment - for the thermometer. Cans can optionally be filled with water,
or this can be done as a separate experiment to demonstrate the higher heat capacity of water compared to
air.
You can either use a flood lamp or a desk lamp (light bulb) to simulate sunlight, as described here, or
you can place the cans on a windowsill (window closed) or other sheltered area in direct sunlight. A flood
lamp will be the most effective option, causing the largest temperature increases.
PROCEDURE
Engagement
Discuss whether dark surfaces (e.g., asphalt) or bright surfaces (e.g., concrete) tend to get hotter in
sunlight. Which would you rather walk on during the day in the summertime? What color are solar cells,
for example those found on some calculators or freeway call boxes?
Explanation: I hope this helps! ∧ ∧
⊂∵→ω←∵⊃
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.
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.
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
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
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.
what is the answer to what does a 150 kg roller coaster traveling at 10m/s has how much kinetic energy
A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exerted by the seats on the passengers is equal to ¼ of their weight.
Answer:
[tex]v=10m/sec[/tex]
Explanation:
From the question we are told that
Radius of vertical r= 8m
Force exerted by passengers is 1/4 of weight
Generally the net force acting on top of the roller coaster is give to be
[tex]F_N+Fg[/tex]
where
[tex]F_N =forceof the normal[/tex]
[tex]Fg= force due to gravity[/tex]
Generally the net force is given to be [tex]FC(force towards center)[/tex]
[tex]F_C =F_N + Fg[/tex]
[tex]F_N =F_C -Fg[/tex]
[tex]F_N=F_C-F_g[/tex]
[tex]F_N=\frac{mv^2}{R} -mg[/tex]
Mathematical we can now derive V
[tex]m_g + \frac{8m}{4}= \frac{mv^2}{8}[/tex]
[tex]\frac{5mg}{4} =\frac{mv^2}{8}[/tex]
[tex]v^2 =\frac{40*10}{4}[/tex]
[tex]v=10m/sec[/tex]
Therefore the speed of the roller coaster is given ton be [tex]v=10m/sec[/tex]
[tex]v=10m/s[/tex]
Given:
Radius of vertical r= 8m
To find:
Force exerted by passengers is 1/4 of weight
Generally the net force acting on top of the roller coaster is give to be
[tex]F_N+F_g[/tex]
where
[tex]F_N=\text{Force of Normal}[/tex] and [tex]F_g=\text{Force due to gravity}[/tex]
Generally the net force is given to be [tex]FC- \text{Force towards centre}[/tex]
[tex]F_C=F_N+F_g\\\\F_N=F_C+F_g\\\\F_N=\frac{mv^2}{r} -mg[/tex]
Mathematically we can now derive V
[tex]mg-\frac{8m}{4} =\frac{mv^2}{8}\\\\\frac{5mg}{4}=\frac{mv^2}{8}\\\\v^2=\frac{40*10}{4} \\\\v=10m/s[/tex]
Therefore, the speed of the roller coaster is given to be 10m/s.
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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.
If an electric iron of 1200W is used for 30minutes every day, find electric energy consumed in the month of November. Also calculate bill amount if the rate of 1 unit = Rs 2
Answer:
Electric energy produced in the month of April is 18 kWh
Explanation:
Given
P = 1200 W = 1.2 kW
Use time per day = 30 minutes = 1 / 2 hrs
We know,
No. of days in April = 30 days
Total time iron being used
T = 1 / 2 * 30 = 15 hrs
We know,
Energy consumed = Pt
Substituting the value of P and T
= 1.2 kW * 15 hr
= 18 kWh
Hence,
Electric energy produced in the month of April is 18 kWh
What is the relationship between the type of energy sublevels present and the
principal energy level?
Do you think antiseptic creams and lotions play an important part in our day
to day life?
Answer:
no, otherwise they just cause a side effect on our lives
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.
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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A hydrogen atom that has lost its electron is moving east in aregion where the magnetic field is directed from south to north. Itwill be deflected:_____________
a. up
b. down
c. not at all
Answer:
a. up
Explanation:
As per the rule of Fleming left hand, the three fingers should be places in a perpendicular manner i.e. mutually also.
The fore finger depicts the field direction
The middle finger depicts the velocity
And, the thumb finger depicts the force direction that experienced on that particle i.e. charged
So the electrons would be deflects to up
Hence, the correct option is a.