Answer:
Explanation:
Fn = m(a + g)
A cart on a horizontal, linear track has a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.22 s to travel a distance of 1.44 m. The mass of the cart plus fan is 350 g. Assume that the cart travels with constant acceleration.
Required:
a. What is the net force exerted on the cart-fan combination?
b. Mass is added to the cart until the total mass of the cart-fan combination is 656 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.63 m now?
Answer:
a
[tex]F = 0.0566 \ N[/tex]
b
[tex]t = 6.147 \ s[/tex]
Explanation:
From the question we are told that
The distance travel in 4.22 s is [tex]s = 1.44 \ m[/tex]
The mass of the cart plus the fan is [tex]m = 350 \ g = 0.35 \ kg[/tex]
Generally from kinematic equation we have that
[tex]s = ut + \frac{1}{2} * a * t^2[/tex]
Here u is the initial velocity with value [tex]u = 0 \ m/s[/tex]
So
[tex]1.44= 0 * t + \frac{1}{2} * a * 4.22^2[/tex]
=> [tex]a = 0.1617 \ m/s^2[/tex]
Generally the net force is
[tex]F = m * a[/tex]
=> [tex]F = 0.35 * 0.1617[/tex]
=> [tex]F = 0.0566 \ N[/tex]
Gnerally the new mass of the cart plus the fan is [tex]M = 656 \ g = 0.656 \ kg[/tex]
The distance considered is [tex]s_1 = 1.63 \ m[/tex]
Generally the new acceleration of the cart is mathematically represented as
[tex]F = M * a_1[/tex]
=> [tex]a_1 = \frac{F}{M}[/tex]
=> [tex]a_1 = \frac{0.0566}{0.656}[/tex]
=> [tex]a_1 = 0.08628 \ m/s^2[/tex]
Gnerally from kinematic equation we have
[tex]s = ut + \frac{1}{2} * a_1 * t ^2[/tex]
Here u is the initial velocity and the value is zero because it started from rest
=> [tex]1.63 = 0 * t + \frac{1}{2} * 0.08628* t ^2[/tex]
=> [tex]t = 6.147 \ s[/tex]
A 15.5 kg box is pushed across the lunch table. The acceleration of the box is 24.2 m/S. What is the net force applied to the box?
Answer:
375.1 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 15.5 × 24.2
We have the final answer as
375.1 NHope this helps you
If a 0.750M solution exerts an osmotic pressure of 22.5atm, what must be the temperature (in Kelvin) of the solution
Answer:
T = 365.58 K
Explanation:
Given that,
The concentration of solution, C = 0.750M
Osmotic pressure, P = 22.5 atm
We need to find the temperature of the solution.
The formula for the osmotic pressure is given by :
[tex]P=CRT[/tex]
Where
R is gas constant, [tex]R=0.08206\ L\ atm/mol-K[/tex]
[tex]T=\dfrac{P}{CR}\\\\=\dfrac{22.5}{0.75\times 0.08206}\\\\=365.58\ K[/tex]
So, the temperature of the solution is 365.58 K.
3.A bridge usually has expansion joints. They allow the bridge to become slightly longer when it
experiences thermal expansion. Look at the diagram below of the bridge joint. When the weather
becomes cool, the "teeth" of the joint move away from each other. When the weather becomes
warm, they move toward each other. Which statement is true about the particles that make up the
bridge?
A.When the sides of the joint are close together, the particles have more kinetic energy
than they do when the sides are farther apart.
B.When the sides of the joint are far apart, the particles have more kinetic energy than
they do when the sides are closer together.
C.The particles contain the same amount of kinetic energy no matter how much the
bridge has expanded.
D. The kinetic energy of the particles changes, but the amount it changes does not
depend on the temperature of the bridge.
Answer:
a) When the sides of the joint are close together, the particles have more kinetic energy than they do when sides are farther apart.
Explanation:
At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is frictionless?
Answer:
The road bank angle is 16.38⁰.
Explanation:
radius of curvature of the road, r = 50 m
allowable speed of car on the road, v = 12 m/s
The bank angle is calculated as;
[tex]\theta = tan^{-1} (\frac{v^2}{gr} )[/tex]
where;
θ is the road bank angle
g is acceleration due to gravity = 9.8 m/s²
[tex]\theta = tan^{-1} (\frac{v^2}{gr} )\\\\\theta = tan^{-1} (\frac{12^2}{9.8 \times 50} )\\\\\theta = tan^{-1} ( 0.2939)\\\\\theta = 16.38 ^0[/tex]
Therefore, the road bank angle is 16.38⁰.
