Answer:
[tex]power = \frac{work \: done}{time \: passed} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
power=2525J/19S
=132.8947368421JS^-1
=132.9W
An ac series circuit contains a resistor of 20 ohms, a capacitor of 0.75 microfarads of 120 x 10-3 H. If an effective (rms) voltage of 120 V is applied, what is the effective (rms) current when the circuit is in resonance
Answer:
The effective (rms) current when the circuit is in resonance is 6 A
Explanation:
Given;
resistance of the resistor, R = 20 ohms
capacitance of the capacitor, C = 0.75 microfarads
inductance of the inductor, L = 0.12 H
effective rms voltage, [tex]V_{rms}[/tex] = 120
At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).
The effective (rms) current, = [tex]V_{rms}[/tex] / R
= 120 / 20
= 6 A
Therefore, the effective (rms) current when the circuit is in resonance is 6 A
A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.7 cm farther apart, the voltage between the plates increases by 100 V.(a) What is the charge Q on the positive plate of the capacitor?_________nC(b) How much does the energy stored in the capacitor increase due to the movement of the plates?_________µJ
Answer:
a) Q = 0.759µCb) E = 39.5µJExplanation:
a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV
C = capacitance of the capacitor (in Farads )
V = voltage (in volts) = 100V
C = ∈A/d
∈ = permittivity of free space = 8.85 × 10^-12 F/m
A = cross sectional area = 600 cm²
d= distance between the plates = 0.7cm
C = 8.85 × 10^-12 * 600/0.7
C = 7.59*10^-9Farads
Q = 7.59*10^-9 * 100
Q = 7.59*10^-7Coulombs
Q = 0.759*10^-6C
Q = 0.759µC
b) Energy stored in a capacitor is expressed as E = 1/2CV²
E = 1/2 * 7.59*10^-9 * 100²
E = 0.0000395Joules
E = 39.5*10^-6Joules
E = 39.5µJ
A) The charge Q on the positive plate of the capacitor is ; 0.759 µC
B) The energy stored in the capacitor increases by : 39.5 µJ
Given data :
Area of plates ( A ) = 600 cm²
Distance between plates ( d ) = 0.7 cm
Voltage across plates = 100 v
∈ ( permittivity of free space ) = 8.85 * 10⁻¹²
A) Determine the Charge on the positive plate of the capacitor
Q = CV --- ( 1 )
where ; C = ∈ * A / d and V = 100 v
∴ C = 8.85 * 10⁻¹² * 600 / 0.7 = 7.59 *10⁻⁹ F
Back to equation ( 1 )
Q = 7.59 *10⁻⁹ * 100
= 0.759 µC
B) Calculate how much The energy stored in the capacitor increases
E = 1/2 * C * V²
= 1/2 * 7.59 *10⁻⁹ * 100²
= 39.5 µJ
Hence we can conclude that The charge Q on the positive plate of the capacitor is ; 0.759 µC, The energy stored in the capacitor increases by : 39.5 µJ.
Learn more : https://brainly.com/question/1457596
Near the end of a marathon race, the first two runners are separated by a distance of 41.6 m. The front runner has a velocity of 3.4 m/s, and the second a velocity of 4.85 m/s.What is the magnitude of the velocity of the second runner relative to the first?
If the front runner is 215 m from the finish line, who will win the race, assuming they run at constant velocity?
By what distance does the winning runner finish ahead of the next runner?
Answer:
v₂₁ = 1.45 m/s
Second runner is the winner.
Δs = 35.1 m
Explanation:
For the relative velocity, we use the formula:
v₂₁ = v₂ - v₁
where,
v₂₁ = relative velocity of second runner with respect to first runner = ?
v₁ = velocity of first runner = 3.4 m/s
v₂ = velocity of second runner = 4.85 m/s
Therefore,
v₂₁ = 4.85 m/s - 3.4 m/s
v₂₁ = 1.45 m/s
Now, for finding the winner, we calculate the time taken by both the runners to reach finish line, by using following equation:
s = vt
t = s/v
for first runner:
t₁ = (215 m)/(3.4 m/s)
t₁ = 63.23 s
for 2nd runner:
t₂ = (215 m + 41.6 m)/(4.85 m/s)
t₂ = 52.9 s
Since, t₂<t₁.
