Answer:
1 m/s
Explanation:
Using law of conservation of momentum
m1v1 + m2v2 = (m1 + m2)vf , where
m1 = mass of object at rest, 20 kg
v1 = initial velocity of object at rest, 0 m/s
vf = final velocity of the bodies
m2 = mass of object in motion, 10 kg
v2 = initial velocity of object in motion, 3 m/s
On substituting, we have
(20 * 0) + (10 * 3) = (20 + 10) vf
0 + 30 = 30 vf
vf = 30 / 30
vf = 1 m/s
Therefore, the velocity of the bodies after hitting each other is 1 m/s
A circuit with a 12 V battery and lamp has a current of 3 A. What is the resistance of the lamp?
derive an expression for torque experiend by an electric dipole placed in a uniform electric field
Answer:
The torque τ on an electric dipole with dipole moment p in a uniform electric field E is given by τ = p × E where the "X" refers to the vector cross product. Ref: Wikipedia article on electric dipole moment.
Explanation:
Tell me the max amount you should owe on this card.
Your credit limit is $1,000.
What is the max you should ever owe on this card?
$
Your credit limit is $2,500.
What is the max you should ever owe on this card?
$
Answer:
the max is 2,500 or less
Explanation:
because you cant owe anymore
If a net horizontal force of 0.8 N is applied to a toy whose mass is 1.2 kg, acceleration is?
Hello!
[tex]\large\boxed{a = \frac{2}{3}m/s^{2}}[/tex]
Use the equation F = m · a to solve. We are given the force (N) and mass (kg), so we can solve for the acceleration by plugging in the given values:
0.8 = 1.2a
0.8 / 1.2 = a
a = 2/3 m/s²
Plants that respond to light are responding
to an:
a. internal stimulus
b. external stimulus
please answer
Answer:
The answer would be a
Explanation:
What was your train of thought as you navigated the picture of the candle?
Answer:
Where is the picture
Explanation:
WHERE IS THE PICTURE
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
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
A thin-walled vessel of volume V contains N particles which slowly leak out of a small hole of area A. No particles enter the volume through the hole. Find the time required for the number of particles to decrease to N/2. Express your answer in terms of A, V, and v.
Answer:
[tex]\frac{V}{2av}[/tex]
Explanation:
From the question we are told that
Volume V
Contains N particles
Leaks from a small hole of area A
Generally the equation for Flow rate is given as
Volume Flow Rate [tex]V_r = A * v[/tex]
Mathematically we find the time taken to flow half way which is given by
[tex]\frac{(V/2)}{A*v}[/tex]
Therefore the time taken is
[tex]\frac{V}{2av}[/tex]
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.
For the following types of electromagnetic radiation, how do the wavelength, frequency, and photon energy change as one goes from the top of the list to the bottom?
a. radio waves
b. infared radiation
c. visible light
d. ultraviolet radiation
e. gamma radiation
Answer:
Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.
Explanation:
Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.
The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
For an object like a planet, with a typical temperature of a few hundred kelvin, what kind of blackbody radiation would it principally emit
Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
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
An ideal gas is confined to a container with constant volume. The number of moles is constant. By what factor will the pressure change if the absolute temperature triples
Answer:
1/3
Explanation:
Gay Lusaac's law states that "the pressure of a given mass of gas is directly proportional with the absolute temperature of the gas, provided that the volume is kept constant."
In formula, we say that
P/T = k
Where
P = pressure at different points
T = temperature at different points
k = constant of proportionality
From the stated formula, if we multiply the temperature by 3, we have
P/3T = k
P * 1/3T = k
And from this, we see the pressure will change by a value of 1/3
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
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.
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.
In a test run, a certain car accelerates uniformly from zero to 20.4 m/s in 2.60 s.
Required:
a. What is the magnitude of the cars acceleration?
b. How long does it take the car to change speed from 10.0 m/s to 20 m/s.
c. Will doubling the time always double the change in speed? why?
Answer:
(a) The acceleration is 7.85 m/s²
(b) It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Doubling the time will double the change in velocity if the acceleration is kept constant.
Explanation:
(a) Acceleration is the physical quantity that measures the rate of change of velocity with time. That is, acceleration relates changes in speed with the time in which they occur, that is, it measures how fast the changes in speed are.
The average acceleration is calculated using the following expression:
[tex]a=\frac{vf-vi}{t}[/tex]
where a is the acceleration, vf is the final velocity, vi is the initial velocity and t is the time.
In this case:
vf= 20.4 m/svi=0 m/st= 2.60 sReplacing:
[tex]a=\frac{20.4 \frac{m}{s} - 0\frac{m}{s} }{2.60 s}[/tex]
a= 7.85 m/s²
The acceleration is 7.85 m/s²
(b) In this case you know:
a= 7.85 m/s²vf= 20 m/svi= 10 m/sReplacing:
[tex]7.85 \frac{m}{s^{2} } =\frac{20 \frac{m}{s} - 10\frac{m}{s} }{t}[/tex]
and solving you get:
[tex]t=\frac{20 \frac{m}{s} - 10\frac{m}{s} }{7.85 \frac{m}{s^{2} } }[/tex]
t=1.27 s
It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Being:
[tex]a=\frac{vf-vi}{t}[/tex]
Then:
a*t= vf - vi
vf - vi represents the change in velocity. You can see that, if a (acceleration) is constant, then (vf - vi) is directly proportional to the time t: therefore, if t doubles, the change in velocity doubles as well.
In other words, doubling the time will double the change in velocity if the acceleration is kept constant.
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.
in how many days the life cycle of honeybee complete
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.
Is Chocolate Milk a compound,mixture or a element
Answer:
mixture
Explanation:
because of many Ingredients it is called a mixture
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
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.
Why do we perform stork stand test
Answer:
umm becuase it is a test and you need them
Explanation:
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!
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
Mental processes refers to
overt actions and reactions.
only animal behavior.
internal, covert processes.
outward behavior.
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
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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].
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