Answer:
B. -20 m/s
Explanation:
Given the following data:
Initial velocity = 30 m/s
Acceleration = 5 m/s²
Deceleration = -5 m/s² (deceleration is the negative of acceleration)
Time = 10 seconds
To find the final velocity, 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.
Substituting into the formula, we have;
[tex] V = 30 + (-5)*10[/tex]
[tex] V = 30 - 50[/tex]
Final velocity, V = -20 m/s
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.
What is the magnitude of a vector that has the following components: x = 32 m y = -59 m
Answer:
Explanation:
Since the x and y components are given
The vectors Magnitude = √32²+(-59)²
=67.12m
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
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The slope at point A of the graph given below is:
WILL MARK BRAINLIEST TO CORRECT ANSWER
RQ/PQ I think
rise/run
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 :)
A 30g bullet at v=900m/s strikes a 1kg soft iron target stopping inside the iron [c=490J/kg°C. How much will the temperature of the iron increase? Ignore the heat that will be shared with the bullet
A) 25°C
B) 24795°C
C) 826°C
D) 82653°C
show your full work please
Answer:
ΔT = 25°C
Explanation:
Given that.
The mass of a bullet, m₁ = 30 g = 0.03 kg
The speed of the bullet, v = 900 m/s
Mass of soft iron, m₂ = 1 k
The specific heat of iron, c=490J/kg°C
We need to find the increase in temperature of iron. using the conservation of energy,
Kinetic energy = heat absorbed
[tex]\dfrac{1}{2}m_1v^2=m_2c\Delta T\\\\\Delta T=\dfrac{\dfrac{1}{2}m_1v^2}{m_2c}\\\\\Delta T=\dfrac{\dfrac{1}{2}\times 0.03\times 900^2}{1\times 490}\\\\=24.79^{\circ} C\\\\or\\\\\Delta T=25^{\circ} C[/tex]
So, the correct option is (A).
A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.
Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
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?
The number of current paths in a series circuit is:
a. one
b. two
C. three
d. four
Answer:
One
Explanation:
In series combination, the circuit follows one path whereas in parallel it follows two or more than two path
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 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.
Learn more about specific heat
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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
The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road
Force required to lift the wheelbarrow
Answer:
The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.
Explanation:
Given that,
The total mass of the wheelbarrow and the road is 80 kg.
The weight of an object is given by :
W = mg
where
g is acceleration due to gravity
So,
W = 80 × 9.8
= 784 N
So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.
1. A skydiver carries a buzzer which emits a steady 1800 Hz tone. A friend on the ground (directly below the skydiver) listens to the doppler shifted frequency. Assume the air is calm and that the speed of sound is independent of altitude. While the skydiver is falling at terminal velocity (the constant speed reached due to the balance of the downward gravitational force and the upward drag force due to air resistance), her friend on the ground hears waves of frequency 2150 Hz. a. How fast is the skydiver falling
Answer:
vs = 55.84 m/s
Explanation:
In this case, the source (skydiver with buzzer) is moving towards the observer (friend). The formula for this case will be:
[tex]f' = \frac{v}{v-v_s} f[/tex]
where,
f' = shifted frequency = 2150 Hz
f = actual frequency = 1800 Hz
v = speed of sound = 343 m/s
vs = speed of skydiver = ?
Therefore,
[tex]2150\ Hz = \frac{343\ m/s}{343\ m/s - v_s}(1800\ Hz)\\\\343\ m/s - v_s = \frac{343\ m/s}{2150\ Hz} (1800\ Hz)\\\\v_s = 343\ m/s - 287.16\ m/s\\[/tex]
vs = 55.84 m/s
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
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
Its Acceleration during the upward Journey ?
If the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change
Answer:
A neutral atom has no net charge and should not be affected by a point charge, and the distance of separation is not a factor.
The force will remain the same and is equal to zero.
We have a point charge and a neutral atom.
We have to investigate if the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change.
State Coulomb's Law of Electrostatic force.The Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically -
[tex]$F=\frac{q _{1}\cdot q _{2}}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]
According to question, we have -
A point charge and a neutral atom.
If initially the distance between the point charge and neutral atom is r meters, then -
q(1) = Q (say)
q(2) = 0 ( Neutral atom has zero charge)
Using Coulomb's law -
[tex]$F=\frac{Q\times 0}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]
F = 0 Newtons.
Now, if you will increase the distance by the factor of six, still the force will remain zero as there is only one point charge and other is neutral atom. There is no electrostatic force between a charged and uncharged particle.
Hence, the force will remain the same and is equal to zero.
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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.
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
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:
An object 1.00cm high is placed 18.0cm from a converging lens, forms a real Image 2.00cm high Calculate the forcal length of to the lens
Answer:
focal length=12cm
Explanation:
object size is equal to 1.00cm
object distance = 18cm
heigh of image = 2.00cm
image distance = ??
but magnification is given by;
M = 2.00/1.00 = 2
but u/v = M
u/18 = 2
u = 36
1/f = 1/u+1/v
1/f = 1/18+ 1/36
1/f = 1/12
f = 12cm
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
dòng điện là gì ?/???????
Answer:
Dòng điện là một dòng các hạt mang điện, chẳng hạn như electron hoặc ion, di chuyển qua vật dẫn điện hoặc không gian. Nó được đo bằng tốc độ thực của dòng điện tích qua một bề mặt hoặc vào một thể tích điều khiển.
Xin lưu ý rằng tôi đã sử dụng một trình dịch để nhập nội dung này, vì vậy có thể có một số từ không hợp lý.
Explanation:
two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
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
a rock with the mass of 10 kg sits at the top of a hill 20 m high. what is the potential energy
Answer:
What is the potential energy? PE= mghPE= hwKE= 1/2mv2
Answer:1960J
Explanation:
The length of a cylindrical axon is 8 cm and its radius of 8μm,and the thickness of the membrane is 0.01μm,dielectric constant ( ε=24.78x10-12 F/m),the capacitance of nerve cell is
Answer:
9.965 nF
Explanation:
The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m
So, C = εA/d
C = ε2πrL/d
Substituting the of the values variables into the equation, we have
C = ε2πrL/d
C = 24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m
C = 9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m
C = 996463 × 10⁻¹⁴ F
C = 9.96463 × 10⁻⁹ F
C = 9.96463 nF
C ≅ 9.965 nF
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]