Convert 15000km/hr into m/s
Answer:
4166.6667m/s
Explanation:
The hour equals 3600 secound, and Km equals to 1000 meter.
Answer:
15000000
Explanation:
formula multiply the lenght value by 1000
A wave with a frequency of 5Hz travels a distance of 40mm in 2 seconds.What is the speed of the wave
Answer:
20mm per second
Explanation:
it is possible that the acceleration and velocity are perpendicular to each other? explain with example
Answer: Ok so We already know that velocity is on the x-axis.
Since acceleration = Force / Mass
Here the Force is downward due to the gravitational pull or we can say it is along y-axis.
Since acceleration is directly proportional to force, so acceleration is also along y-axis. This means that velocity & acceleration are perpendicular to each other.
Example:
Let us assume that an aeroplane is flying parallel to the horizontal plane. The aeroplane will experience the acceleration in several directions. One of them here is the gravitational pull which is perpendicular to the the apparent velocity. So the net velocity & its direction will depend upon the vector sum total of all the forces/acceleration acting on it. Also because of this gravitational pull the aeroplane rotates along with the earth, which is a proof that the force/g experienced by it does not go waste.
Hope this helps have a awesome day/night❤️✨Explanation:
A 40-pF capacitor is charged to a potential difference of 500 V. Its terminals are then connected to those of an uncharged 10-pF capacitor. Calculate: (a) the original charge on the 40-pF capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.
(a) The original charge on the 40-pF capacitor is [tex]2 .0 \ \times \ 10^{-8} \ C[/tex].
(b) The charge on each capacitor after the connection is made is [tex]4 .0 \ \times \ 10^{-9} \ C[/tex].
(c) The potential difference across the plates of each capacitor after the connection is 100 V and 400 V.
Original charge of the capacitorThe original charge on the 40-pF capacitor is calculated as follows;
[tex]Q = CV\\\\Q = 40 \times 10^{-12} \times 500\\\\Q = 2 .0 \ \times \ 10^{-8} \ C[/tex]
Charge on each capacitor[tex]C = \frac{C_1C_2}{C_1 + C_2} \\\\C = \frac{10 \times 10^{-12} \times 40 \times 10^{-12}}{10\times 10^{-12} \ + \ 40 \times 10^{-12}} \\\\C = 8 \times 10^{-12} \ F[/tex]
[tex]Q = Q_1 = Q_2\\\\Q = 8 \times 10^{-12} \ \times \ 500\\\\Q = 4 \times 10^{-9} \ C[/tex]
Potential differenceThe potential difference across the plates of each capacitor after the connection is calculated as follows;
[tex]V = \frac{Q}{C} \\\\V_1 = \frac{Q}{C_1} \\\\V_1 = \frac{4 \times 10^{-9}}{40 \times 10^{-12}} \\\\V_1 = 100 \ V\\\\V_2 = \frac{Q}{C_2} \\\\V_2 = \frac{4 \times 10^{-9}}{10 \times 10^{-12} } \\\\V_2 = 400 \ V[/tex]
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Which of the following types of models could be used to represent a cell?
a.
idea model
b.
physical model
c.
computer model
d.
all of the above
Answer:
d. all of the above
a ball is spun around in circular motion such that it completes 50 rotations in 25 s.
1) What is the period of its rotation?
2) what is the frequency of its rotation?
Answer:
(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.
Explanation:
Given that,
a ball is spun around in circular motion such that it completes 50 rotations in 25 s.
(1). Let T be the period of its rotation. It can be calculated as follows :
[tex]T=\dfrac{25}{50}\\\\T=0.5\ s[/tex]
(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,
[tex]f=\dfrac{50}{25}\\\\f=2\ Hz[/tex]
Hence, this is the required solution.
what is dimensional analysis
Answer:
The formula used to convert from the metric system outside of it. So like converting kilograms into pounds. The Formula is as follows:
# unit x (#unit/#unit) = # unit
^ ^ ^
I I I
given Conversion Answer
factor
*Note: Italic "units" are the same. Bold "units" are the same.
Example:
One thousand eighty kilometers is how many miles? Set it up dimensionally.
1080 km (1 mi/1.61 km) = 670.81 mi
*This is because 1080 x 1 = 1080, but then you divide 1080 by 1.61
Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit around its star. The astronomers measure each planet's orbital period, as shown in the table.
