Answer:
Doubling the amount of charge on one of the particles.
Explanation:
The force between two charges is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Where
r is the distance between charges
or
[tex]F\propto \dfrac{1}{r^2}[/tex]
On doubling the charge on one of the particle,
F' = 2F
So, the force gets doubled. Hence, the correct option is (d).
A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car arrives at the position of the stop-light 7.5 s after the light had turned green. If t = 0 when the light turns green, at what time does the green car catch the blue car if the green car maintains the slowest constant speed necessary to catch up to the blue car?
Answer:
After 15 seconds, the green car will catch up with the blue car
Explanation:
Let the time for the green car to catch up with the blue car be T
When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal
Distance covered by blue car after time T is given by: s = ut + 0.5 at²
Where u = 0, a = 0.2 m/s², t = T
S = 0.5 × 0.2 × T² = 0.1 T²
Velocity of blue car, v = u+ at
v = 0.2T
Distance covered by green car at T is given as: S = Velocity × time
Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)
S = 0.2T (T - 7.5)
S = 0.2 T² - 1.5T
Equating the distance covered by the two cars
0.2T² - 1.5T = 0.1T²
0.1T² - 1.5T = 0
T(0.1T - 1.5) = 0
T = 0 or
T = 1.5/0.1 = 15 secs
Therefore, after 15 seconds, the green car will catch up with the blue car
Diwn unscramble the word
OUM I THINK IS WIND
CORRECT ME IF IM WRONG
#CARRYONLEARNINGUse a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is directly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied
Complete question is;
Use a variation model to solve for the unknown value.
The stopping distance of a car is directly proportional to the square of the speed of the car.
a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.
b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?
Answer:
A) d = 333.2 ft
B) 60 mph
Explanation:
Let the stopping distance be d
Let the speed of the car be v
We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;
d ∝ v²
Therefore, d = kv²
Where k is constant of variation.
A) Speed is 50 mph and stopping distance of 170 ft.
v = 50 mph
d = 170 ft = 0.032197 miles
Thus,from d = kv², we have;
0.032197 = k(50²)
0.032197 = 2500k
k = 0.032197/2500
k = 0.0000128788
If the car is now travelling at 70 mph, then;
d = 0.0000128788 × 70²
d = 0.06310612 miles
Converting to ft gives;
d = 333.2 ft
B) stopping distance is now 244.8 ft
Converting to miles = 0.046363636 miles
Thus from d = kv², we have;
0.046363636 = 0.0000128788(v²)
v² = 0.046363636/0.0000128788
v² = 3599.99658
v = √3599.99658
v ≈ 60 mph
A 9V battery is connected to two light bulbs. If the current through the circuit is 1.5 A, what is the resistance of each light bulb (assume both light bulbs are identical)?
Answer:
3 ohm
Explanation:
Given :
V=9V
And according to given question same current is flowing in both resistance that means resistance will connected in series
So,
R= R+R=2R
Now,
Applying ohm's law
[tex]V=IR\\9=1.5*2R\\9=3R\\R=\frac{9}{3} \\R= 3ohm[/tex]
Therefore, answer is 3 ohm
(15 PTS) An observer riding on the platform measures the angle q that the thread supporting the light ball makes with the vertical. There is no friction anywhere. If you can vary m1 and m2, find the largest angle q you could achieve.
Solution :
Given :
Angle q = angle between the thread supporting the ball with the vertical.
Let mass [tex]$m_1 >>>m_2$[/tex].
Then [tex]$m_1+m_2=m_1$[/tex]
In this case, acceleration can be found out by applying Newton's law of motion.
Thus,
Acceleration, [tex]$a=\frac{m_1}{m_1+m_2}. g$[/tex]
[tex]$a=\frac{m_1}{m_1}. g$[/tex]
[tex]$a=g$[/tex]
Therefore, [tex]$\tan \theta =\frac{a}{g}$[/tex]
or [tex]$\tan \theta =\frac{a}{a}$[/tex]
or [tex]$\tan \theta =1$[/tex]
[tex]$\theta = \tan ^{-1}(1)$[/tex]
[tex]$\theta = 45^\circ$[/tex]
Therefore the largest angle q is [tex]$\theta = 45^\circ$[/tex]
Leslie incorrectly balances an equation as 2C4H10 + 12O2 → 8CO2 + 10H2O.
Which coefficient should she change?
Answer:
13 behind o2
Explanation:
answer is in photo above
Answer:
12
Explanation:
The temperature of a quantity of an ideal gas is a. one measure of its ability to transfer thermal energy to another body. b. proportional to the average molecular kinetic energy of the molecules. c. proportional to the internal energy of the gas. d. correctly described by all the statements above. e. correctly described only by the first two statements above.
