Answer:
the correct one is b
the difference between the final moment and the initial moment
Explanation:
The momentum is related to the moment
I = ΔP
∫ F dt = p_f - p₀
where p_f and p₀ are the final and initial moments, respectively
When checking the different answers, the correct one is b
the difference between the final moment and the initial moment
1. A child slide down an inclined plane of length 10 m at an angle of 45°. If the coefficient friction between the child and the plane is 0.1, evaluate The velocity just before touching the bottom of the plane.
Answer:
The speed at the bottom is 11.2 m/s.
Explanation:
length, s = 10 m
Angle, A = 45 degree
coefficient of friction = 0.1
let the velocity is v.
The acceleration is given by
[tex]a = g sin A - \mu g cos A \\\\a = 9.8 (sin 45 - 0.1 cos 45)\\\\a = 6.24 m/s^2[/tex]
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 6.24 \times 10 \\\\v = 11.2 m/s[/tex]
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?
Answer:
The correct answer is - low pitch
Explanation:
Now for the case it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
if frequency increases then pitch will be increase as well as pitch depends on frequency.
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.
As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the x-axis does the third piece move
Answer:
Explanation:
There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.
momentum of first piece = .320 kg x 2 m/s
= 0.64 kg m/s along x -axis.
momentum of second piece = .355 kg x 1.5 m/s
= 0.5325 kg m/s along y- axis .
Let the velocity of third piece be v and it is making angle of θ with x -axis .
Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ
vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ
For making total momentum in the plane zero
.1 v cosθ = 0.64 kg m/s
.1 v sinθ = 0.5325 kg m/s
Dividing
Tanθ = .5325 / .64 = .83
θ = 40⁰.
The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.
Answer:
8.3 m/s, 2196 degree from + X axis
Explanation:
m = 320 g , u = 2 m/s along X axis
m' = 355 g, u' = 1.5 m/s along Y axis
m'' = 100 g, u'' = v
Let the speed of the third piece is v makes an angle A from the X axis.
use conservation of momentum along X axis
0 = 320 x 2 + 100 x v cos A
v cos A = - 6.4 ..... (1)
Use conservation of momentum along Y axis
0 = 355 x 1.5 + 100 x v sin A
v sinA = - 5.3 ... (2)
Squaring and adding
[tex]v^2 = (-6.4)^2 +(-5.3)^2\\\\v= 8.3 m/s[/tex]
The angle is given by
[tex]tan A = \frac{-5.3}{-6.4}\\\\A = 219.6 degree[/tex] from + X axis
Physics question on picture
Answer:
B. according to Newton's Third Law of Motion, the force of the Moon on the Earth and the force of the Earth on the Moon are equal in magnitude and opposite in direction
What is the current in milliamperes produced by the solar cells of a pocket calculator through which 9.00 C of charge passes in 8.50 h
Answer:
Current = 0.000294 A
Explanation:
Below is the given values:
Given the charge = 9.00 C
Time = 8.50 h
Use the below formula to find the current:
Current = Q / t
Now plug the values:
Current = 9 / (8.5 x 3600)
Current = 0.000294 A
A glass block in air has critical angle of 49. What will happen to a ray of light coming through the glass when it is incident at and angle of 50 at the glass air boundary? Illustrate with a diagram
Answer:
b
Explanation:
A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with the axis. On the left and right faces, the field has a strength and , respectively. The field along the front and back faces has strengths and . The field at the bottom and top faces has strengths and , respectively. What is the total charge enclosed by the cube
Complete Question
Complete Question is attached below
Answer:
[tex]q=1.558*10^{-9}c[/tex]
Explanation:
From the question we are told that:
Side length s=1.13m
Left field strength [tex]E_l=784.75N/m[/tex]
Right field strength [tex]E_r=776.38 N/m[/tex]
Front field strength [tex]E_f=725.5 N/m[/tex]
Back field strength [tex]E_b=749.54 N/m[/tex]
Top field strength [tex]E_t=944.95 N/m[/tex]
Bottom field strength [tex]E_{bo}=1082.58 N/m[/tex]
Generally, the equation for Charge flux is mathematically given by
[tex]\phi=EAcos\theta[/tex]
Where
Theta for Right,Left,Front and Back are at an angle 90
[tex]cos 90=0[/tex]
Therefore
[tex]\phi =0[/tex] with respect to Right,Left,Front and Back
Generally, the equation for Charge Flux is mathematically also given by
[tex]\phi=\frac{q}{e_o}[/tex]
Where
[tex]Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2[/tex]
Therefore
[tex]Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t[/tex]
[tex]Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)[/tex]
[tex]Q_{net}=176N/C m^2[/tex]
Giving
[tex]q=\phi*e_0[/tex]
[tex]q=176N/C m^2*1.558*10^{-12}c[/tex]
[tex]q=1.558*10^{-9}c[/tex]
The index of refraction for a vacuum is 1.00000. The index of refraction for air is 1.00029. 1) Determine the ratio of time required for light to travel through 1000 m of air to the time required for light to travel through 1000 m of vacuum. (Express your answer to six significant figures.)