Therefore, second runner is the winner.
Now, for the difference between runners at the time of winning, we first calculate the distance covered by first runner at that time. Using second equation of motion:
s = (3.4 m/s)(52.9 s)
s = 179.9 m
So, the distance by which the second runner finishes ahead of the first runner is given as follows:
Δs = 215 m - 179.9 m
Δs = 35.1 m
Objects floating in the water, like buoys, only bob up and down when waves pass. Why do they not get pushed all the way to wherever the wave goes
Answer:
Because as the waves propagates, the particles of the medium (molecules of water) vibrates perpendicularly (upward and downward) about their mean position and not in the direction of the waves.
Explanation:
A wave is a phenomena which causes a disturbance in a medium without any permanent deformation to the medium. Examples are; transverse wave and longitudinal wave. Waves transfer energy from one point in the medium to another.
The waves generated by water are transverse waves. Which are waves in which the vibrations of the particles of the medium is perpendicular to the direction of propagation of the waves.
Thus as the waves propagates, the molecules of water vibrates up and down and not along the direction of propagation of the waves. So that the floating objects do not get pushed in the direction of the waves every time.
A generator is connected to a resistor and a 0.049-H inductor in series. The rms voltage across the generator is 7.9 V. When the generator frequency is set to 100 Hz, the rms voltage across the inductor is 2.8 V. Determine the resistance of the resistor in this circuit
Answer:
56.04 ohmsExplanation:
The voltage across the inductor VL = IXL
I is the total current flowing in the circuit and XL is the inductive reactance.
First we need to get the current flowing in the circuit.
From the expression above;
I = VL/XL
Since XL = 2πfL
I = VL/ 2πfL
Given VL = 2.8V, frequancy f = 100Hz and inductance L = 0.049-H
I = 2.8/2π*100*0.049
I = 2.8/30.79
I = 0.091A
Also;
Vrms = VL + VR
VR is the voltage across the resistor.
VR = Vrms - VL
VR = 7.9 - 2.8
VR = 5.1V
Then we can calculate the resistance of the resistor
According to ohms law VR = IR
Since the inductance and resistance ar connected in series, the same current will flow through them.
R = VR/I
R = 5.1/0.091
R = 56.04 ohms
Hence the resistance of the resistor in this circuit is 56.04 ohms
7. Which statement is true about teens that are in Marcia’s final state of identity formation?
Answer:
D. All of the above
Explanation:
The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.
James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure, moratorium, and achievement.
His eyes are 1.64 m above the floor and the top of his head is 0.14 m higher. Find the height (in m) above the floor of the top and bottom of the smallest mirror in which he can see both the top of his head and his feet.
Answer:
Height above the floor from the bottom is 0.82m
Form the top is 1.71m
Explanation:
See attached file
When a force of 20.0 N is applied to a spring, it elongates 0.20 m. Determine the period of oscillation of a 4.0-kg object suspended from this spring.
Answer:
1.26 secs.
Explanation:
The following data were obtained from the question:
Force (F) = 20 N
Extention (e) = 0.2 m
Mass (m) = 4 Kg
Period (T) =.?
Next, we shall determine the spring constant, K for spring.
The spring constant, K can be obtained as follow:
Force (F) = 20 N
Extention (e) = 0.2 m
Spring constant (K) =..?
F = Ke
20 = K x 0.2
Divide both side by 0.2
K = 20/0.2
K = 100 N/m
Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:
Mass (m) = 4 Kg
Spring constant (K) = 100 N/m
Period (T) =..?
