Answer:
The weight of the combined system is 1989.8 N.
Explanation:
The weight, W, of an object can be expressed as;
W = mg
where m is the mass of the object, and g is the acceleration due to gravity.
Weight of the combined system = weight of tank + weight of water
mass of the plastic tank = 3 kg
weight of the plastic tank = m x g
= 3 x 9.8
= 29.4 N
Weight of the plastic tank is 29.4 N
But,
density = [tex]\frac{mass}{volume}[/tex]
mass = density x volume
volume of water = 0.2 [tex]m^{3}[/tex], density of water = 1000 kg/[tex]m^{3}[/tex].
mass = 1000 x 0.2
= 200
mass of water = 200 kg
weight of water = 200 x 9.8
= 1960 N
Thus,
combined weight of the system = 29.4 + 1960
= 1989.8 N
The weight of the combined system is 1989.8 N.
A 151 kg crate is pulled along a level surface by an engine. The coefficient of kinetic friction between the crate and the surface is 0.399. How much power must the engine deliver to move the crate at a constant speed of 6.04 m/s
Answer:
3566.26 Watts
Explanation:
From the question given above, the following data were obtained:
Mass (m) of crate = 151 kg
Coefficient of kinetic friction (μ) = 0.399
Velocity (v) = 6.04 m/s
Power (P) =?
Next, we shall determine the normal reaction. This can be obtained as follow:
Mass (m) of crate = 151 kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (R) = mg
Normal reaction (R) = 151 × 9.8
Normal reaction (R) = 1479.8 N
Next, we shall determine the force applied to move the crate. This can be obtained as follow:
Coefficient of kinetic friction (μ) = 0.399
Normal reaction (R) = 1479.8 N
Force (F) =?
F = μR
F = 0.399 × 1479.8
F = 590.44 N
Finally, we shall determine the power used to move the crate as follow:
Velocity (v) = 6.04 m/s
Force (F) applied = 590.44 N
Power (P) =?
P = F × v
P = 590.44 × 6.04
P = 3566.26 Watts
Therefore, the power used to move the crate is 3566.26 Watts.
A satellite of mass m orbits a moon of mass M in uniform circular motion with a constant tangential speed of v. The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit that is 2R from the center of the moon. What is the gravitational field strength of the moon at this new distance
The satellite is moved to a new circular orbit that is 2R from the center of the moon, then the gravitational field strength of the moon at this new distance would be one-fourth of the initial gravitational field.
What is gravity?It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another.
As given in the problem A satellite of mass m orbits a moon of mass M in a uniform circular motion with a constant tangential speed of v. The gravitational field strength at a distance R from the center of the moon is gR. The satellite is moved to a new circular orbit that is 2R from the center of the moon.
The gravitational field strength is inversely proportional to the square of the distance from the center of the planet.
Thus, the gravitational field strength of the moon at this new distance would be one-fourth of the initial gravitational field.
To learn more about gravity here, refer to the link given below ;
brainly.com/question/4014727
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A/An _____ is described as a device used to store electrical energy, usually two conductors separated by an insulator.
Answer: A capacitor.
Explanation:
The capacitor is a passive element that is used in electronics to store electrical energy maintaining an electrical field. The simpler case of a capacitor is the parallel plates capacitor.
It consists of two parallel metal plates separated by a distance D, in this case, the air between the plates works as a dielectric, as the plates do not touch each other and are separated by a dielectric, the charge is stored in the surface plates.
There are a lot of other types of capacitors, the most used in actuality may be the cylindrical one, where instead of parallel plates, it uses two concentric cylinders, and the space between the cylinders is filled with a dielectric/insulator.
g Calculate the electric potential at the center of the square with a side of 1 meter, formed by the four charged particles.
Answer:
The potential at the center of square due to four identical charges each at the corner is 5.09 q x [tex]10^1^0[/tex] Volts, where q is the charge.