A 5-kg block slides down a plane inclined at 30 0 to the horizontal. Find
a. The acceleration of the block if the plane is frictionless.
b. The acceleration if the coefficient of kinetic friction is 3 2 .
Answer:
a) 4.9m/s²
b) 2.18m/s²
Explanation:
a) According to Newton's second law of motion
\sum Fx = ma
Fm-Ff = ma
Fm is the moving force = Wsin theta
Ff is the frictional force = 0N (frictionless plane)
m is the mass
a is the acceleration
Substituting into the formula
Fm -Ff = ma.
Wsintheta = ma
mgsintheta = ma
gsintheta = a
a = 9.8sin30°
a = 9.8(0.5)
a = 4.9m/s²
Hence the acceleration of the block if the plane is frictionless is 4.9m/s²
b) Let the coefficient if friction given be 0.32
From the formula
Fm-Ff = ma
mgsintheta - nmgcostheta = ma
gsintheta - ngcostheta = a
9.8(sin30)-0.32(9.8)cos30 = a
4.9-2.72 = a
a = 2.18m/s²
Hence the acceleration if the coefficient of kinetic friction is 0.32 is 2.18m/s²
What is the maximum speed that a 2200 kg car can go around a level circular track with a radius of 30.0 m without slipping if the coefficient of static friction between the tires and the road is 0.900?
Answer:
The maximum speed of turn on the given circular track is 16.27 m/s.
Explanation:
Given;
mass of the car, m = 2200 kg
radius of the track, r= 30 m
coefficient of static friction between the tires and the road, μ = 0.9
The net vertical force on the car = N = mg
The net horizontal force on the car = Centripetal force
The coefficient of static friction is given as;
[tex]\mu = \frac{F_c}{N} \\\\[/tex]
[tex]F_c = \mu N\\\\\frac{mv^2}{r} = \mu mg\\\\ \frac{v^2}{r}= \mu g\\\\v^2 = \mu gr\\\\v = \sqrt{\mu gr}[/tex]
where;
v is the maximum speed of turn
[tex]v = \sqrt{\mu gr} \\\\v = \sqrt{0.9 \times 9.8 \times 30 } \\\\v = 16.27 \ m/s[/tex]
Therefore, the maximum speed of turn on the given circular track is 16.27 m/s.
A force of 100 newtons Is applled to a box at an angle of 36° with the horizontal. If the mass of the box Is 25 kilograms, what Is the horizontal
POR FAVOR AYUDENME A COMLETARLO
Answer:
English please?
Explanation:
A 260-kg glider is being pulled by a 1,940-kg jet along a horizontal runway with an acceleration of a= 2.20 m/s^2 to the right. Find the following:
a. the magnitude of the thrust provided by the jet's engines.
b. the magnitude of the tension in the cable connecting the jet and glider
Answer:
a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.
b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.
Explanation:
a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:
Jet
[tex]\Sigma F = F - T = m_{J}\cdot a[/tex] (1)
Glider
[tex]\Sigma F = T = m_{G}\cdot a[/tex] (2)
Where:
[tex]F[/tex] - Thrust of jet engines, measured in newtons.
[tex]T[/tex] - Tension in the cable connecting the jet and glider, measured in newtons.
[tex]m_{G}[/tex], [tex]m_{J}[/tex] - Masses of the glider and the jet, measured in kilograms.
[tex]a[/tex] - Acceleration of the glider-jet system, measured in meters per square second.
If we know that [tex]m_{G} = 260\,kg[/tex], [tex]m_{J} = 1,940\,kg[/tex] and [tex]a = 2.20\,\frac{m}{s^{2}}[/tex], then the solution of this system of equations:
By (2):
[tex]T = (260\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)[/tex]
[tex]T = 572\,N[/tex]
By (1):
[tex]F = T+m_{J}\cdot a[/tex]
[tex]F = 572\,N+(1,940\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 4840\,N[/tex]
The magnitude of the thrust provided by the jet's engines is 4840 newtons.
b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.
O D. Both objects won't move at all. They will just stay where they were released.
