Answer:
13.3
Explanation
work= force/area
2000/150
Suppose you are standing on a scale in an elevator that is accelerating upward. Will the scale read your weight as larger for smaller than the weight it reads when you are stationary?
Answer:
following are the solution to this question:
Explanation:
When I stand at such a scale in an elevated that's already rising upwards, its scale would appear to also be 0 because of free fall and would often reveal that weight whenever the lift is stable.
In this, the free fall is also known as the object, that is influenced exclusively by gravity, and an object operating only through the influence of gravity is said to be in a free-fall state.
A 495-kg dragster accelerates from rest to a final speed of 105 m/s in 395 m, during which it encounters an average frictional force of 1400 N. show answer No Attempt 50% Part (a) What is its average power output of the dragster, in watts, if this takes 8.2 s
Answer:
Explanation:
According to energy conservation which states that the workdone is equal to change in the system
Workdone = change in kinetic energy + (frictional force * distance)
Workdone = ΔK + fd
Workdone = kf-Ki + fd
Workdone = = 1/2(m(v-u)^2) + fd
Given
Mass m = 495kg
final velocity v = 105m/s
initial velocity = 0m/s
Force f= 1400N
distance d = 395m
Substitute
Workdone = 1/2(495(105-0)^2) + 1400(395)
Workdone = 2,728,687.5+553000
Workdone = 3,281,687.5 Joules
Time = 8.2secs
Power output = Workdone/Time
Power output = 3,281,687.5/8.2
Power output = 885,766.768
Power output = 8.858 * 10^5 watts
You release a ball from rest at the top of a ramp. 6 s later it is moving at 4.0
m/s. What is the acceleration? (in meters per second squared) *
Your answer
[tex]a = \frac{vf - vi}{t} [/tex]
here initial velocity vi=0 as ball release from rest
the final velocity is vf=4.0
time is t=6
so putting all these values in above equation
[tex]a = \frac{ 4.0- 0}{6} [/tex]
[tex]a = 0.6667m \s {}^{2} [/tex]
A Carnot engine operates with a cold reservoir at a temperature of 455 K and a hot reservoir at a temperature of 619 K. What is the net entropy change (in J/K) as it goes through a complete cycle? Round your answer to the nearest whole number.
Answer:
Assuming that 4000 J of heat transfer occurs from it the change in entropy will be=2.33j/k
Explanation:
Given data
Th=619 K
Tc= 455K
We are going to assume that 4000 J of heat transfer occurs from it, since it was not specified in the question.
we know that the change in entropy is given as
ΔStotal=ΔShot+ΔScold .
ΔShot=−Q/Th=−4000J/619K=−6.46J/K
For the cold reservoir,
ΔScold=Q/Tc=4000J/455K=8.79J/K
therefore the total is
ΔStotal=ΔShot+ΔScold
=(−6.46+8.79)J/K
=2.33j/k
Match the variables with its corresponding unit
Answer:
1. s
2. m
3. start m/s
4. end m/s
5. m/s^2
Explanation:
1. t is the suffix for the time and it is related to seconds [s]
2. d is the suffix for distance and it is related to meters [m]
3. Vi is the suffix for the initial velocity and it is related to meters per second [m/s]
4. Vf is the suffix for the final velocity and it is related to meters per second [m/s]
5. a is the suffix for the acceleration and it is related to meters per squared second [m/s^2]
The elements in Groups 3 through 12 of the periodic table are the ______.
Answer:
Transition Metals
Explanation:
__________________________
A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.64 s. What was the average acceleration during this run?
Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
HELP URGENTTTTTTTTTTTTT
Answer:
Gravitational Potential Energy
B Gravitational potential energy
Shoot the cannon velocity of 25 m/s and observe the velocity vector on the simulator and make a statement about your observation . In other words what is happening to the projectile as it travel its parabolic trajectory ?
Answer:
Its speed remains the same but its velocity keeps chaging (in direction)
Explanation:
how much ume does it take to use 300 W of power to do 1800 J of Work?
