Answer:
The speed at B is 16.18 ft/s .
Explanation:
Speed at A, u = 11 ft/s
Speed at C, v' = 18 ft/s
Time from A to C = 5 s
Time from B to C = 1.3 s
Let the speed of car at B is v.
Let the acceleration is a.
From A to B
Use first equation of motion
v = u + a t
18 = 11 + a x 5
a = 1.4 ft/s^2
Let the time from A to B is t' .
t' = 5 - 1.3 = 3.7 s
Use first equation of motion from A to B
v = 11 + 1.4 x 3.7 = 16.18 ft/s
Hai quả cầu kim loại nhỏ giống nhau, mang điện tích q1 = 2.10-8 C; q2 = 6.10-8 C, đặt cách
nhau một đoạn r trong không khí thì chúng đẩy nhau bằng một lực là 18.10-5 N. Cho hai quả cầu
tiếp xúc nhau rồi đưa về khoảng cách cũ thì lực tương tác giữa hai quả cầu là:
Answer:
fgggggffgggcffghhhjjkuuu of to ok with queen size of your yyyygtyttttttyyyhgggghhhhfrghjjkkkExplanation:
jjjgggyuuuuiiii hhjiiihuyyuuugggyujjhhhhggghhhjjhhhhjhhui
a. Do the waves made by the two faucets travel faster than the waves made by just one faucet?
b. How do you know this? Describe how the two-faucet wave pattern compares with the one-faucet pattern.
c. Describe what happens to the two-faucet wave pattern as the separation of the faucets is increased.
Answer:
asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd
Explanation:
Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed
Answer:
c.
magnitude of displacement
ai là người phát hiện trái đất hình cầu đầu tiên ?
Answer:
Can't understand the language
A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-life of the radioisotope?
11.54 minutes
Explanation:
The decay rate equation is given by
[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]
where [tex]\lambda[/tex] is the half-life. We can rewrite this as
[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]
Taking the natural logarithm of both sides, we get
[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]
Solving for [tex]\lambda[/tex],
[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]
[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]
[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]
How much amount of water can be decomposed
through electrolysis by passing 2 F charge?
Answer:
So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.
Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]
Answer:
Gap left = Change in length on heating
Gap=Initial length×Coefficient of linear expansion×change in temperature
Gap=10×0.000012×15m
⟹Gap=0.0018 m
this is an example u have to put your equation in it
A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released. Assuming no friction and no external force, the natural frequency W (measured in radians per unit time) for the system is? (Recall that the acceleration due to gravity is 32ft/sec2).
a) None of the other alternatives is correct.
b) W = v2 3
c)w=212
d) w = 4/6
e) w=213
Answer:
4√6 rad/s
Explanation:
Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.
So, W = F
mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft
making k the spring constant subject of the formula, we have
k = mg/x
substituting the values of the variables into the equation, we have
k = mg/x
k = 4 lb × 32 ft/s² ÷ 1/3 ft
k = 32 × 4 × 3
k = 384 lbft²/s²
Now, assuming there is no friction and no external force, we have an undamped system.
So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb
So, substituting the values of the variables into the equation, we have
ω = √(k/m)
ω = √(384 lbft²/s² ÷ 4 lb)
ω = √96
ω = √(16 × 6)
ω = √16 × √6
ω = 4√6 rad/s
Một loa phát ra với cường độ âm là 40 (W/m2
). Mức cường độ âm của loa thuộc phạm vi?
Answer:ew
Explanation:
qeeqw
Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters above the ground, how much work do you do
Answer:
100 Joules
Explanation:
Applying,
W = mgh................... Equation 1
Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.
From the question,
Given: mg = 10 newtons, h = 10 meters.
Substitute these values into equation 1
W = 10×10
W = 100 Joules.
Hence the amount of workdone is 100 Joules
Use the DC Construction kit to build a simple circuit to perform the following task:
You are asked to use a single resistor and a 110 V DC battery for the purpose of boiling a litter of water (4,184 Joule/Kg*degree Celsius), with a starting temperature of 20 C, in exactly 4 minutes.
Answer:
The resistance is 8.7 ohm.
Explanation:
Voltage, V = 110 V
mass, m = 1 kg
change in temperature, T = 100 - 20 = 80 C
time, t = 4 min = 4 x 60 = 240 s
specific heat, c = 4184 J/kg C
let the resistance is R.
The heat generated by the heater is used to the heat the water.
[tex]\frac{V^2}{R} t = m c T \\\\\frac{110^2}{R}\times 240 = 1\times 4184\times 80\\\\R = 8.7 ohm[/tex]
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88
is the correct answer
The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the gravitational force between them ? ans. 2.668× 10-9N
[tex]F = 5.93×10^{13}\:\text{N}[/tex]
Explanation:
Given:
[tex]m_1= 2×10^{16}\:\text{kg}[/tex]
[tex]m_2= 4×10^{22}\:\text{kg}[/tex]
[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]
Using Newton's universal law of gravitation, we can write
[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]
[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]
[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]
A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?
A, 0
B, 16 J
C, 72 J
D, 450 J
E, 90 J
=F×s×cosa=2×g×0,8×cos90°= 0
The work done by gravity on a ball of 2 kg which is moving with a constant speed of 6 meter per second is zero. Thus, the correct option is A.
What is Work?Work is the energy transfer to or from an object through the application of force along with the displacement. For a constant force aligned with the direction of motion, the work done is equal to the product of the force strength which is applied and the distance traveled by the object.
Work = Force × Displacement
Force = Mass × Acceleration
Acceleration of the ball is zero as it is moving with a constant speed. Therefore, the work done by the gravity is zero.
