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.
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.
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!
what is general relativity
Explanation:
General relativity is a theory of space and time. ... The central idea of general relativity is that space and time are two aspects of spacetime. Spacetime is curved when there is matter, energy, and momentum resulting in what we perceive as gravity. The links between these forces are shown in the Einstein field equations.
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.
Bob, of mass m, drops from a tree limb at the same time that Esther, also of mass m, begins her descent down a frictionless slide. If they both start at the same height above the ground, which of the following is true about their kinetic energies as they reach the ground?
A) Bob's kinetic energy is greater than Esther's.B) Esther's kinetic energy is greater than Bob's.C) They have the same kinetic energy.D) The answer depends on the shape of the slide.
Answer:
Explanation:
They have the same kinetic energy
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
For more details Force, visit
brainly.com/question/13691251
#SPJ2
Question 18 of 20
Which situation results in the least attraction between two magnets?
A. The south pole of one magnet is near the south pole of the other
magnet
B. The north pole of one magnet is near the south pole of the other
magnet
C. The north pole of one magnet is far away from the north pole of
the other magnet
D. The north pole of one magnet is far away from the south pole of
the other magnet.
Answer:D
Explanation:
Answer:
Explanation: the answer is D.
Is Chocolate Milk a compound,mixture or a element
Answer:
mixture
Explanation:
because of many Ingredients it is called a mixture
a fisherman moves from one end of a boat to another
Answer:
That isnt a question so no one will know the answer to what you are talking about. I suggest adding a sceenshot or picture of the question.
What acceleration will you give a 22.4 kg box if you push it with a force of 83.1N
Answer:
mass =22.4kg
force=83.1N
a=?
f=ma
a=f/m
a=83.1/22.4
a=3.70m/s^2
If the net work done on a particle is zero, which of the following statements must be true?
A. The speed is unchanged.
B. The velocity is zero.
C. The velocity is unchanged.
D. More information is needed.
E. The velocity is decreased.
Answer:
A. The speed is unchanged.
Explanation:
In the case when the work is to be done on a particle i.e. zero so the change made in KE of the particle would be zero. This represent the work energy theroem. But when the KE remains same or does not change so it should be the same and the particle speed would also the same
Therefore as per the given statement, the first option is correct
And rest of the options are wrong
Mental processes refers to
overt actions and reactions.
only animal behavior.
internal, covert processes.
outward behavior.
What us the difference in the ways objects move at a speed of a car and an object mkvinf close to the speed of light?
Answer:
The difference is in who or what is observing the speed.
Explanation:
Giving that speed is relative between the objects and the reference point from which it is being observed.
It is concluded that speed alone has no direct effect on a moving object, hence it is just a determining unit for the difference in distance between two objects.
Therefore, in this case, the difference is in who or what is observing the speed.
A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at this instant?
KEi = _________________J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
KEf/KEi=______________
Answer:
(a) the runner's kinetic energy at the given instant is 308 J
(b) the kinetic energy increased by a factor of 4.
Explanation:
Given;
mass of the runner, m = 64.1 kg
speed of the runner, u = 3.10 m/s
(a) the kinetic energy of the runner at this instant is calculated as;
[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J[/tex]
(b) when the runner doubles his speed, his final kinetic energy is calculated as;
[tex]K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J[/tex]
the change in the kinetic energy is calculated as;
[tex]\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4[/tex]
Thus, the kinetic energy increased by a factor of 4.
in how many days the life cycle of honeybee complete
Please HElp!!
Name five conductors and insulators
Answer:
Conductors- copper, aluminum, gold, and silver.
Insulators- glass, air, plastic, rubber, and wood.
Explanation:
Answer: 5 conductors are copper, aluminum, gold, iron, and silver and 5 insulators are glass, air, plastic, rubber, and wood.
A basketball is rolling rightward onto the court with a speed of 4.0 m/s and slows down with a consta leftward acceleration of magnitude 0.50 m/s^2 over 14 m.
What is the velocity of the basketball after rolling for 14 m?
a. 2.0 m/s
b. 5.5 m/s
c. 7.5 m/s
d. 1.4 m/s
Answer:
d. 1.4 m/s
Explanation:
Given;
initial velocity of the `basketball, u = 4.0 m/s
acceleration of the basketball, a = -0.5 m/s² (leftward)
distance the basketball traveled, d = 14 m
the final velocity of the basketball, v = ?