Answer:
P=300 W
W=1800 J
P=W/t
t=W/P
t=6 s
If you borrow your mom's car to drive 128 m to the store, but you go cruising
instead and drive a total distance of 1256 m in 3600s. What was your speed during the
trip? Round answer to 2 significant figures.
Answer:
v = 0.349 m/s
Explanation:
If you borrow your mom's car to drive 128 m to the store, but you go cruising instead and drive a total distance of 1256 m in 3600s.
We need to find your speed during the trip. It is equal to the total distance divided by time taken.
He has done cruising instead and drive a total distance of 1256 m.
So,
[tex]v=\dfrac{d}{t}\\\\v=\dfrac{1256\ m}{3600\ s}\\\\v=0.349\ m/s[/tex]
So, the speed of the person during the trip is 0.349 m/s.
The earth has a mass of, MEarth=6.4 X 1024 kg and a radius of REarth= 6.0 X 106meter and it has acceleration due to gravity, gE = 9.8 . One of Jupiter’s moon also has MJmoon = ½ * MEarth =3.2 x 1024 kg with a radius of, RJmoon = ½ * REarth =3.0 x 106 m and gJmoon value unknown. From the information, what is acceleration due to gravity of Jupiter’s moon, gjmoon, comparing to the earth’s acceleration due to gravity, gEarth?
Answer:
gjmoon = 19.6 m/s2
Explanation:
Applying the Universal Law of Gravitation, the force exerted by the Earth on a mass m, can be written as follows:[tex]F_{gE} = m* \frac{G*M_{E} }{R_{E} ^{2} } = m*g_{E}[/tex]
In the same way, the force due to gravity of Jupiter's moon, can be written as follows:[tex]F_{gjmoon} = m* \frac{G*M_{Jmoon} }{R_{Jmoon} ^{2} } = m*g_{jmoon} (1)[/tex]
Now, we know the following:[tex]M_{Jmoon} = \frac{M_{E} }{2} (2)[/tex]
[tex]R_{Jmoon} = \frac{R_{E} }{2} (3)[/tex]
Replacing (2) and (3) in (1), we get:[tex]F_{gjmoon} = m* \frac{G*M_{E} }{2*\frac{R_{E} ^{2}}{4} } = m*g_{jmoon} (1) = m* 2* \frac{G*M_{E} }{R_{E} ^{2}} = 2* m*g_{E}[/tex]
⇒ [tex]g_{jmoon} = 2* g_{E}[/tex]
gjmoon = 19.6 m/s2 (in magnitude).
Which part of the brain plays a vital role in sensory information processing?
Occipital Lobe
Cerebral Cortex
Pons
Parietal Lobe
Answer:
Parietal Lobe
Explanation:
What measurement do geologist use to find the absolute age of once living things
Answer:
Radiocarbon dating
Explanation:
Find the distance along an arc on the surface of the earth that subtends a central angle of 1 minutes (1 minute = 1/60 degree). The radius of the earth is 3960 miles. Round to the thousandths. (3 decimal places)
Answer:
1.152 miles
Explanation:
Given: central angle = 1 minute = [tex](\frac{1}{60}) ^{o}[/tex]
radius of the earth = 3960 miles
The length of an arc = [tex]\frac{\alpha }{360^{o} }[/tex] 2[tex]\pi[/tex]r
where: [tex]\alpha[/tex] is the central angle, and r is the radius.
Thus,
Distance along the arc = [tex]\frac{\alpha }{360^{o} }[/tex] 2[tex]\pi[/tex]r
Distance along the arc = [tex]\frac{(\frac{1}{60}) ^{o} }{360^{o} }[/tex] x 2 x [tex]\frac{22}{7}[/tex] x 3960
= [tex]\frac{(\frac{1}{60}) ^{o} }{360^{o} }[/tex] x 24891.4286
= 1.1524
The required distance along an arc is 1.152 miles.
my mom is dum and anoying asl, what should i do?