T = 2π√(m/K)
T = 2π√(4/100)
T = 2π x √(0.04)
T = 2π x 0.2
T = 1.26 secs.
Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.
Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge q3=20μCq3=20μC situated there?
Answer:
a) E = 2.7x10⁶ N/C
b) F = 54 N
Explanation:
a) The electric field can be calculated as follows:
[tex] E = \frac{Kq}{d^{2}} [/tex]
Where:
K: is the Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
d: is the distance
Now, we need to find the electric field due to charge 1:
[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]
The electric field due to charge 2 is:
[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]
The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):
[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]
Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.
b) The force on a charge q₃ situated there is given by:
[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]
[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]
Therefore, the force on a charge q₃ situated there is 54 N.
I hope it helps you!
(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].
(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.
The answer can be explained as follows.
Electric FieldGiven that the two charges are;
[tex]q_1 = 50\times 10^{-6}\,C[/tex] and [tex]q_2 = -25\times 10^{-6}\,C[/tex](a) At the midpoint; [tex]r = 0.5\,m[/tex].
We know that the electric field due to charge [tex]q_1[/tex].
[tex]E_1 = k\,\frac{q_1}{r^2}[/tex]Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]
[tex]E_1 = (9\times 10^9) \times\frac{(50 \times 10^{-6})}{(0.5)^2}=1.8\times 10^6N/C[/tex]The electric field due to charge [tex]q_2[/tex] is given by;
[tex]E_2 = (9\times 10^9) \times\frac{(-25 \times 10^{-6})}{(0.5)^2}=-9\times 10^5\,N/C[/tex]Therefore, the net electric field in the midpoint is given by;
[tex]E_{net} =E_2+E_1[/tex][tex]\implies E_{net}=1.8 \times 10^6 N/C + 9 \times 10^5\,N/C=2.7\times 10^6\,N/C[/tex]The direction is towards the right side.
Electrostatic Force(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.
So the force on the charge is ;
[tex]F=E_{net} \times q_3=(2.7 \times 10^6\,N/C) \times (20\times 10^{-6}\,C)=54\,N[/tex]Find out more about electrostatic force and fields here:
https://brainly.com/question/14621988
What must the charge (sign and magnitude) of a 1.60 g particle be for it to remain balanced against gravity when placed in a downward-directed electric field of magnitude 680 N/C
Answer:
Explanation:
The charge must be negative so that force in a downward electric field will be upward so that its weight is balanced .
Let the charge be - q .
force on charge
= q x E where E is electric field
= q x 680
weight = 1.6 x 10⁻³ x 9.8
so
q x 680 = 1.6 x 10⁻³ x 9.8
q = 1.6 x 10⁻³ x 9.8 / 680
= 23 x 10⁻⁶ C
- 23 μ C .
Run 2 17. Set # of slits to 2 18. Set Wave Length to 400nm 19. Set Slit width to 1600 nm 20. Set Slit spacing to 5000nm In row 18 21. Record distance to 1st bright fringe 22. Record distance to 2nd bright fringe 23. Record distance to 3rd bright fringe Knowing the screen distance to be 1m 24. Calculated the measured angle to 1st bright fringe 25. Calculated the measured angle to 2nd bright fringe 26. Calculated the measured angle to 3rd bright fringe Using sin(θ)=mλ/d 27. Calculate θ for 1st bright fringe
Answer:
a) m=1, y₁ = 0.08 m , θ₁ = 4.57º , b) m=2, y₂ = 0.16 m , θ₂ = 9.09º , c) m=3, y₃ = 0.24 m , θ₃ = 13.5º
Explanation:
After reading your strange statement, I understand that this is an interference problem, I transcribe the data to have it more clearly. Number of slits 2, distance between slits 5000 nm, wavelength 400 nm, distance to the screen 1 m.
They ask us to calculate the angles for the first, second and third interference, they also ask us to write down the distance from the central maximum.
