Answer:
The velocity of the particle = -1.92 m/s
The speed of the particle = 5.72 m/s
Explanation:
Given equation of motion;
[tex]f(t) = 18 \ + \ \frac{48}{t} \ + \ 1[/tex]
Velocity is defined as the change in displacement with time.
[tex]V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s[/tex]
The distance traveled by the particle in 5 s:
[tex]s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m[/tex]
The speed of the particle when t = 5s
[tex]Speed = \frac{28.6}{5} = 5.72 \ m/s[/tex]
Its Acceleration during the upward Journey ?
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 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
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.
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
In electronic circuits:______.
a. the power used by a circuit is the resistance times the current squared.
b. electric and magnetic fields are transporting the energy.
c. electrons are transporting the energy.
d. the power used by a circuit is the voltage times the current squared.
e. the power used by a circuit is the current times the voltage.
Answer:
(a), (c) and (e) s correct.
Explanation:
a. the power used by a circuit is the resistance times the current squared.
The power is given by P = I^2 R, so the statement is correct.
b. electric and magnetic fields are transporting the energy.
false
c. electrons are transporting the energy.
The energy is transferred by flow of electrons. It is correct.
d. the power used by a circuit is the voltage times the current squared.
The power is given by P = V I, the statement is wrong.
e. the power used by a circuit is the current times the voltage.
The power is given by P = V I, the statement is correct.
A 285-kg load is lifted 22.0 m vertically with an acceleration a=0.160g by a single cable. Determine
(a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the
load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started
from rest.
Answer:
a) T = 2838.6 N, b) W = 1003.2 J, c) W = 6.22 10⁴ J, d) W = 2.79 10³ J
e) v_f = 2.65 m / s
Explanation:
a) To find the tension of the cable let's use Newton's second law
T - W = m a
T = W + ma
T = m (g + a)
let's calculate
T = 285 (-9.8 - 0.160)
T = 2838.6 N
b) net work is stress work minus weight work
W = F d
W = (T-W) d
W = (m a) d
W = (285 0.160) 22
W = 1003.2 J
c) the work done by the cable
W = T d cos 0
W = 2838.6 22.0
W = 6.22 10⁴ J
d) The work done by the weight
the displacement is upwards and the weight points downwards, so the angle is 180º
W = F. d
W = F d cos 180
W = -285 22.0
W = 2.79 10³ J
e) the final speed of the load. Let's use the relationship between work and the change in kinetic energy
W = ΔK
as part of rest K₀ = 0
W = ½ m v_f²
v_f = [tex]\sqrt{ \frac{2W}{m} }[/tex]
v_f = [tex]\sqrt{\frac{2 \ 1003.2}{285} }[/tex]
v_f = 2.65 m / s
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 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:
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.
If we will be conducting the experiments in tanks with dimensions of roughly 10 in by 10 in. Estimate the volume of oil you would need to cover half the total area. Again assume a molecule is 10 nm in size. Show your work.
Answer:
1. 10³ m²
2. 32.4 ml
3. 1.91 × 10²¹ molecules
Explanation:
Here is the complete question
1. Estimate the size of a one-molecule-thick oil film formed by spreading 1 ml of oil on the surface of the water. Assume that an oil molecule is roughly 10 nm in size. Show your work.
2. If we will be conducting the experiments in tanks with dimensions of roughly 10 in by 10 in. Estimate the volume of oil you would need to cover half the total area. Again assume a molecule is 10 nm in size. Show your work.
3. Estimate the number of molecules in 1 ml of oil assuming they’re 10 nm in size. What assumptions do you have to make? Show your work.
Solution
1. Since the film would cover an area, A, and would have a height which is the thickness of the molecule, h = 10 nm = 1 × 10⁻⁹ m, its volume is V = Ah. This volume also equals the volume of the oil. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, V = 1 × 10⁻⁶ m³.
The size of the oil drop is its area. So, A = V/h
= 1 × 10⁻⁶ m³ ÷ 1 × 10⁻⁹ m
= 10³ m²
2. The area of the tank is 10 in by 10 in = 100 in². Since we want to cover half the area, we require 100 in²/2 = 50 in² = 50 in² × (0.0254)² m²/in² = 0.0324 m².
If the thickness of oil is one molecule thick which is 10nm = 1 × 10⁻⁹ m, the volume of oil is then thickness × area = 1 × 10⁻⁹ m × 0.0324 m²
= 0.0324 × 10⁻⁹ m³
= 32.4 × 10⁻⁶ m³
= 32.4 ml since 1 × 10⁻⁶ m³ = 1 ml
3. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, we need to find the volume of one molecule. Since it is assumed to be a sphere, its volume is V' = πd³/6 where d = size of oil molecule = 10 nm = 1 × 10⁻⁹ m.
Let n be the number of molecules present in 1 ml, then nV' = 1 ml = 1 × 10⁻⁶ m³. So, n = 1 × 10⁻⁶ m³/V' = 1 × 10⁻⁶ m³ ÷ πd³/6 = 6 × 10⁻⁶ m³/πd³
Substituting d into the equation, we have
n = 6 × 10⁻⁶ m³/π(1 × 10⁻⁹ m)³
n = 6 × 10⁻⁶ m³/π × 10⁻²⁷ m³
n = 1.91 × 10²¹ molecules
The slope at point A of the graph given below is:
WILL MARK BRAINLIEST TO CORRECT ANSWER
RQ/PQ I think
rise/run
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]
Diwn unscramble the word
OUM I THINK IS WIND
CORRECT ME IF IM WRONG
#CARRYONLEARNINGA 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
Use 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
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?