Answer:
[tex]\frac{t_{air}}{t_{vaccum}}[/tex] = 1.00029
Explanation:
The refractive index is defined
n = c / v
v = c / n
the speed of light per se wave is constant, so we can use the relations of uniform motion
v = x / t
t = x / v
we substitute
t = x n / c
let's calculate the time
vacuum
t₁ = 1000 1/3 10⁸
t₁ = 3.333333 10⁻⁶ s
air
t₂ = 1000 1.00029 / 3 10⁸
t2 = 3.3343 10⁻⁶ s
the relationship between these times is
t₂ / t₁ = 3.3343 / 3.3333333
t₂ / t₁ = 1.00029
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge
Answer:
I think it is of science is it true na i knew it bro dont take tension
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
C
Explanation:
Similarity
The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.
Required:
a. What is the net force on the side of the container?
b. How long does it take and how much the water level will drop till water no longer comes out of the faucet?
Answer:
a) F = 2.66 10⁴ N, b) h = 1.55 m
Explanation:
For this fluid exercise we use that the pressure at the tap point is
Exterior
P₂ = P₀ = 1.01 105 Pa
inside
P₁ = P₀ + ρ g h
the liquid is water with a density of ρ=1000 km / m³
P₁ = 0.85 1.01 10⁵ + 1000 9.8 5
P₁ = 85850 + 49000
P₁ = 1.3485 10⁵ Pa
the net force is
ΔP = P₁- P₂
Δp = 1.3485 10⁵ - 1.01 10⁵
ΔP = 3.385 10⁴ Pa
Let's use the definition of pressure
P = Fe / A
F = P A
the area of a circle is
A = pi r² = [i d ^ 2/4
let's reduce the units to the SI system
d = 100 cm (1 m / 100 cm) = 1 m
F = 3.385 104 pi / 4 (1) ²
F = 2.66 10⁴ N
b) the height for which the pressures are in equilibrium is
P₁ = P₂
0.85 P₀ + ρ g h = P₀
h = [tex]\frac{P_o ( 1-0.850)}{\rho \ g}[/tex]
h = [tex]\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}[/tex]
h = 1.55 m
To get maximum current in a circuit, the resistance should be in _____
1)series
2)parallel
Answer:
no parallel is the correct answer
A measurement was made of the magnetic field due to a tornado, and the result was 13.00 nT to the north. The measurement was made at a position 8.90 km west of the tornado. What was the magnitude (in A) and direction of the current in the funnel of the tornado? Assume the vortex was a long, straight wire carrying a current.
Answer:
4
Explanation:
The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.
Answer:
A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J
Explanation:
In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.
In this case let's start by finding the equivalent capacitance.
A) Let's solve the part where C1 and C3 are. These two capacitors are in serious
[tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex] (you has an mistake in the formula)
[tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]
[tex]\frac{1}{C_{eq1}}[/tex] = 0.1 10⁶
[tex]C_{eq1}[/tex] = 10 10⁻⁶ F
capacitors C₂, C₄ and C₅ are in series
[tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]
[tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]
[tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶
[tex]C_{eq2}[/tex] = 5 10⁻⁶ F
the two equivalent capacitors are in parallel therefore
C_{eq} = C_{eq1} + C_{eq2}
C_{eq} = (10 + 5) 10⁻⁶
C_{eq} = 15 10⁻⁶ F
B) the energy stored in C₃
The charge on the parallel voltage is constant
is the sum of the charge on each branch
Q = C_{eq} V
Q = 15 10⁻⁶ 6
Q = 90 10⁻⁶ C
the charge on each branch is
Q₁ = Ceq1 V
Q₁ = 10 10⁻⁶ 6
Q₁ = 60 10⁻⁶ C
Q₂ = C_{eq2} V
Q₂ = 5 10⁻⁶ 6
Q₂ = 30 10⁻⁶ C
now let's analyze the load on each branch
Branch C₁ and C₃
In series combination the charge is constant Q = Q₁ = Q₃
U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]
U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]
U₃ = 3 J
In Branch C₂, C₄, C₅
since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅
U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]
U₄ = 0.5 J
A police car travels towards a stationary observer at a speed of 15m/s. the siren on the car emits a sound of frequency 250Hz. Calculate the observer frequency. the speed of sound is 340m/s
Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)
= 250 x (340/( 340-15))
= 261.54 Hz
Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến
95oC for 20 minutes.