The expression for constructive interference for two slits is
d sin θ = m λ
where d is the distance between the slits, λ is the wavelength used, m is an integer representing the order of interference
Let's use trigonometry to find the distance from the central maximum
tan θ = y / L
in all interference experiments the angle is small,
tan θ = sin θ / cos θ = sin θ
sint θ = y / L
let's replace
d y / L = m λ
y = m λ L / d
let's calculate
distance to the first maximum m = 1
y₁ = 1 400 10⁻⁹ 1/5000 10⁻⁹
y₁ = 0.08 m
distance to second maximum m = 2
y₂ = 2 400 10⁻⁹ 1/5000 10⁻⁹
y₂ = 0.16 m
distance to the third maximum m = 3
y₃ = 3 400 10⁻⁹ 1/5000 10⁻⁹
y₃ = 0.24 m
with these values we can search for each angle
tan θ = y / L
θ = tan⁻¹ y / L
for m = 1
θ₁ = tan⁻¹ (0.08 / 1)
θ₁ = 4.57º
for m = 2
θ₂ = tan⁻¹ (0.16 / 1)
θ₂ = 9.09º
for m = 3
θ₃ = tan⁻¹ (0.24 / 1)
θ₃ = 13.5º
We can see Objects because of
A) reflection
B) refraction
C) transmission
D) diffraction
Please help a friend
Answer:
I believe it's A.)
Explanation:
Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.
Hope this helps you out : )
What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep, if you assumed it was a closed tube
Answer:
f = 357.29Hz
Explanation:
In order to calculate the fundamental frequency in the closed tube, you use the following formula:
[tex]f_n=\frac{nv}{4L}[/tex] (1)
n: order of the mode = 1
v: speed of sound = 343m/s
L: length of the tube = 24cm = 0.24m
You replace the values of the parameters in the equation (1):
[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]
The fundamental frequency of in the tube is 357.29Hz
An airplane propeller is 2.18 m in length (from tip to tip) with mass 97.0 kg and is rotating at 2600 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod.What is its rotational kinetic energy?
Answer:
1.4*10^6 J
Explanation:
Given that
Length of the propeller, l = 2.18 m
Mass of the propeller, m = 97 kg
Speed of the propeller, w = 2600 rpm
The formula for finding rotational Kinetic energy, K is = ½Iw²
Where, I is the moment of Inertia, and is given as 1/12 * m * l²
I = 1/12 * 97 * 2.18²
I = 8.083 * 4.7524
I = 38.41 kgm²
w = 2600 rpm, converting to rad/s, we have
w = 2600 * 2π rad/s
w = 272.31 rad/s
Now, Kinetic Energy, K is
K = ½Iw²
K = ½ * 38.41 * 272.31²
K = 19.205 * 74152.7361
K = 1424103.3 J
K = 1.4 MJ or 1.4*10^6 J
Thus, the rotational Kinetic Energy is 1.4*10^6 J
in a certain region of space, the gravitational field is given by -k/r,where r=distance,k=const.if gravitational potential at r=r0 be v0,then what is the expression for the gravitational potential v?
options
1)k log(r/ro)
2)k log(ro/r)
3)vo+k log(r/ro)
4)vo+k log(ro/r)
plz help me out
I will mark u as brainliest if u answer correct
Answer:
The correct answer is option 3 .
Please check the answer once :)
A ball is thrown at an angle 40.00 above the horizontal with an initial velocity of 22.0 m/s. What is the range of the ball?
Answer:
48.64 m
Explanation:
From the question,
Range(R) = (U²Sin2Ф)/g.................. Equation 1
Where U = initial velocity, Ф = Angle to the horizontal, g = acceleration due to gravity.
Given: U = 22 m/s, Ф = 40°, g = 9.8 m/s².