(this final velocity should be less than the initial velocity since the ball is slowing down at a constant rate)
Apply the following the kinematic equation;
v² = u² + 2ad
v² = 4² + 2(-0.5)14
v² = 16 - 14
v² = 2
v = √2
v = 1.41 m/s
Therefore, the final velocity of the basketball is 1.4 m/s.
A solid is 5 cm tall, 3 cm wide, and 2 cm thick. It has a mass of 129 g. What is its
density?
Answer:
4.3 g/cm³ or 4.3g/cc
Explanation:
Volume(V) = Height × Length × Width
= 5cm × 3cm × 2cm
= 30cm³
Mass(m) = 129gram
So,
Density = m/V
= 129g/30cm³
= 4.3g/cc or 4.3g/cm³
An astronaut weighing 190 lbs on Earth is on a mission to the Moon and Mars.
Required:
a. What would he weigh in newtons when he is on the Moon?
b. How much would he weigh in newtons when he is on Mars, where the acceleration due to gravity is 0.38 times that on Earth?
Answer:
The weight is defined as:
W = m*g
where:
m = mass
g = gravitational acceleration.
We know that in Earth the astronaut weights 190 lb-f (this is force, not mass, the correct unit here is 190 lb*m/s^2)
then:
190 lb*m/s^2 = m*9.8m/s^2
(190 lb*m/s^2)/(9.8m/s^2) = 19.39 lb
Now we know the mass of the astronaut.
a) wieght on the moon in Newtons.
Newtons uses kilograms as the units of mass, then we need to rewrite the mass of the astronaut in kg.
we know that 1lb = 0.454 kg
Then 19.39 lb is equal to: 19.39*0.454 kg = 8.8 kg
We know that the acceleration due to gravity on the Moon is one-sixth that on Earth.
then: g = (9.8m/s^2)/6
And the weight of the astronaut in the moon will be:
W = 8.8 kg*(9.8m/s^2)/6 = 14.37 N
b) The weight on mars, where the acceleration due to gravity is 0.38 times that on Earth, we have:
g = (9.8m/s^2)*0.38
then the weight will be:
W = 8.8kg*(9.8m/s^2)*0.38 = 32.77 N
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.
At a distance of 10 km from a radio transmitter, the amplitude of the E-field is 0.20 volts/meter. What is the total power emitted by the radio transmitter
Answer:
The total power is [tex]P = 6.665 *10^{4} \ W[/tex]
Explanation:
From the question we are told that
The distance is [tex]r = 10 \ km = 1000 \ m[/tex]
The amplitude of the electric field is [tex]E = 0.20 \ volt/meter[/tex]
Generally the average intensity of the electromagnetic field from the radio transmitter is mathematically represented as
[tex]I = \frac{E^2}{ 2 \mu_o * c }[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
[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]I = \frac{0.2^2}{ 2 * 4\pi *10^{-7} * 3.0*10^{8} }[/tex]
=> [tex]I = 5.307 *10^{-5} \ W/m^2[/tex]
Generally this intensity can also be mathematically represented as
[tex]I = \frac{P }{ 4 \pi r^2 }[/tex]
=> [tex]P = I ( 4 \pi r^2 )[/tex]
=> [tex]P = 5.307 *10^{-5} ( 4 * 3.142 * 1000^2 )[/tex]
=> [tex]P = 6.665 *10^{4} \ W[/tex]
The total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].
The given parameters;
amplitude of the electric field, E = 0.2 V/mdistance of the transmitter, = 10 km = 10,000 mThe intensity of the radio wave from the transmitter is calculated as follows;
[tex]I = \frac{E^2}{2\mu_0 c} \\\\I = \frac{0.2^2 }{2\times 4\pi \times 10^{-7} \times 3\times 10^8} \\\\I = 5.305 \times 10^{-5} \ W/m^2[/tex]
The total power emitted by the radio transmitter is calculated as follows;
[tex]I = \frac{P}{A} \\\\P = IA\\\\P = I \times 4\pi r^2\\\\P = (5.305\times 10^{-5} )\times 4\pi \times (10,000)^2\\\\P = 6.67\times 10^{4} \ W[/tex]
Thus, the total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].
Learn more here:https://brainly.com/question/19340071
When gasoline is burned, it gives off 46,000 J/gram of heat energy. If an automobile uses 13.0 kg of gasoline per hour with an efficiency of 21%, what is the average horsepower output of the engine
Answer:
The correct output will be "46.76 h.p.".