Explanation:
From the question it is given that
side of square = a = 1 m
let the charge of each square is q
potential at center due to 1 charge is V = [tex]\frac{q}{4\pi \epsilon r}[/tex] , where
[tex]\epsilon[/tex] is electrical permittivity of space
r is the distance between charge and point.
since the charge is present at at the corner so the distance between charge and point is the half the length of diagonal of square.
⇒ distance between charge and point = r =[tex]\frac{a}{\sqrt{2}}[/tex] = [tex]\frac{1}{\sqrt{2} }[/tex] = 0.707 m
thus, V= [tex]\frac{q}{4\pi \epsilon r}[/tex] , on substituting the respected values of r = 0.707m and [tex]\frac{1}{4\pi \epsilon}[/tex] = 9 x [tex]10^9[/tex] we get,
V = 1.272 q x [tex]10^1^0[/tex] V
thus potential due to all 4 charges is
V = 1.272 q x [tex]10^1^0[/tex] x 4 = 5.09 q x [tex]10^1^0[/tex] Volts
Help pleaseeee need the answers ASAP
Answer:
- 670 kg.m/s
Explanation:
Newton's third law states that to every action, there is equal and opposite reaction force. Since the force will be same but different in direction and acted in the same time then the impulses ( force multiply by time) of the two car be same in magnitude but different in direction - 670 kg.m/s
Q2) A Ferrari moves at 240 km/h for 1.2 h. How far it will go.
Answer:
Explanation:
Multiply the two numbers to get the distance. 240 x 1.2 = 288km
What happens to the dewpoint temperature of a decreasing mass of air?
Answer:What happens to the dew point temperature of a descending mass of air? As air sinks, it becomes warmer. (Warmer air expands and can hold more water, therefore the dew point may increase.) ... The hot water causes condensation of water vapor as it touches the cooler mirror.
Explanation:
Imagine that you were having a conversation with a behaviorist and an ethologist. What would you say to help them meet in the middle of the nature/nurture debate?
Answer:my points back
Explanation:
During conversations with behaviorists and ethologists, it is mandatory to have a speech of interest to meet in the middle of the debate.
What do mean by conversation?The mode of communicating any topic or an issue to others by means of a set of words and speeches is known as a conversation.
In order to meet in the middle of the conversation, one needs to be deliverable with topics related to the domain of interest. Here the domain of interest means anything that is related to the profession.
For example, a behaviorist's role is to create plans for managing the behaviors of students. Similarly, an ethologist is a scientist, who studies animal behaviors. So, one to talk only related to their professional kinds of stuff for sake of clear understanding.
Thus, we can conclude that during conversations with behaviorists and ethologists, it is mandatory to have a speech of interest to meet in the middle of the debate.
Learn more about the fundamental of conversation here:
https://brainly.com/question/9311228
Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shifted green to you. Take red light to have a wavelength of 650 nm and green to have a wavelength of 550 nm.
Answer:
The doppler effect equation is:
[tex]f' = \frac{v +v0}{v - vs}*f[/tex]
In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:
v = f*λ
then:
f = v/λ
and v is the speed of light, then:
f = c/λ
where:
f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm
f is the real frequency, in this case, is (3*10^17nm/s)/650 nm
vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.
v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s
v0 is your speed, this is what we want to find.
Replacing those quantities in the equation, we get:
(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm
(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)
1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)
1.182*(3*10^8 m/s) - (3*10^8 m/s) = v0 = 54,600,000 m/s
So your speed was 54,600,000 m/s, which is a lot.
Estimate the net force needed to accelerate a 1000-kg car at 3m/s/s from rest.
FILL IN THE BLANKS.
If this force acts on this car for 4 seconds, then the car will increase its speed by ————————- m/s a second each second reaching a final speed of ————————- m/s. The distance traveled during this motion is ——————- meters.
Answer:
First Blank. 30,000
Second Blank. 12,000
Third Blank. 20.0
Explanation:
Helppp!!!!!!!!!!!!?!!??
PLZ HELP ASAP
With number 10 above
Answer:
P = 1500 Watt
Explanation:
Mechanical Work and Power
Mechanical work is the amount of energy transferred by a force.