Substitute these values into equation 1
R = 22²Sin(40×2)/9.8
R = 484×0.9848/9.8
R = 48.64 m
Hence the range of the ball is 48.64 m
us
A 13.3 kg box sliding across the ground
decelerates at 2.42 m/s2. What is the
coefficient of kinetic friction?
(No unit)
Answer:
0.242
Explanation:
m = 13.3 kg
a =d= 2.42 m/s²
g = 10 m/s²
from the laws of friction F = ¶R
===> ¶ = F/R = ma/mg = a/g
¶ = a/g = 2.42/10 = 0.242
I WILL MARK YOU AS BRAINLIEST!!! An object is launched straight up into the air with an initial velocity of 40 meters per second, from a height 30 m above the ground. Assuming that gravity pulls it down, changing its position by about 4.9 /2, after how many seconds will the object hit the ground? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
8.9 seconds
Explanation:
The height of the object at time t is:
y = h + vt − 4.9t²
where h is the initial height, and v is the initial velocity.
Given h = 30 and v = 40:
y = 30 + 40t − 4.9t²
When y = 0:
0 = 30 + 40t − 4.9t²
4.9t² − 40t − 30 = 0
Solving with quadratic formula:
t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)
t = [ 40 ± √(1600 + 588) ] / 9.8
t = 8.9
It takes 8.9 seconds for the object to land.
The position of an object moving along the x
axis is given by x(t) = 2t^2+t^3 +1, where x is
in meters and t in seconds. The acceleration of
the object at t = 2 seconds is:
4m/s?
Answer: 16 meters per second
Explanation:
The derivative of the position is the velocity.
The derivative of the velocity is the acceleration.
x(t) = 2t² + t³ + 1
x'(t) = v(t) = 4t + 3t²
x''(t) = v'(t) = a(t) = 4 + 6t
a(2) = 4 + 6(2)
= 4 + 12
= 16
In one of the classic nuclear physics experiments performed by Ernest Rutherford at the beginning of the 20th century, alpha particles (helium nuclei) were shot at gold nuclei and their paths were substantially affected by the Coulomb repulsion from the nuclei. If the energy of the (doubly charged) alpha nucleus was 5.1 MeV, how close to the gold nucleus (79 protons) could it come before being deflected? r =
Answer:
r = 3.8 × 10 ⁻¹⁴ m
Explanation:
given data
alpha nucleus = 5.1 MeV
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C
Kinetic energy of the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) = 9.564 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
solution
we know that when its kinetic energy is equal to the potential energy than alpha particle will deflect \
so
Kinetic energy = potential energy = k q₁q₂ ÷ r ..................1
here r is close distance the alpha particle
so r will be put here value
r = ( 9 × 10⁹ × 3.2 × 10⁻¹⁹ × 1.264 × 10⁻¹⁷ ) ÷ ( 9.564 × 10⁻¹³ )
r = 3.8 × 10 ⁻¹⁴ m
A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matter it passes through. One such particle, initially traveling at 2.40 ✕ 106 m/s in a straight line, decreases in speed to 1.56 ✕ 106 m/s over a distance of 1.22 km.
(a) What is the magnitude of the force experienced by the muon?
(b) How does this force compare to the weight of the muon?
|F|/Fg =______
Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
"On earth, you have a pendulum of length L that oscillates with period T. Your friend lives on a planet where the acceleration of gravity is four times as big as it is on the earth. What should be the length of a pendulum on your friend s planet so that it also oscillates with the same period T
Answer:
4L
Explanation:
Data provided in the question according to the question is as follows
Length = L
Gravity = G
For friend
Length = ?