Answer:
d. correctly described by all the statements above.
Explanation:
Kinetic molecular theory of gases states that gas particles exhibit a perfectly elastic collision and are constantly in motion.
According to the kinetic-molecular theory, the average kinetic energy of gas particles depends on temperature.
This ultimately implies that, the average kinetic energy of gas particles is directly proportional to the absolute temperature of an ideal gas. Thus, an increase in the average kinetic energy of gas particles would cause an increase in the absolute temperature of an ideal gas.
Temperature can be defined as a measure of the degree of coldness or hotness of a physical object. It is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).
Generally, the temperature of a quantity of an ideal gas is;
a. a measure of the ability of an ideal gas to transfer thermal energy to another body.
b. the average kinetic energy of gas particles is directly proportional to the absolute temperature of an ideal gas
c. proportional to the internal energy of the gas.
A boxer punches a sheet of paper in midair from rest to a speed of 20 m/s in 0.05 s. If the mass of the paper is 0.01 kg, the force of the punch on the paper is
A) 0.08 N.
B) 4.0 N.
C) 8.0 N.
D) 40 N.
Record the lengths of the sides of ABC and ADE.
what is effort arm
don't say the answer of gogle
Answer:
effort arm mean the use of any work by using your hand force motion or by hand power
A prece of cotton is measured between two points on a ruler
cotton
1
2.
3
4
5
6
7
8
9
10
11
12
13
14
15
16
When the length of cotton is wound closely around a pen, it goes round six times.
pen
six turns of cotton
What is the distance once round the pen?
A 2.2 cm
B 2.6 cm
с
13.2 cm
D
15.6cm
Answer:
13.2
Explanation:
I would say this cause this is a reasonable answer
The length of cotton is wound closely around a pen, it goes around six times. The distance once around the pen is 13.2 cm. The correct option is c.
What is the distance?Distance is the sum of an object's movements, regardless of direction. The distance can be defined as the amount of space an object has covered, regardless of its starting or ending position.
Measuring units are units that are used to measure the magnitude, amount, or quantity of any object. There are 7 basic measuring units that are used in worldwide and everyday life.
Given, the piece of cotton is tied around the pen, it goes around six times around the pen. The diameter of the pen will the length of the cotton. The actual length can be for six rounds is 13.2 cm.
Therefore, the correct option is с. 13.2 cm.
To learn more about length, refer to the link:
https://brainly.com/question/16188698
#SPJ2
Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sample of pharmacy assistants.
Lower Bound Upper Bound
6.51 8.50
50 8.51
10.50 18
10.51 12.50
42 12.51
14.50 20
14.51 16.50
(a) standard deviation = σ = 4.9996
(b) variance = σ² = 24.996
Explanation:Given frequency table (find attached as Table 1);
(a) To find the sample standard deviation and sample variance, follow these steps;
i. Calculate the mid-point c for each group by using the mid-point formula;
c = (lower bound + upper bound) / 2
=> c = (6.51 + 8.50) / 2 = 7.505
=> c = (8.51 + 10.50) / 2 = 9.505
=> c = (10.51 + 12.50) / 2 = 11.505
=> c = (12.51 + 14.50) / 2 = 13.505
=> c = (14.51 + 16.50) / 2 = 15.505
So the new table becomes (find attached as Table 2);
ii. Calculate the total number of samples (n) which is the sum of all the frequencies.
n = 50+18+42+20+46
n = 176
iii. Calculate the mean (M)
This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).
M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176
M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176
M = [2012.88] / 176
M = 11.44
iv. Find the variance (σ²);
The variance is calculated using the following formula
σ² = [Σ(f x c²) - (n x M²)] / (n - 1) ------------(i)
Where;
f = frequency of each boundary data point
=> Let's first calculate Σ(f x c²).
This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)
Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]
Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]
Σ(f x c²) = 24708.1244
=> Now calculate (n x M²)
n x M² = 176 x 11.44²
n x M² = 23033.7536
=> Now substitute these values into equation (i) to calculate the variance
σ² = [Σ(f x c²) - (n x M²)] / (n - 1)
σ² = [24708.1244 - 23033.7536] / (176 - 1)
σ² = [4374.3708] / (175)
σ² = 24.996
Therefore, the variance is 24.996
v. Find the standard deviation (σ)
The standard deviation is the square root of the variance. i.e
σ = √σ²
σ = √24.996
σ = 4.9996
Therefore, the standard deviation is 4.9996
If this guy is really faster than a speeding bullet (v=700m/s) and he has a mass of 100kg. How much force is behind him? *
A) 70000N
B) 9800N
C) 6860000N
D) We need his acceleration, not speed, to calculate this
show your work please
Answer:
if we want to find force by using newton's law equation ( f = ma ) we have to use mass and acceleration not velocity ,but in this question they did not mention about acceleration but speed so the answer is D
help asap please I will give you 5stars
Explanation:
In the parallel combination, the equivalent resistance is given by :
[tex]\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....[/tex]
4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :
[tex]\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega[/tex]
5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,
[tex]\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega[/tex]
Hence, this is the required solution.