3. A 46-gram golf ball and a 400-gram soccer ball were released at the same time from the top of a tall tower. What is the best explanation of this picture?
I
O A Heavier objects always fall faster than lighter objects.
O B. Lighter objects always fall faster than heavier objects
O C. The soccer ball was affected by air resistance less than the golf ball
O D. The soccer ball was affected by air resistance more than the golf ball.
Answer:
Do not see a picture or graph but suspect it would show the golf ball falling faster and striking the ground slightly before the soccer ball.
Probably D: Soccer ball was affected by air resistance more than the golf ball.
Explanation:
Even though heavier, friction loss of the greater surface area soccer ball will counter pull of gravity more than the compact golf ball.
In a vacuum, (no friction) both objects fall at the same rate regardless of mass.
A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.60 m/s. How much energy was lost to air friction during this jump?
kJ
process and answer please
Answer:
Efriction = 768.23 [kJ]
Explanation:
In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.
We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.
The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.
[tex]E_{pot}-E_{friction}=E_{kin}[/tex]
where:
[tex]E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}[/tex]
m = mass = 56 [kg]
h = elevation = 1400 [m]
v = velocity = 5.6 [m/s]
[tex](56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ][/tex]
Bradley drops a rock in a well. It falls for 12 seconds. How deep is the well?
A student is performing an experiment that involves the charge on a metal sphere that is attached to a charged electroscope. A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more. The rod is then removed, and the leaves return to their initial separated position. The student repeats the procedure, but this time the electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves again end up separated. What can be concluded about the charge on the separated leaves of the electroscope
Answer:
The leaves have a charge in each experiment, but the sign of the charge cannot be determined.
Explanation:
In the first experiment, A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more.
Thus indicates that there are charges involved. Now, like charges would repel like what is happening here but we don't know if they are both positive or negative because in both cases, they will still repel.
Now for the second experiment, electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves end up being separated again.
Similar to the first time, it's clear there are charges but the charges repel. Thus, they are the same sign charges but we don't know if they are both positive or negative.
Thus, in both cases we can conclude that the leaves have charges but we don't know their signs.
Starting at t = 0 a net external force in the +x-direction is applied to an object that has mass 5.00 kg. A graph of the force as a function of time is a straight line that passes through the origin and has slope 5.00 N/s.
If the object is at rest at t = 0 what is the magnitude of the force when the object has reached a speed of 9.00 m/s?
Express your answer with the appropriate units.
Answer:
15√2 N
Explanation:
The acceleration is given by ...
a = F/m = 5t/5 = t . . . . meters/second^2
The velocity is the integral of acceleration:
v = ∫a·dt = (1/2)t^2
This will be 9 m/s when ...
9 = (1/2)t^2
t = √18 . . . . seconds
And the force at that time is ...
F = 5(√18) = 15√2 . . . . newtons
Please help I’m confused on this
Answer:
inversely proportional to
The Earth has a diameter of 12,800 km and a globe has a diameter of 35.5 cm. What is the scale of the globe
Answer:
1:36,056,338
Explanation:
First thing we do is convert the diameter of the earth from kilometers into centimeters. Thus, we have
12800 km = 12800 * 100000 cm
12800 km = 1280000000 cm
Then we have diameter of the globe to be 35.5 cm.
To get the scale, what we do is divide the diameter of the earth by that of the globe, and as such we have
1280000000 / 35.5 =
36056338.
Therefore, the scale of the globe is 1:36,056,338
For every 1 cm on the globe, it is 36,056,338 cm or 360.6 km on earth
The scale of this globe is equal to 1 : 36,056,338.
Given the following data:
Earth diameter = 12,800 kmGlobe diameter = 35.5 cmIn Science, a scale can be defined as a ratio of the distance on a map to the actual (corresponding) distance on planet Earth.
A globe refers to a scale model of planet Earth that accurately depicts various geographic information such as distance, circumference, area, etc.
To determine the scale of this globe:
First of all, we would convert the value of Earth diameter in kilometers to centimeters as follow:
Conversion:
1 kilometer = 100,000 centimeter
12,800 kilometer = [tex]12800 \times 10^5 = 128 \times 10^7\;centimeters[/tex]
Now, we can calculate the scale of the globe by using this formula:
[tex]Distance = \frac{Earth\;diameter}{Globe\;diameter}[/tex]
Substituting the given parameters into the formula, we have;
[tex]Distance = \frac{128 \times 10^7}{35.5}[/tex]
Distance = 36,056,338.03 ≈ 36,056,338 centimeters.