Answer:
stop being an annoying child and appreciate everything your mother does for you because she cares for you
Explanation:
its the right thing to do
Be thankful for everything your mom does bc one day she wont be here any more! Love her while you can
Vertical acceleration of the stone thrown upwards at the top of the trajectory (where its velocity vanishes) is equal to
A. g
B. g/2
C. 0
D. nine of the above
Answer:
A. g
Explanation:
A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside the solenoid near its center
Answer:
The value is [tex]B = 0.0452 \ T [/tex]
Explanation:
From the question we are told that
The number of turns is N = 1000
The length is L = 50 cm = 0.50 m
The radius is r = 2.0 cm = 0.02 m
The current is I = 18.0 A
Generally the magnetic field is mathematically represented as
[tex]B = \mu_o * \frac{N }{L} * I[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = 4\pi * 10^{-7} * \frac{1000}{0.50} * 18.0[/tex]
=> [tex]B = 0.0452 \ T [/tex]
Question #4
4. Anthony walks to the pizza place for lunch. He walk 4 km
east, he realized he passed it and then walked 1 km west.
What distance did he cover? What was his displacement?
Answer:
Distance 5 km, Displacement 3 km east
Explanation:
The distance covered by Anthony is 5 km, while his displacement is 3 km.
Distance and displacement:The distance is defined as the total length of space covered during motion between the starting point and end point, not necessarily a straight line. Whereas displacement is defined as the minimum distance between two points in space, that is a straight line.
Suppose, if you start from one point and walk for 100 meters then come back at the same point that you started your walk, you would have traveled 2 times 100 meters that is 200 meters. But your displacement will be zero because your starting point and the end point are the same.
The given question clarifies the difference between distance and displacement:
Since Anthony walks 4 km east and then walks 1 km west, the distance covered by him is:
distance = 4 + 1 = 5 km
but as he walks west, he comes closer to the starting point, and his displacement is the distance between the starting point and the end point.
displacement = 4 - 1 = 3 km
Referto the figure below for more information.
Learn more about distance and displacement:
https://brainly.com/question/24522401?referrer=searchResults
Colored lights are called additive colors. Why do you think this is so?
Lab report on motion 6th grade
Answer:
I'm in 10th grade
Explanation:
If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s than the recoil speed of the cannon will be *
A. 0.2 m/s
B. 2 m/s
C. 4 m/s
D. 10 m/s
Answer:
D. 10 m/s
Explanation:
PLEASE HELP WILL MARK BRAINLIEST
Answer:
also choose D and E i belive those are also correct
8. What is the mass of an object if a force of 34 N produces an acceleration of 4 m/s/s?
F = 34 N
F
34
m-
m-
m = 8.5 kg
a
4
a - 4 m/s
Answer:
The answer is 8.5 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
where
f is the force
a is the acceleration
So we have
[tex]m = \frac{34}{4} = \frac{17}{2} \\ [/tex]
We have the final answer as
8.5 kgHope this helps you
What do all elements in a column in the periodic table have in common? (2 points)
Oa
Their atoms have the same number of valence electrons.
Ob
Their atoms have the same atomic mass.
Their atoms have the same number of protons and electrons.
Od
Their atoms have the same number of neutrons.
Answer:
You mean in periods.
They have the same number of elctron shells.
But if they are in the same group then they have the sane number oifValence electrons.
Explanation:
You are studying circular motion by placing pennies on a turntable and then turning it on so that it will spin. You keep increasing the speed until one of the pennies slips off. You repeat this procedure and observe that the penny close to the outer edge always slip off first. What is the best inference?
1.The penny near the edge has a greater tangential velocity than the one in the center, so it experiences more air resistance. It’s the effect of the air “blowing” it off.
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
3.The centrifugal force acting on the pennies is stronger on the one near the edge than the one near the center.
Answer:
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
Explanation:
centripetal force = m ω² R
here m is mass , ω is angular velocity and R is distance of penny from centre
So this force depends upon R
penny on the outer edge will require greater centripetal force to move in circular path .
The centripetal force will be provided by frictional force of table which is same for both the coin . Hence the penny on the outer edge will slip off first the moment , frictional force reach its maximum value for it . But it will be sufficient to keep in balance the penny nearer to the centre .