Being F the magnitude of the force vector and s the distance, the work is calculated as:
W=F.s
Power is the amount of energy transferred or converted per unit of time. In the SI, the unit of power is the watt, equal to one joule per second.
The power can be calculated as:
[tex]\displaystyle P=\frac {W}{t}[/tex]
Where W is the work and t is the time.
The force to be considered is the weight of the mass of m=100 kg, [tex]g= 10\ m/s^2[/tex]:
F = 100 * 10 = 1000 N
The distance is s=3 m, thus the work done by the weight lifter is:
W = 1000 N * 3 m
W = 3000 J
Finally, the power is:
[tex]\displaystyle P=\frac {3000}{2}[/tex]
P = 1500 Watt
short essay your teacher have provided you
An ostrich with a mass of 141 kg is running to the right with a velocity of 13m/s. Find the momentum of the ostrich. Answer in units of Kg.m/s
Answer:
1833 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 141 × 13
We have the final answer as
1833 kg.m/sHope this helps you
A zone plate is found to give series of images of a point source on its axis. If the strongest and the second strongest images are at distances of 0.30m and 0.06 m respectively from the zone plate (both from the same side remote from the source), calculate the distance of the source from the zone plate, principal focal length and the radius of the second zone. Assume λ = 6 x 10-7m.
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
the following ATTENTION：using the energy/work formulae only: 3.The kinetic energy of the Rock half way down ？ 4.the speed of the Rock half way down？
5.The speed of the Rock as it hits the ground？
Answer:
See the answers below.
Explanation:
In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].
[tex]E_{A}=E_{B}[/tex]
The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.
So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.
[tex]E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}[/tex]
At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.
[tex]E_{B}=m*g*h+\frac{1}{2}*m*v^{2}[/tex]
Therefore we will have the following equation:
[tex](6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s][/tex]
The kinetic energy can be easily calculated by means of the kinetic energy equation.
[tex]KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J][/tex]
In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.
[tex]E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s][/tex]
Why is 1/2kx^2=gym not a linear equation
Four charges of equal magnitude, q = +1 uC, are placed at the four corners of a square with a side length of 1 meter located in the xy-plane. A small ball with a known charge +2 uC and an unknown mass is placed 2 meters above the center of the square (so along the z-axis).
Required:
a. What is the total force on the small ball from the four charges?
b. If the ball is in equilibrium where it was placed and we assume standard earth gravity, 9.8 m/s^2, what must the mass of the ball be?
Answer:
a) F _total = 15.08 10⁻³ N , b) m = 1,539 10⁻³ kg
Explanation:
a) To solve this problem we can use Coulomb's law to find the force created by each charge
F =[tex]k \frac{\ q_{1} q_{2} }{r_{12}^{2} }[/tex]
and the total force is
F = F₁₅ + F₂₅ + F₃₅ + F₄₅
The bold are vectros, In the exercise they indicate the charges
q₁ = q₂= q₃ = q₄ = 1 10⁻⁶ C
q₅ = 2 10⁻⁶ C
Let's set a reference system where the sphere is on the z axis and the other charges on the xy plane, let's write the coordinates of card charge
sphere (subscript 5)
x₅ = 0
y₅ = 0
z₅ = 2 m
charge 1