Scale = 1 : 36,056,338.
Read more on globe here: https://brainly.com/question/5659485
13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there in the condenser
Answer:
12°F
Explanation:
Calculation for how much subcooling is there in the condenser
Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula
Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature
Let plug in the formula
Amount of Condenser subcooling=120°-108f
Amount of Condenser subcooling=12°F
Therefore the amount of subcooling that is there in the condenser will be 12°F
Describe how the force between two charged objects changes as the objects move closer to each other
Answer:
the electrical force between two charged objects is inversely related to the distance of separation between the two objects. Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance.
ASAP PLEASE HELP!
What is nuclear potential energy?
Answer:Nuclear potential energy is the potential energy of the particles inside an atomic nucleus
Please halp
Why does a suspended magnet always rest in the North-South direction?
Explanation:
because magnet attract opposite sides. like north and south.
Answer:
Its because the magnet has both north and south pole, so when suspended it turns its south pole towards southern hemisphere and the north pole towards northern hemisphere
Hope it helped u,
pls mark as the brainliest
^_^
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
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
Answer:
That is not meant to be red, it is the bottom of the beaker
That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.
Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
(was not able, but I don't know the reason)
Answer:
It will not hit.
Explanation:
In this problem, we need to tell that was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s.
When it will hit the target, final velocity, v = 0
Using third equation of motion as follows :
[tex]v^2-u^2=2as[/tex]
Here, a =-g
[tex]s=\dfrac{u^2}{2g}\\\\s=\dfrac{(20)^2}{2\times 9.8}\\\\=20.40\ m[/tex]
The target is in a straight line 50m away on the x-axis. As 50m is far from 20.40 m and that’s why it won’t hit.
If the Earth were flat, then the shadows of two towers at two different places on the Earth would:__________.
A. be different lengths with the one further South being longer
B. be different lengths with the one further North being longer
C. be different lengths with the one further East being longer
D. be the exact same length
If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 29.1 m/s?
A. 54 m
B. 50 m
C. 64 m
D. 60 m
Answer:
54m is the answer......PLEASE HELP
If your mass is 120 Kg how many Newtons do you weigh on the moon? (gravity on moon is 1.63m/s2)
The correct answer is 195.6 N
Explanation:
Different from the mass (total of matter) the weight is affected by gravity. Due to this, the weight changes according to the location of a body in the universe as gravity is not the same in all planets or celestial bodies. Moreover, this factor is measured in Newtons and it can be calculated using this simple formula W (Weight) = m (mass) x g (force of gravity). Now, leps calculate the weigh of someone whose mass is 120 kg and it is located on the moon:
F = 120 kg x 1.63 m/s2
F= 195.6 N
Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from west to east, while along the second track, train B moves with a speed of 60 m/s from east to west . calculate speed of Bw.r.t. A
Answer:
(i) Relative velocity of B w.r.t A= Sum of speeds of trains
=54+90
=144kmph
(ii)Relative velocity of B w.r.t Ground(G)=v
B/G
=−90kmph
v
G
=0
Relative velocity of ground(G) w.r.t B =v
G/B
=v
G
−v
B/G
v
G/B
=0−(−90)
v
G/B
=90kmph
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
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
Answer: It is the color of the container
Explanation:
The red at the bottom is the color of the bottom of the container. It is not part of the experiment and is not liquid.
The red liquid that is part of the experiment does indeed have the lowest density of the liquids which is why it is floating at the top of al the liquids with the only thing floating on top of it being the blue cube which has a lower density than it.
Answer: That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.
Is Florida an example of an isthmus or a peninsula? Both Peninsula Neither Isthmus
Answer:
Peninsula
Explanation:
Just did the Ed puzzle and got it right
why are curved roads banked
Answer:
So that vehicles do not topple or skid off the road.
Explanation:
As a vehicle is negotiating a curve, two equal and opposite forces act on the vehicle; centripetal force which keeps it moving and centrifugal force which tends to throw it out of the path.
In order to avoid skidding off the road curved roads are banked. Banking a curved surface provides the centripetal force that points towards the center of the road hence the vehicle or car does not skid off or topple.