Therefore, the correct option is A.
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In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.
Answer:
The distance between mirror and you is 2 ft.
Explanation:
diameter, d = 8 ft
radius of curvature, R = 4 ft
magnification, m = 0.5
focal length, f = R/2 = 4/2 = 2 ft
let the distance of object is u and the distance of image is v.
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]
Use the formula of magnification
[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]
A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]
Let d is the distance moved in 2.25 s. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]
So, it will move 6.32 m from rest in 2.25 seconds.
how interfacial angles are determined using contact goinometer
Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.
Answer:
Explanation:
Since energy is conserved:
2
mu
2
=
2
mv
2
+mgh
⇒u
2
=v
2
+2gh
⇒(3)
2
=v
2
+2(9.8)(0.5−0.5cos60)
⇒v=2m/s
Acceleration of the simple pendulum is 2.62 m/s².
What is meant by a simple pendulum ?When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.
Here,
Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.
Let T be the tension acting on the string.
As, the bob passes through the angle,
The weight of the bob becomes equal to the vertical component of the tension.
mg = T cos15°
Also, the horizontal component of the tension,
T sin15° = ma
By solving these two equations, we get that,
Acceleration of the simple pendulum,
a = g tan15°
a = 9.8 x 0.267
a = 2.62 m/s²
Hence,
Acceleration of the simple pendulum is 2.62 m/s².
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Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate
Answer:
[tex]\omega=3.135rad/s[/tex]
Explanation:
From the question we are told that:
initial Speed [tex]V_1=2.50[/tex]
Mass [tex]m=70.0kg[/tex]
Center of mass [tex]d=0.0.800m\[/tex]
Generally the equation for angular velocity is is mathematically given by
[tex]\omega=\frac{v}{r}\\\\\omega=\frac{2.50}{0.0800}[/tex]
[tex]\omega=3.135rad/s[/tex]
Assume the speed of sound is 343 m/s. You are sitting 150 m away from home plate at a baseball game. How much time in seconds elapses between the batter hitting a home run and the moment you actually hear the batter hitting the ball
Answer:
t = 0.437 s
Explanation:
Sound is a wave so its speed is constant
v = x / t
t = x / v
indicates that the distance is x = 150 m
t = 150/343
t = 0.437 s
this is the time it takes to hear the hit
To see the blow it is almost instantaneous since the speed of light is much greater c = 3 10⁸ m / s
The current through each resistor in the two-resistor circuit is _________ the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is ___________ the voltage across the resistor in the one-resistor circuit.
Answer:
Serial circuit. the current is constant.
the voltage across a given resistor is half the rating in a one-resistor circuit.
Parallel circuit the voltage is constant
the current is half the value of the current with a single resistor
Explanation:
To answer exactly this exercise, you need the diagram or the indication of the type of circuit being used, since there are two possibilities, let's consider the results of each one.
Serial circuit.
The two resistance are one after the other.
In this case the current in the resistor sides is the same, that is, the current is constant in the circuit.
The voltage is proportional to the value of each resistor and if the two resistors are equal, the voltage across a given resistor is half the rating in a one-resistor circuit.
Parallel circuit
the two resistance is next to each other.
In this case the voltage is constant, that is, the voltage across the two resistors is the same as in the case of a single resistor.
The current is inversely proportional to the value of the resistance
i₁ = V / R₁
i₂ = V / R₂
for a single resistance
I = V / R
these currents are related
i = i₁ + i₂
if the two resistors have the same value the current is half the value of the current with a single resistor
5. A bicyclist is finishing her repair of a flat tire when a friend rides by at a constant velocity of
3.5 m/s. Three seconds later, the bicyclist hops on her bike and accelerates at 3.6 m/s² until she
catches her friend.
a. How much time does it take until she catches her friend?
b. How far has she traveled in this time?
c. What is her speed when she catches up?
Answer:
a) t = 3.6 s
b) d = 23 m
c) v = 13 m/s
Explanation:
Let t be the time the accelerating rider rides
the distance she travels is
d = ½3.6t²
the distance for the other cyclist is
d =3.5(t + 3)
½3.6t² = 3.5(t + 3)
1.8t² - 3.5t - 10.5 = 0
quadratic formula, positive answer
t = (3.5 + √(3.5² - 4(1.8)(-10.5))) / (2(1.8))
t = 3.575786...
d = ½(3.6)(3.575786²) = 23.015...
v = 3.6(3.575786) = 12.8728...
a volcano that may erupt again at some time in the distant future is
what is environment in world
Answer:
World Environment Day has been celebrated every year on 5 June, engaging governments, businesses and citizens in an effort to address pressing environmental issues
Explanation:
Answer:
The natural environment or natural world encompasses all living and non-living things occurring naturally, meaning in this case not artificial. The term is most often applied to the Earth or some parts of Earth.
when a boron is added to a pure semi conducter it becomes
Answer:
it becomes a p-type conductor
Explanation:
answer from gauth math
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
Question
Name and write briefly on the international body
that Introduced si units .
Answer:
International System of Units. it was established in 1960 by the 11the General Conference on Weights and Measures
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.
Answer:
0.5849Weber
Explanation:
The formula for calculating the magnetic flus is expressed as:
[tex]\phi = BAcos \theta[/tex]
Given
The magnitude of the magnetic field B = 3.35T
Area of the loop = πr² = 3.14(0.24)² = 0.180864m²
angle of the wire loop θ = 15.1°
Substitute the given values into the formula:
[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]
Hence the magnetic flux Φ through the loop is 0.5849Weber
on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]
Answer:
The correct answer is - 8.99N/C
Explanation:
[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]