Planet; Orbital Period (Earth days)
Planet W; 10
Planet X; 640
Planet Y; 80
Planet Z; 270
To determine the distance each planet is from the star, astronomers applied one of Kepler's three laws.
Kepler's first law: The path of each planet around a star is an ellipse, with the star at one focus. Kepler's second law: A planet sweeps out equal areas in equal amounts of time as it revolves around the star.
Kepler's third law: The square of the time for one revolution of a planet is proportional to the cube of the radius of its orbit.
Based on the table, identify the planet that is the farthest distance from the star, and indicate which of Kepler's three laws can be used to justify your answer. Enter your answer in the box provided.
Answer:
planet that is farthest away is planet X
kepler's third law
Explanation:
For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets
T² = ([tex]\frac{4\pi ^2}{ G M_s}[/tex] a³ = K_s a³
Let's apply this equation to our case
a = [tex]\sqrt[3]{ \frac{T^2}{K_s} }[/tex]
for this particular exercise it is not necessary to reduce the period to seconds
Plant W
10² = K_s [tex]a_{w}^3[/tex]
a_w = [tex]\sqrt[3]{ \frac{100}{ K_s} }[/tex]
a_w = [tex]\frac{1}{ \sqrt[3]{K_s} }[/tex] 4.64
Planet X
a_x = [tex]\sqrt[3]{ \frac{640^3}{K_s} }[/tex]
a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3
Planet Y
a_y = [tex]\sqrt[3]{ \frac{80^2}{K_s} }[/tex]
a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6
Planet z
a_z = [tex]\sqrt[3]{ \frac{270^2}{K_s} }[/tex]
a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8
From the previous results we see that planet that is farthest away is planet X
where we have used kepler's third law
PLS HELP
SHOW WORK
A car starting from rest accelerates at a rate of 1.5 m/s ^ 2 What is its final speed at the end of 18.0 seconds ?
Answer:
27 ms^-1
Explanation:
by using v= u + at
u = 0 ( because the object id starting from rest)
v= 0 + 1.5 x 18
v = 27 ms^-1
Why does a person feel cooler under a rotating fan?
What a fan does is create a wind chill effect. ... By blowing air around, the fan makes it easier for the air to evaporate sweat from your skin, which is how you eliminate body heat. The more evaporation, the cooler you feel.
Select the correct answer.
A pair of pliers is an example of a simple machine with two levers. Which part of the pliers is the fulcrum?
A. A
B. B
C. C
D. D
E. E
Answer:
There isnt enough in your question to answer the question bro, like we need a picture or something bro.
Explanation:
You don't have a image attached
PLEASE HELP 15 POINTS AND BRAINIEST!!!
Reporting fake answers
Time Dilation. A group of scientists discover a new, rare isotope and are able to store a small amount of it. They determine that the isotope is unstable and half of their sample will decay in 13.0 monthsmonths . The scientists need a new laboratory to properly conduct measurements on the isotope. If laboratory is built before half the sample decays there will still be enough of the isotope available for experiments. However, it will take 21.0 monthsmonths to build the new lab. Fortunately, the scientists can quickly start construction and they have access to a spaceship that can travel at speeds approaching the speed of light c = 3.00
This question is incomplete, the complete question is;
Time Dilation. A group of scientists discover a new, rare isotope and are able to store a small amount of it. They determine that the isotope is unstable and half of their sample will decay in 13.0 months . The scientists need a new laboratory to properly conduct measurements on the isotope. If laboratory is built before half the sample decays there will still be enough of the isotope available for experiments. However, it will take 21.0 months to build the new lab. Fortunately, the scientists can quickly start construction and they have access to a spaceship that can travel at speeds approaching the speed of light c = 3.00 x 10⁸ m/s.
If they place their sample of the isotope on the spaceship, at what speed must it travel in order for the new laboratory to be completed on Earth by the time half of the isotope on the spaceship decays? Assume that it took one month for the scientists to actually start construction and launch the spaceship so half the sample will remain in 12.0 months but it will still take 21.0 months to build the new lab.