Answer:
P = 2439.5 W = 2.439 KW
Explanation:
First, we will find the mass of the water:
Mass = (Density)(Volume)
Mass = m = (1 kg/L)(10 L)
m = 10 kg
Now, we will find the energy required to heat the water between given temperature limits:
E = mCΔT
where,
E = energy = ?
C = specific heat capacity of water = 4182 J/kg.°C
ΔT = change in temperature = 95°C - 25°C = 70°C
Therefore,
E = (10 kg)(4182 J/kg.°C)(70°C)
E = 2.927 x 10⁶ J
Now, the power required will be:
[tex]Power = P = \frac{E}{t}[/tex]
where,
t = time = (20 min)(60 s/1 min) = 1200 s
Therefore,
[tex]P = \frac{2.927\ x\ 10^6\ J}{1200\ s}[/tex]
P = 2439.5 W = 2.439 KW
Catching a wave, a 77-kg surfer starts with a speed of 1.3 m>s, drops through a height of 1.65 m, and ends with a speed of 8.2 m>s. How much nonconservative work was done on the surfer
Answer:
W = 2523.67 J
Explanation:
Given that,
The mass of surfer, m = 77 kg
He starts with a speed of 1.3 m/s
It drops through a height of 1.65 m and ends with a speed of 8.2 m/s.
We know that, the work done is equal to the change in kinetic energy. So,
[tex]W=\dfrac{1}{2}m(v_2^2-v_1^2)\\\\=\dfrac{1}{2}\times 77\times (8.2^2-1.3^2)\\W=2523.67\ J[/tex]
So, the required work done is equal to 2523.67 J.
The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases uniformly to 36.5 rad/s in 5.75 seconds. What is the magnitude of the fan's angular acceleration
Answer: [tex]1.79\ rad/s^2[/tex]
Explanation:
Given
Initial angular speed is [tex]\omega_1=26.2\ rad/s[/tex]
Final angular speed is [tex]\omega_2=36.5\ rad/s[/tex]
Time period [tex]t=5.75\ s[/tex]
Magnitude of the fan's acceleration is given by
[tex]\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}[/tex]
Insert the values
[tex]\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2[/tex]
Thus, fan angular acceleration is [tex]1.79\ rad/s^2[/tex]
Answer:
The angular acceleration is given by 1.8 rad/s^2.
Explanation:
initial angular speed, wo = 26.2 rad/s
final angular velocity, w = 36.5 rad/s
time, t = 5.75 seconds
The first equation of motion is
[tex]w = wo + \alpha t\\\\36.5 = 26.2 + 5.75\alpha\\\\\alpha = 1.8 rad/s^2[/tex]
pha của dao động làm hàm
Answer:
pha của dao động là hàm bậc nhất của thời gian.
During the Moment of Inertia experiment, a group of students decided to mount a solid sphere on the top of the horizontal disk. The sphere and disk have the same mass, M, and radius, R. They recorded the values for the initial and final positions for the falling mass, and the height above the floor when falling mass is at its lowest position. These values are respectively 90cm, 68.5 cm, and 23.5cm. The radius of the pulley, r, was 1/10 of the radius of the sphere. Furthermore, the mass of the disk, M, was 1.4kg, what was the mass, m, for the falling mass
Answer:
u are the very good person I know if u will do itself u will becam3 a rising star
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop
Answer:
The correct solution is "122.2211".