x = 0.5 m
y = 0.5 m
z = 0
charge 2
x = -0.5 m
y = 0.5 m
z = 0
charge 3
x = -0.5 m
y = -0.5 m
z = 0
charge 4
x = 0.5 m
y = -0.5 m
z = 0
With these values we can calculate the distance between each charge and the sphere
charge 1 and sphere
r₁₅² = (x₅ -x₁)² + (y₅ - y₁)² + (z₅ -z₁)²
substitute
r₁₅² = (0- 0.5)² + (0 - 0.5)² + (2 -0)²
r₁₅² = 0.5² + 0.5² + 2²
r₁₅² = 4.5
we can see that for the other charges the result is the same since being squared always gives positive
r₁₅ = r₂₅ = r₃₅ = r₄₅
the force created by the card charge on the sphere is the projection on the Z axis of the total force
Let's find the angle with respect to the Z axis
tan φ = r / z
where r is the magnitude of the vector in the xy plane
r = [tex]\sqrt{0.5^{2} + 0.5^{2} }[/tex]
r = 0.7071 m
φ = tan⁻¹ r / z
φ = tan⁻¹ (0.7071 / 2)
φ = 19.5
consequently the total force is
F_total = 4 F cos 19.5
F _total = [tex]4 \frac{k \ q_{1} q_{5} }{r_{15}^{2}}[/tex] cos 19.5
let's calculate
F_toal = 4 9 10⁹ 1 10⁻⁶ 2 10⁻⁶ /4.5 cos 19.5
F _total = 15.08 10⁻³ N
b) For this part indicate that the sphere is in equilibrium with the weight
F_total - W = 0
W = mg
F_total = mg
m = F_total / g
m = 15.08 10³ / 9.8
m = 1,539 10⁻³ kg
the earth is broken into smaller subsystems including the atmosphere the biosphere and the hydrosphere true or false
Answer:
true
Explanation:
these are not the only parts of the atmosphere, i dont know the full list but i know these arent the only parts
Deshaun Watson launches a football at a speed of 24.7 ms and an angle of 33° above the horizontal How far down
the football field does the football land? What is the max height the football reaches during flight?
Show work
Answer:
9.23m
Explanation:
Max height = u²sin²theta/2g
u is the speed = 34.7m/s
theta is the angle of elevation = 33°
g is the acceleration due to gravity = 9.8m/s²
Substitute into the formula
Max height = 24.7²sin²33/2(9.8)
Max height = 610.09sin²33/2(9.8)
Max height = 610.09(0.29663)/19.6
Max height = 180.97/19.6
Max Height = 9.23m
Hence the max height the football reaches during flight is 9.23m
What is the speed of a commercial jet which travels form New York to Los Angeles (4800) in 6 hours
Answer:
[tex]Speed = 800km}/hr[/tex]
Explanation:
Given
[tex]Distance = 4800km[/tex]
[tex]Time = 6hr[/tex]
Required
Determine the speed of the jet
The speed is calculated as:
[tex]Speed = \frac{Distance}{Time}[/tex]
Substitute 4800 km for Distance and 6hr for Time
[tex]Speed = \frac{4800km}{6hr}[/tex]
[tex]Speed = 800km}/hr[/tex]
Hence, the speed of the commercial jet is 800km/hr
You are sending waves down a spring. You send a small amplitude wave down the spring. Then you
send a large amplitude wave. The large amplitude wave is...
A.slower than the low amplitude wave
B.the same speed as the low amplitude wave
C.faster than the low amplitude wave
Answer:
a is the correct answer hope it will help you
( I will give a brainliest )
What must be changed, temperature or heat energy, during condensation?
Answer:
The answer is temperature lol
Explanation:
:)
if a train has 1,450 kg of momentum and is traveling at 40 m/s ¿what is the mass of tren?
Help plis I have this wrong
Answer:
P=mv
1x450,200=m40
m= 36,255kg
Help! Help!
Alcohol abuse has...
A. only physiological aspects.
B. only psychological aspects.
C. physiological and psychological aspects.
Answer
I feel the answer is C because it could cause mental and physical trauma
Explanation:
If a ball with a momentum of 40 kgm/s has a velocity of 2 m/s, what is its mass?