Answer:
the required speed is 2.4618 x 10⁸ m/s
Explanation:
Given the data in the question;
Time taken to complete a lab = 21 months
it took them one one to actually start so
Time remaining so that sample only decays by 1/2 = (13 - 1) = 12 months
we need to find at what speed the spaceship should travel so that time on earth ( = 12 months) become time on spaceship ( = 21 months)
we make use of time dilation equation;
t' = t/√( 1 - v²/c² )
where t is time in rest frame = 12 and t' is time in moving frame = 21
so we substitute
21 = 12/√( 1 - v²/c² )
21√( 1 - v²/c² ) = 12
√( 1 - v²/c² ) = 12/21
we square both side
( 1 - v²/c² ) = ( 12/21 )²
( 1 - v²/c² ) = 0.3265
v²/c² = 1 - 0.3265
v²/c² = 0.6735
v² = 0.6735 × c²
v = √0.6735 × √c²
v = 0.8206 × c
given that speed of light c = 3 x 10⁸ m/s
v = 0.8206 × 3 x 10⁸ m/s
v = 2.4618 x 10⁸ m/s
Therefore, the required speed is 2.4618 x 10⁸ m/s
What is the chemical formula for magnesium sulfide?
Answer:
MgS
Explanation:
How effective are child care programs for children compared to at-home care ?
Answer:
in my opinion i would say that at home cares of more effective because half of the time your child is safe and sound but in child care programs sometimes you have no idea what is going on when you are not around
Explanation:
3. A woman drove her car from home to her daughter's school. The odometre on her dashboard says she travelled 4.5 km to do this. She then immediately drove back home, using a different route, which was 5.5 km long. The whole journey took 30 minutes.
a. What distance did she travel?
b. What was her displacement?
C. What was her average speed during the journey?
Answer:
Look Below -->
Explanation:
a. She traveled 10 km, add 4.5 km + 5.5 km = 10 km (Distance is the total units travelled, so just add them all up :) )
b. Her displacement is 0 km because she went back home. (Displacement is the difference between the end and starting points)
c. 3 km/hr (30 minutes / 10 km)
Two teams of nine members each engage in a tug of war. Each of the first team's members has an average mass of 64 kg and exerts an average force of 1350 N horizontally. Each of the second team's members has an average mass of 69 kg and exerts an average force of 1367 N horizontally. (a) What is the acceleration (in m/s2 in the direction the heavy team is pulling) of the two teams
Answer:
[tex]a=0.13m/s^2[/tex]
Explanation:
From the question we are told that
Mass of first team man [tex]m_1=64kg[/tex]
Force of man first team man [tex]F_1=1350[/tex]
Mass of second team man [tex]m_2=69kg[/tex]
Force of man second team man [tex]F_2=1367N[/tex]
Generally the equation for net force F_n is mathematically given by
[tex]F_n=9(m_1+m_2)a[/tex]
[tex]9(m_1+m_2)a=9(f_2-f_1)[/tex]
[tex]9(64+69)a=9(1367-1350)[/tex]
[tex]a=\frac{9(1367-1350)}{9(64+69)}[/tex]
[tex]a=0.127819m/s^2[/tex]
Therefore the acceleration is given by
[tex]a=0.13m/s^2[/tex]
determine the value of a and K when 0.51 is written is scientific nation
Answer:
The value 0.51
Explanation:
If 155 g of sugar can be dissolve in 100g of water at 20 degree than how much sugar will be dissolved in 300 g of water at same temperature if someone do I will make you brainly
Answer:
465 grams of sugar.
Explanation:
I'm not sure this is true: is the relationship linear?
If it is then 300 grams of water should be able told 3*155 grams of sugar
3 * 155 = 465 grams of sugar in 300 grams of water.
A 125 kg mail bag hangs by a vertical rope 3.3 m long. A postal worker then displaces the bag to a position 2.2 m sideways from its original position, always keeping the rope taut.
1) What horizontal force is necessary to hold the bag in the new position?
2) As the bag is moved to this position, how much work is done by the rope?
3) As the bag is moved to this position, how much work is done by the worker?