A homeowner has a new oil furnace which has an efficiency of 60%. For every 100 barrels of oil used to heat his house, how much (in barrels of oil) goes up the chimney as waste heat?
Answer:
below
Explanation:
here did you walk? What did you find most enjoyable while walking: listening to music, listening to an audio book, or nothing? How did your body react to this introductory amount of exercise? Was it more exercise or less exercise than you are used to? If you did not walk, what other type of physical movement did you do?
Part A
Calculate the work done when a force of 6.0 N moves a book 2.0 m
Express your answer with the appropriate units.
Answer:
Work done applied = 12 newton-meter
Explanation:
Given examples:
Force applied = 6 newton
Distance of book = 2 meter
Find from the given data:
Work done
Computation:
The equation can be used to compute work.
Work done applied = Force applied x Distance of book
Work done applied = Force x Distance
Work done applied = 6 x 2
Work done applied = 12 newton-meter
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.
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:
A train is moving at a constant
speed of 55.0 m/s. After 5.00
seconds, how far has the train
gone?
cara
(Units = m)
Answer:
Distance = speed * time
55*5
275 meters.
The train would have covered a distance of 275 m
What is distance ?
We can define distance as to how much ground an object has covered despite its starting or ending point.
Distance = speed * time
given
speed= 55 m/s
time = 5 sec
Distance = 55 * 5 = 275 m
The train would have covered a distance of 275 m
learn more about distance
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Elapsed Time
(s)
Cart Speed
(Low fan speed)
(cm/s)
Cart Speed
(Medium fan speed)
(cm/s)
Cart Speed
(High fan speed)
(cm/s)
0
1
16.4
23.0
31.7
2
31.5
64.0 cm/s
3
54.0 cm/s
36.0
89.8
4
118,81
5
120.0 cm/s
6
128.2
96.0
7
145.8
Table B
Answer:
cm/s
6
128.2
96.0
7
145.8
Table B
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:
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
he cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.8 m/s in 2.50 s and can continue to accelerate to reach a top speed of 28.8 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. 1) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.) mih 2) Starting from a crouched position, how long does it take a cheetah to reach its top speed
Answer:
a) the cheetah's top speed is 64.4 miles/hr
b) time taken to reach top speed is 3.3 seconds
Explanation:
Given the data in the question;
Cheetahs can accelerate to a speed of 21.8 m/s in 2.50 s.
They can continue to accelerate to reach a top speed of 28.8 m/s.
a) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.)
The cheetah's top speed = 28.8 m/s = ( 28.8 × 2.237 ) miles/hr
= 64.4256 ≈ 64.4 miles/hr
Therefore, the cheetah's top speed is 64.4 miles/hr
b) Starting from a crouched position, how long does it take a cheetah to reach its top speed.
given that
v₁ = 21.8 m/s and t₁ = 2.50 s
let a represent the acceleration of the cheetah
From the First Equation of Motion;:
v = u + at
we substitute
21.8 = 0 + ( a × 2.50 )
21.8 = a × 2.50
a = 21.8 / 2.50
a = 8.72 m/s²
Now, let the time taken by cheetah to reach top speed ( 28.8 m/s ) be t
so from the first equation of motion;
v = u + at
we substitute
28.8 = 0 + ( 8.72 × t )
t = 28.8 / 8.72
t = 3.3 seconds
Therefore, time taken to reach top speed is 3.3 seconds
The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specific heat of the solid is 390 J/kgK and its heat of fusion is 4000 J/kg.
Answer and I will give you brainiliest
Hey again!
Ok..
Now... The melting Point of this solid is 90°C.
Meaning That as soon as it gets to this temp... It STARTS Melting.
So at that temp... It still has some solid parts in it.
You can say its a Solid Liquid Mixture.
Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.
After Fusion...It'd then Be a Pure Liquid with no solids in it.
So
Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c
Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.
So
Q= Q' + Q"
Q= mc∆0 + ml
∆0 = 90°c - 30°c = 60°c
Q= 2.5(390)(60) + (2.5)(4000)
Q=6.9 x 10⁴Joules
The heat required to change 2.5 kg of the solid at 30.0C to a liquid is 6.9 x 10⁴J.
What is specific heat?The specific heat is the amount of heat energy required to change the temperature of 1kg of object by 1°C.
The heat needed to change the solid's temperature from 30°C - 90°C is
Q' = mC∆T
The heat used to change the phase solid-liquid phase .i.e.
Q'' =mL where L =latent heat of fusion
The total heat required is
Q= Q' + Q"
Q= mc∆T + ml
Q= 2.5(390)(90 - 30) + (2.5)(4000)
Q=6.9 x 10⁴Joules
Thus, the heat required to change the solid to liquid is 6.9 x 10⁴J.
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A computer monitor uses 200 W of power. How much energy does it use in
10 seconds?
Answer: see below explanation, should be straight forward from there? ;)
Explanation: 1 watt = 1 joule per second
Watt is a measure of energy over time
So 10 seconds... u got this :)
(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]