A projectile is launched at ground level with an initial speed of 49.5 m/s at an angle of 40.0° above the horizontal. It
strikes a target above the ground 3.50 seconds later. What are the x and y distances from where the projectile was
launched to where it lands?
x distance
m
y distance
m
Answer:
x = 132.7 m
y = 51.34 m
Explanation:
Given :
Initial speed, u = 49.5 m/s²
Angle of projection, θ = 40°
Time, t = 3.50 seconds
The distance, x = horizontal component ;
Distance = speed * time
Distance = uCosθ * 3.50
Distance = 49.5 * Cos40° * 3.50
Distance = 49.5 * Cos40° * 3.50
Horizontal distance = 132.7 m
Vertical distance, y :
Sy = ut + 1/2gt²
Sy = Vertical distance ; g = 9.8 m/s²
Sy = 49.5 * sin40 * 3.5 - (0.5 * 9.8 * 3.5²)
Sy = 111.36295 - 60.025
Sy = 51.33795 m
x = 132.7 m
y = 51.34 m
How many atoms of carbon, C, are in 0.020 g of carbon?
Answer:
9.6352× 10²⁰ C atoms
Explanation:
From the given information,
The molar mass of Carbon = 12 g/mol
number of moles = 0.020g/ 12 g/mol
number of moles = 0.0016 mol
If 1 mole of C = 6.022 × 10²³ C atoms
∴
0.0016 mol of C = (6.022 × 10²³ C atoms/ 1 mol of C)×0.0016 mol of C
= 9.6352× 10²⁰ C atoms
Hence, the number of carbon atoms present in 0.020 g of carbon = 9.6352× 10²⁰ C atoms
An object with a mass of 5 kg is swung in a vertical circle by a rope with a length of 0.67 m. The tension at the bottom of the circle is 88 Newtons. What is the tension, in Newtons, at the side of the circle, halfway between the top and bottom if the speed of the mass is the same at the bottom and side
Answer:
[tex]T_2=39.5N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=5kg[/tex]
Length [tex]L=0.67m[/tex]
Tension [tex]T=88N[/tex]
Generally the equation for Tension is mathematically given by
[tex]T = m * ( g + v^2 /l)[/tex]
Therefore
[tex]T_1 = m * ( g + \frac{v^2}{l})[/tex]
[tex]88 = 5 * ( 9.8 + \frac{v^2}{0.67})[/tex]
[tex]v^2=5.2[/tex]
[tex]v=2.4m/s[/tex]
The uniform velocity is
[tex]v=2.4m/s[/tex]
Therefore
The tension at the side of the circle halfway between the top and bottom is
[tex]T_2=5*\frac{2.3^2}{0.67}[/tex]
[tex]T_2=39.5N[/tex]
Question 18/55 (2 p.)
A vibrating object produces ripples on the surface of a liquid. The object completes 20 vibrations
every second. The spacing of the ripples, from one crest to the next, is 3.0 cm.
What is the speed of the ripples?
D
C 60 cm/s
120 cm/s
A 0.15cm/s
B 6.7 cm/s
Answer:
the correct answer is C v = 60 cm / s
Explanation:
The speed of a wave is related to the frequency and the wavelength
v = λ f
They indicate that the object performs 20 oscillations every second, this is the frequency
f = 20 Hz
the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength
λ = 3 cm = 0.03 m
let's calculate
v = 20 0.03
v = 0.6 m / s
v = 60 cm / s
the correct answer is C
A system can only achieve a lower energy state by:_______
A horse gallops a distance of 10 kilometers in a time of 30 minutes. Its average speed is
Answer:
Explanation:
Distance=10km
Time=30min=0.5hr
So,speed=10/0.5
So,speed=20km/hr
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s.
Required:
a. Compute the magnitude and direction of the velocity of the stone after it is struck.
b. Is the collision perfectly elastic?
Answer:
(a)Magnitude=28.81 m/s
Direction=33.3 degree below the horizontal
(b) No, it is not perfectly elastic collision
Explanation:
We are given that
Mass of stone, M=0.150 kg
Mass of bullet, m=9.50 g=[tex]9.50\times 10^{3} kg[/tex]
Initial speed of bullet, u=380 m/s
Initial speed of stone, U=0
Final speed of bullet, v=250m/s
a. We have to find the magnitude and direction of the velocity of the stone after it is struck.