Answer:
1) the required horizontal force F is 1095.6 N
2) W = 0 J { work done by rope will be 0 since tension perpendicular }
3) work is done by the worker is 1029.4 J
Explanation:
Given that;
mass of bag m = 125 kg
length of rope [tex]l[/tex] = 3.3 m
displacement of bag d = 2.2 m
1) What horizontal force is necessary to hold the bag in the new position?
from the figure below; ( triangle )
SOH CAH TOA
sin = opp / hyp
sin[tex]\theta[/tex] = d / [tex]l[/tex]
sin[tex]\theta[/tex] = 2.2/ 3.3
sin[tex]\theta[/tex] = 0.6666
[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )
[tex]\theta[/tex] = 41.81°
Now, tension in the string is resolved into components as illustrated in the image below;
Tsin[tex]\theta[/tex] = F
Tcos[tex]\theta[/tex] = mg
so
Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg
sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg
we know that; tangent = sine/cosine
so
tan[tex]\theta[/tex] = F / mg
F = mg tan[tex]\theta[/tex]
we substitute
Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )
F = 1225 × 0.8944
F = 1095.6 N
Therefore, the required horizontal force F is 1095.6 N
2) As the bag is moved to this position, how much work is done by the rope?
Tension in the rope and displacement of mass are perpendicular,
so, work done will be;
W = Tdcos90°
W = Td × 0
W = 0 J { work done by rope will be 0 since tension perpendicular }
3) As the bag is moved to this position, how much work is done by the worker
from the diagram in the image below;
SOH CAH TOA
cos = adj / hyp
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex]
we substitute
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex] = 1 - h/[tex]l[/tex]
cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]
h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]
h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
now, work done by the worker against gravity will be;
W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
we substitute
W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )
W = 4042.5 × ( 1 - 0.745359 )
W = 4042.5 × 0.254641
W = 1029.4 J
Therefore, work is done by the worker is 1029.4 J
Which of the following requires constant agonist-antagonist muscle contraction
Answer: Dynamic balance
Explanation: Dynamic balance movements are movements in which constant agonist-antagonist muscle contractions occur in order to maintain a certain position or posture. ISSA pg 121
amount of force times the distance it moves object
work=force×distance this is the formula
A girl with a mass of 22 kg is playing on a swing. There are three main forces acting on her at any time: gravity, force due to centripetal acceleration, and the tension in the swing's chain (ignore the effects of air resistance). At the instant shown in the image below, she is at the bottom of the swing and is traveling at a constant speed of 2 m/s. What is the tension in the swing's chain at this time? (Recall that g = 9.8 m/s^2)
A. 97.5N
B. 180.9N
C. 237.6N
D. 117.4N
Answer:
A is the answer I think pls check it
If an object is placed at distance of 16cm from a plane mirror, How far would it be from it's image?
Explanation:
A plane mirror always creates an image with the same distance to the mirror as the object, only in the other direction. So both of them have a distance of 10cm, one is 10cm to the left, one 10cm to the right, thus their mutual distance is 20cm
A spring has a spring constant of 65.5 N/m and it is
stretched with a force of 15.3 N. How far will it stretch?
If you increase the frequency of a sound wave four times, what will happen to its speed?
А
The speed will increase four times.
B.
The speed will decrease four times,
C. The speed will remain the same.
D.
The speed will increase twice.
E.
The speed will decrease twice.
A custodian pulls a vacuum 13.5 m
with a 33.9 N force at a 55.0°
angle, against a 14.2 N friction
force. Find the total work done on
the vacuum.
(Unit = J)
Answer:
Horizontal component of pull = (cos 55 x 33.9) = 19.4N.
Net horizontal force = (19.4 - 14.2) = 5.2N.
Work = (fd) = (5.2 x 13.5) = 70.2 Joules.
Rounded to 1 decimal place throughout.
Explanation:
The total work done on the vacuum is 70.2 J.
What is work done?
Work done is equal to product of force applied and distance moved.
Work = Force x Distance
Given is a custodian pulls a vacuum 13.5 m with a 33.9 N force at a 55.0° angle, against a 14.2 N friction force.
Horizontal component of pull = (cos 55 x 33.9) = 19.4N.
Net horizontal force = (19.4 - 14.2) = 5.2N.
Work done by vacuum will be
Work =5.2 x 13.5
Work =70.2 J
Thus, the total work done on the vacuum is 70.2 J.
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OWA OWA
Answer:
I think it might be (e
Explanation:
OWA OWA
14
How many
electrons
are in the
atom
pictured?
9 p*
10 nº
Answer:
9 electrons
Explanation:
The structure has 9 protons and hence the number of electrons equals the number of protons that's why it is said to be electrical neutral
The sun warms Earth through the process of _________
conduction
convection
insulation
radiation