Explanation:
The given values are:
Gasoline's energy content,
= 46000 J/g
= 4.6 x10⁷ J/kg
Fuel usage,
= 13 kg/h
= [tex]\frac{13}{3600} \ kg/s[/tex]
Fuel efficiency,
= 21%
= 0.21
Now,
The average output will be:
= [tex](Energy \ content \ of \ gasoline)\times (fuel \ usage)\times (fuel \ efficiency)[/tex]
On substituting the values, we get
= [tex]4.6\times 10^7\times (\frac{13}{3600} )\times 0.21\times (\frac{1}{746}) h.p.[/tex]
= [tex]46.76 \ h.p.[/tex]
Answer:
the average horsepower output of the engine is 46.76 hp
Explanation:
The computation of the average horsepower output of the engine is shown below
The output of the engine is
= (Energy content of gasoline) × (usage of the fuel ) × (Fuel efficiency)
As we know that
1 hp = 746 W
where,
Energy content of gasoline = 46000 J / g = 4.6 × 107 J/kg
Usage of the fuel = 13 kg / h = 13 ÷ 3600 kg/s
Fuel efficiency = 21% = 0.21
Therefore, the output of the engine is
= 4.6 × 107 × (13 ÷ 3600) × 0.21 × (1 ÷ 746) h.p.
= 46.76 h.p.
hence, the average horsepower output of the engine is 46.76 hp
A circuit with a 12 V battery and lamp has a current of 3 A. What is the resistance of the lamp?
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
A 10 kg block is attached to a light cord that is wrapped around the pulley of an electric motor, as shown above. If the motor raises the block from the floor to a height of 8.0 meters in 5 seconds what was the motor’s power output?
Answer:
156.8 Watts
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 10 kg
Height (h) = 8 m
Time (t) = 5 s
Power (P) =?
Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:
Mass (m) = 10 kg
Height (h) = 8 m
Acceleration due to gravity (g) = 9.8 m/s²
Energy (E) =?
E = mgh
E = 10 × 9. 8 × 8
E = 784 J
Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:
Time (t) = 5 s
Energy (E) = 784 J
Power (P) =?
P = E/t
P = 784 / 5
P = 156.8 Watts
Therefore, the power output of the motor is 156.8 Watts
7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent
beam 12cm from P. At what point does the beam converges if the lens is a) a convex lens of
focal length 20cm and b) a concave lens of focal length 16cm.
Answer:
Lens at a distance = 7.5 cm
Lens at a distance = 6.86 cm (Approx)
Explanation:
Given:
Object distance u = 12 cm
a) Focal length = 20 cm
b) Focal length = 16 cm
Computation:
a. 1/v = 1/u + 1/f
1/v = 1/20 + 1/12
v = 7.5 cm
Lens at a distance = 7.5 cm
b. 1/v = 1/u + 1/f
1/v = 1/16 + 1/12
v = 6.86 cm (Approx)
Lens at a distance = 6.86 cm (Approx)
1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea's tangential speed is which of the following?a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.2) When the merry go round is rotating at a constant angular speed, Andrea's tangential speed is:______.a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.
Explanation:
The tangential speed of Andrea is given by :
[tex]v=r\omega[/tex]
Where
r is radius of the circular path
ω is angular speed
The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'
r' = 2r
New angular speed,
[tex]v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v[/tex]
New angular speed is twice that of the Chuck's speed.
My dad gifted me a calculator. I have observed that very small cells are used in a calculator. What are these cells called and what are their main advantage?
Show that a 2,500,000-J change in kinetic energy occurs for an airplane that is moved 500 m in takeoff by a sustained force of 5000 N.
Answer:
The answer to your question is given below
Explanation:
To solve this problem, we'll assume that the plane is initially at rest.
Hence, the kinetic energy of the plane at rest is zero i.e Initial kinetic energy (KE₁) = 0
Next, we shall determine the final kinetic energy of the plan when the force was applied. This can be obtained as follow:
Force (F) = 5000 N
Distance (s) = 500 m
Energy (E) =?
E = F × s
E = 5000 × 500
E = 2500000 J
Since energy an kinetic energy has the same unit of measurement, thus, the final kinetic energy (KE₂) of the plane is 2500000 J
Finally, we shall determine the change in the kinetic energy of the plane. This can be obtained as follow:
Initial kinetic energy (KE₁) = 0
Final kinetic energy (KE₂) = 2500000 J
Change in kinetic energy (ΔKE) =?
ΔKE = KE₂ – KE₁
ΔKE = 2500000 – 0
ΔKE = 2500000 J
Hence, the change in the kinetic energy of the plane is 2500000 J.