Using conservation of momentum
[tex]mu+ MU=mv+ MV[/tex]
Substitute the values
[tex]9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V[/tex]
[tex]3.61i=2.375j+0.150V[/tex]
[tex]3.61 i-2.375j=0.150V[/tex]
[tex]V=\frac{1}{0.150}(3.61 i-2.375j)[/tex]
[tex]V=24.07i-15.83j[/tex]
Magnitude of velocity of stone
=[tex]\sqrt{(24.07)^2+(-15.83)^2}[/tex]
|V|=28.81 m/s
Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s
Direction
[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
=[tex]tan^{-1}(\frac{-15.83}{24.07})[/tex]
[tex]\theta=tan^{-1}(-0.657)[/tex]
=33.3 degree below the horizontal
(b)
Initial kinetic energy
[tex]K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2[/tex]
[tex]K_i=685.9 J[/tex]
Final kinetic energy
[tex]K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2[/tex]
=[tex]\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2[/tex]
[tex]K_f=359.12 J[/tex]
Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.
if an object is moving with a velocity of 24 and has an acceleration of -4 square root how long will it take it to stop
Answer:
Time, t = 8 seconds.
Explanation:
Given the following data;
Initial velocity = 24 m/s
Final velocity = 0 m/s (since the object is coming to a stop).
Acceleration = -4 m/s²
To find how long it will take the object to stop, we would use the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity. U is the initial velocity. a is the acceleration. t is the time measured in seconds.Making time, t the subject of formula, we have;
[tex] t = \frac{V - U}{a}[/tex]
Substituting into the equation, we have;
[tex] t = \frac{0 - 24}{-4}[/tex]
[tex] t = \frac{-24}{-4}[/tex]
Time, t = 8 seconds.
ohm's law is not applicable to
Answer:
Ohm's law is not applicable to semi-conductors and insulators.
Explanation:
Is this what you want?
Hạt mang điện q > 0 chuyển động trong từ trường của một dòng điện thẳng dài có cường độ I = 10A như hình. Hạt mang điện chuyển động song song với dây dẫn và cách dây một khoảng 5cm. Vẽ hình và:
a. Xác định cảm ứng từ do dòng điện gây ra tại điểm mà hạt mang điện đi qua.
b. Hạt mang điện chuyển động với tốc độ 104m/s, lực Lorentz tác dụng lên hạt là 8.10-4N. Tính độ lớn của điện tích.
Answer:
I dnt know that language
Explanation:
What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?
53.0 kg
52.0 kg
51.0 kg
54.1 kg
Answer:
52.006 Kilograms
.............................................
The mass of an object that experience a gravitional force of 510 N near earths surface in 52.0 kg
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
Answer:
First of all the formula is F= uR,( force= static friction× reaction)
mass= 5+25=30
F= 50
R= mg(30×10)=300
u= ?
F=UR
u= F/R
u= 50/300=0.17N
Malar is playing with a toy car and track set. She has made a hill to let the car roll down, followed by a series of small hills and then a short flat section on the floor. The car starts at the top of the hill which is 137 cm above the short flat section. How much energy was lost due to friction (between the car and the sides of the track, and the car’s axels) during the entire trip? You don’t have any details about the time while the car is going down the hill or through the loop, so you don’t know how fast it is going at the top of the loop.
Mass is 150 g
Answer:
E = 2.01 J
Explanation:
The energy lost by the car due to friction through the entire trip must be equal to the potential energy of the car at the starting point.
Energy Lost = E = Potential Energy
E = mgh
where,
m = mass of car = 150 g = 0.15 kg
g = acceleration due to gravity = 9.81 m/s²
h = height of the track = 137 cm = 1.37 m
Therefore,
E = (0.15 kg)(9.81 m/s²)(1.37 m)
E = 2.01 J
A car accelerates from 0-30m/s in 5 s. It has a mass of 1200kg. What force does the engine produce? *
A) 36000N
B) 7200N
C) 60000N
D) 2000N
include explanation please
Answer:
a = ?
u = 0
v = 30
by using v = u + at equation we can find " a "
30 = 0 + 5a
6 m/s = a
by using f = ma equation we can find force produce by engine ,
f = ?
a = 6
m = 1200
f = 1200 × 6
f = 7200 N
so the answer is "B"
A 2 kg stone is dropped from a height of 100 m. How far does it travel in the third second? take g = 9.8 m/s2
Answer:
S = 1/2 gt² = 1/2 × 9.8 × 3² = 4.9×9 = 44.1 m
Explanation:
What does a model of a light wave tell us about
brightness and color?
Answer:
A wave model of light is useful for explaining brightness,color, and the frequency-dependent bending of light at a surface between media. However, because light can travel through space, it cannot be a matter wave, like sound or water waves.