An airline employee tosses two suitcases in rapid succession with a horizontal velocity of 7.2 ft/s onto a 50-lb baggage carrier which is initially at rest. Problem 14.003.a Conservation of momentum: two colliding suitcases Knowing that the final velocity of the baggage carrier is 4.8 ft/s and that the first suitcase the employee tosses onto the carrier has a weight of 30 lb, determine the weight of the other suitcase. (You must provide an answer before moving on to the next part.) The weight of the other suitcase is lb.
Answer:
m₁ = 70 lb
Explanation:
Here we will use the law of conservation of momentum:
m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃
where,
m₁ = mass of first suitcase = ?
m₂ = mass of second suitcase = 30 lb
m₃ = mass of baggage carrier = 50 lb
u₁ = initial speed of first suitcase = 7.2 ft/s
u₂ = initial speed of second suitcase = 7.2 ft/s
u₃ = initial speed of baggage carrier = 0 ft/s
v₁ = Final speed of first suitcase = 4.8 ft/s
v₂ = Final speed of second suitcase = 4.8 ft/s
v₃ = Final speed of baggage carrier = 4.8 ft/s
because after collision all three will have same speed
Therefore,
(m₁)(7.2 ft/s) + (30 lb)(7.2 ft/s) + (50 lb)(0 ft/s) = (m₁)(4.8 ft/s) + (30 lb)(4.8 ft/s) + (50 lb)(4.8 ft/s)
(m₁)(7.2 ft/s) + (216 lb ft/s) + (0 lb ft/s) = (m₁)(4.8 ft/s) + (144 lb ft/s) + (240 lb ft/s)
(m₁)(7.2 ft/s) - (m₁)(4.8 ft/s) = 168 lb ft/s
m₁ = (168 lb ft/s)/(2.4 ft/s)
m₁ = 70 lb
What is psychoanalysis
What is the relation between the weight of a body and acceleration due to gravity?
Answer:
The weight W, or gravitational force, is then just the mass of an object times the gravitational acceleration. Since the gravitational constant (g) depends on the square of the distance from the center of the earth, the weight of an object decreases with altitude.
Explanation:
please mark as brainlist
You and some friends visit the Florida State Fair and decide to play a game. To play the game you must slide a metal hockey-type puck up a wooden ramp so that it drops through a hole at the top of the ramp. Your prize, if you win, is a large, stuffed gorilla. You realize the secret to winning is giving the puck just enough velocity at the bottom of the ramp to make it to the hole. You estimate the distance from the bottom of the ramp to the hole at about 3 m, and the ramp appears to be inclined with an angle of 10o from the horizontal. You just got out of physics class and recall the coefficient of static friction between steel and wood is 0.1 and the coefficient of kinetic friction between steel and wood is 0.08. The mass of the puck is about 1 kg. You decide to impress your friends by sliding the puck at the precise speed on the first try so as to land it in the hole. You slide the puck at 2.0 m/sec. Do you win the stuffed poodle? Let g = 10 m/s2. Note: A slightly too hard throw will win since your aim is so good the puck would drop in the hole.
Answer:
No you didn't win the stuffed poodle
Explanation:
From the question we are told that
The distance of the ramp to the hole is d = 3 m \
The angle of inclination is [tex]\theta = 10^o[/tex]
The coefficient of static friction is [tex]\mu_s = 0.1[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.08[/tex]
The mass of the puck is m = 1 kg
The velocity of the first slide is [tex]v_1 = 2.0 m/s[/tex]
Generally the kinetic energy at the bottom of the ramp is equal the energy loss due to friction and this can be mathematically represented as
[tex]\frac{1}{2} m * v^ 2 = \mu_k * [m * g] * cos (theta ) * d[/tex]
=> [tex]\frac{1}{2} * v^ 2 = 0.08 * 9.8 * cos (10 ) * 3[/tex]
=> [tex] v= 2.17 \ m/s [/tex]
Comparing the relative velocity obtained and the velocity of your first throw we can see that you didn't win the stuffed animal