momentum (m)=40
velocity (v)=2
now,
momentum (m)=m×v
40=m×2
m=40÷2
m=20kg
The wavelength of a particular color of yellow light is 579 nm. The energy of this wavelength of light is
Answer:
3.44× 10⁻¹⁹Joules
Explanation:
Energy of the wavelength is expressed using the formula:
E = hc/λ
h is the Planck constant
c is the velocity of light
λ is the wavelength
Given
h = 6.63 × 10^-34 m² kg / s
c = 3×10⁸ m/s
λ = 579nm = 579 × 10⁻⁹m
λ = 5.79× 10⁻⁷m
Substitute the given values into the formula
E = hc/λ
E = (6.63 × 10⁻³⁴× 3×10⁸)/5.79× 10⁻⁷
E = 19.89× 10⁻³⁴⁺⁸/5.79× 10⁻⁷
E = 19.89× 10⁻²⁶/5.79× 10⁻⁷
E = 3.44× 10⁻²⁶⁺⁷
E = 3.44× 10⁻¹⁹Joules
Hence the energy of this wavelength of light is 3.44× 10⁻¹⁹Joules
A horizontal force of 5.0-N accelerates a 4.0-kg mass, from rest, at a rate of 0.50 m/s^2 in the positive direction. What friction force acts on the mass
Answer:
3N
Explanation:
The frictional force always acts directly opposite to the force of motion, that is it opposes motion. According to the Newton second law of motion:
Sum of horizontal forces = 0
Hence:
5N - 4(0.5) - F = 0
5 - 2 = F
F = 3N
A wrench is accidentally dropped at the top of an elevator shaft in a tall building. (a) How many meters does the wrench fall in 1.5s
Answer:
11.25m
Explanation:
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ = 0 \times 1.5 + \frac{1}{2} \times 10 \times {1.5}^{2} \\ = 5 \times 2.25 \\ = 11.25[/tex]
Approximately, What is the value of the Hubble Constant, as measured by scientists? Hypothetically, if the value of the Hubble Constant were 700 km/s/Mpc, what would this imply about the age of our universe?
Answer:
The current value of the Hubble's constant = 73 km/sec/Mpc.
t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc
Explanation:
The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.
Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.
Hence,
v is directly proportional to d
where, v = apparent velocity
d = distance
if we equate velocity and distance then there comes Hubble constant.
v = [tex]H_{0}[/tex] x d
[tex]H_{0}[/tex] = 73 km/sec/Mpc
where, Mpc = Mega Parsec = 1 Mpc = 3.086 x [tex]10^{19}[/tex] km
We can use Hubble constant to tell the age of universe.
t = d/v
t = d/( [tex]H_{0}[/tex] xd)
t = 1/[tex]H_{0}[/tex]
Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.
Now, if we hypothetically change the value of Hubble constant,
from [tex]H_{0}[/tex] = 73 km/sec/Mpc to [tex]H_{0}[/tex] = 700 km/sec/Mpc
then the age of universe will be:
t = 1/[tex]H_{0}[/tex]
first convert the units of new [tex]H_{0}[/tex] into 1/s
[tex]H_{0}[/tex] = (700) x (/3.08 x [tex]10^{19}[/tex] )
[tex]H_{0}[/tex] = 227.27 x[tex]10^{-19}[/tex] = 2.27 x [tex]10^{-21}[/tex] 1/s
So,
Age of universe will be:
t = 1/[tex]H_{0}[/tex] = 1/2.27x[tex]10^{-21}[/tex] 1/s
t = 2.27 x [tex]10^{21}[/tex] s
t = 71.9 trillion years
t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc
Given value:
Hubble Constant,
[tex]H_o = \frac{700 \ km/s}{Mpc}[/tex]We know,
[tex]Mpc = 3.086\times 10^{19} \ km[/tex]By substituting the value of "Mpc" in Hubble constant, we get
→ [tex]H_o = \frac{700}{3.086\times 10^{19}}[/tex]
[tex]= 227\times 10^{-19} \ 1/s[/tex]
[tex]= 2.27\times 10^{-21} \ 1/s[/tex]
The Hubble's time will be:
→ [tex]H_o = \frac{1}{t}[/tex]
or,
→ [tex]t = \frac{1}{H_o}[/tex]
[tex]= \frac{1}{2.27\times 10^{-21}}[/tex]
[tex]= 4.4\times 10^{20} \ seconds[/tex]
Thus the above approach is right.
Learn more about Hubble Constant here:
https://brainly.com/question/13691927?