Explanation:
Given:
deceleration,
a = 22 ft/sec²
Initial velocity,
[tex]V_i=50 \ m/h[/tex]
Now,
[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]
[tex]=73.333 \ ft/sec[/tex]
Now,
Final velocity,
[tex]V_f=0[/tex]
Initial velocity,
[tex]V_{initial} = 73.333 \ ft/sec[/tex]
hence,
⇒ [tex]V_f^2=V_i^2+2aD[/tex]
By putting the values, we get
[tex]0=(73.333)^2+2\times( -22) D[/tex]
[tex]44D=(73.333)^2[/tex]
[tex]D=\frac{(73.333)^2}{44}[/tex]
[tex]=122.2211[/tex]
NEED AN ANSWER QUICKLY PLEASE!!
If the length and number of turns of a solenoid are doubled strength of magnetic field will :
(a) Be doubled (b) become half (c) not change d) be four time
Answer:
c). It wouldn't change.
Explanation:
[tex]{ \bf{F = \frac{ \ \gamma _{o}NI }{l} }}[/tex]
An oscillating dipole antenna 1.73 m long with a maximum 36.0 mV potential creates a 500 Hz electromagnetic wave. (a) What is the maximum electric field strength created
Answer:
[tex]E_0=0.021v/m[/tex]
Explanation:
From the question we are told that:
Length [tex]l=1.73m[/tex]
Voltage [tex]V=36.0mV[/tex]
Frequency [tex]F=500Hz[/tex]
Generally the equation for maximum Electric Field is mathematically given by
[tex]E_0=\frac{\triangle}{r}[/tex]
[tex]E_0=\frac{36*10^{-3}}{1.73}[/tex]
[tex]E_0=0.021v/m[/tex]
Question 15
Calculate the velocity of a body if its total energy is three times its rest energy
OA 0.54c
OB. 0.760
OC0.94c
OD.C
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Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned between the electric and magnetic fields?
1. The energy density of an electromagnetic wave is 25% in the magnetic field and 75% in the electric field.
2. The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
3. The energy density of an electromagnetic wave is entirely in the magnetic field.
4. The energy density of an electromagnetic wave is 25% in the electric field and 75% in the magnetic field.
5. The energy density of an electromagnetic wave is entirely in the electric field
Answer:
Option (2) is correct.
The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Explanation:
An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.
So, the correct option is (2).
The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.
The given problem is based on the concept and fundamentals of electromagnetic waves. The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.
In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.
Learn more about the electromagnetic waves here:
https://brainly.com/question/25559554
The reason why a teacher is more important then a farmer
Answer:
A teacher is more important than a famer.
Explanation:
A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.
Work-Energy Theorem & Power
A 0.5 kg mass sitting on smooth ice is accelerated from rest by a force until is
acquires a speed of 8 m/s. The force acts while the mass moves through a
displacement of 2 m.
A. Calculate the kinetic energy of the mass after the force acts.
B. Calculate the work done by the force.
C. Calculate the magnitude of the force that accelerated the mass.
Answer:
A. 16 J
B. 16 J
C. 8 N
Explanation:
A. Determination of the kinetic energy.
Mass (m) = 0.5 Kg
Velocity (v) =. 8 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 8²
KE = ½ × 0.5 × 64
KE = 0.5 × 32
KE = 16 J
B. Determination of the Workdone by the force.
Kinetic energy (KE) = 16 J
Workdone =.?
Workdone and kinetic energy has the same unit of measurement. Thus,
Workdone = kinetic energy
Workdone = 16 J
C. Determination of the force.
Workdone (Wd) = 16 J
Displacement (s) = 2 m
Force (F) =?
Wd = F × s
16 = F × 2
Divide both side by 2
F = 16 / 2
F = 8 N
A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.
1. How many revolutions do the kids make before the constant operational speed is reached ?
2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.
3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?
Answer:
we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer
The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars
Answer:
Explanation:
I’ll help
An initially motionless test car is accelerated uniformly to 105 km/h in 8.43 s before striking a simulated deer. The car is in contact with the faux fawn for 0.635 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
acceleration before collision:
3.45
m/s2
What is the magnitude of the average acceleration of the car during the collision?
average acceleration during collision:
19.68
m/s2
What is the magnitude of the average acceleration of the car during the entire test, from when the car first begins moving until the collision is over?
105 km/h ≈ 29.2 m/s
60.0 km/h ≈ 16.7 m/s
Before the collision the test car has an acceleration a of
a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²
During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is
a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²
i.e. with magnitude about 19.7 m/s².
Overall, the car has an average acceleration of
a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²