Growth = 4G
Moreover,
[tex]T_1 = T_2[/tex]
Based on the above information ,
Now we have to apply the simple pendulum formula which is shown below:
[tex]T = 2\pi \frac{L}{G}[/tex]
Now equates these equations in both sides
[tex]2\pi \frac{L}{G} = 2\pi \frac{L}{4G}[/tex]
So after solving this, the length of the pendulum is 4L
Answer:
the length of a pendulum on your friend s planet should be 4 times than that on earth
Explanation:
We know that time period of simple pendulum is given by
[tex]T= 2\pi\sqrt{\frac{L}{g} }[/tex]
L= length of pendulum
g= acceleration due to gravity
therefore, Let T_1 and T_2 be the time period of the earth and other planet respectively.
[tex]\frac{T_1}{T_2} =\sqrt(\frac{L_1}{L_2}\times\frac{g_2}{g_1})[/tex]
ATQ
T_1=T_2=T, g_2=4g_1
Putting the values in above equation and solving we get
[tex]\frac{L_1}{L_2} =\frac{1}{4}[/tex]
A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evporate
Answer:
Gain heat and evaporate
Explanation:
As water molecules are exposed to sunlight, they begin to heat up. This means that the molecules begin to jiggle faster, and as such take up less space. Since they take up less space they are less dense, and therefore more bouyant. This means that they begin to rise into the air, and evaporate. Hope this helps!
A motorcycle has a constant acceleration of 3.49 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a)29.0 to 39.0 m/s, and (b)59.0 to 69.0 m/s?
Answer:
(a)2.865 s
(b)2.865 s
Explanation:
We are given that
Acceleration,a=[tex]3.49 m/s^2[/tex]
a.Initial speed,u=29 m/s
Final speed,v=39 m/s
We know that
[tex]t=\frac{v-u}{a}[/tex]
Using the formula
[tex]t=\frac{39-29}{3.49}=2.865 s[/tex]
b.Initial speed,u=59 m/s
Final speed,v=69 m/s
Again using the formula
[tex]t=\frac{69-59}{3.49}=2.865 s[/tex]
A 15-m rope is pulled taut with a tension of 140 N. It takes 0.545 s for a wave to propagate along the rope. What is the mass of the rope
Answer:
Mass of the rope = 2.8 kg
Explanation:
The speed of waves travelling through a rope with linear density (μ) and under tension T is given as v = √(T/μ)
The speed of waves in the rope is also calculated as
v = (d/t)
d = L = length of the rope = 15 m
t = time taken for the wave to move through the rope = 0.545 s
Speed = v = (15/0.545) = 27.523 m/s
Speed = v = √(T/μ)
T = tension in the rope = 140 N
μ = linear density = ?
27.523 = √(140/μ)
27.523² = (140/μ)
(140/μ) = 757.512
μ = (140/757.512) = 0.1848155556 = 0.1848 kg/m
Linear density = μ = (m/L)
m = mass of the rope = ?
L = length of the rope = 15 m
0.1848 = (m/15)
m = 0.1848 × 15 = 2.77 kg = 2.8 kg to 1 d.p.
Hope this Helps!!!
f the mass of the block is 2 kg, the radius of the circle is 0.8 m, and the speed of the block is 3 m/s, what is the tension in the string at the top of the circle
Answer:
the size are components relative to the whole.
Explanation:
they are particularly good at showing percentage or proportional data
. A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens must the film be
Answer:
The film must be 0.051 m away from the lensExplanation:
Using the lens formula 1/f = 1/u + 1/v
f = focal length of the lens
u = object distance
v = image distance
Given focal length f = 50.0mm = 50/1000 = 0.05m
Note that a camera uses a convex lens and the focal length of a convex lens is positive.
object distance u = 5.00m
To know how far from the lens must the film be, we need to calculate the image distance v.
from the formula above;
1/v = 1/f - 1/u
1/v = 1/0.05 - 1/3
1/v = 20-0.33
1/v = 19.67
v = 1/19.67
v = 0.051m
Hence, the film must be 0.051 m away from the lens
In Einstein's Thoery of Relativity. What did he believe was the relationship between energy and malter?
Explanation:
Einsteins theory of relativity explains how space and time are linked for objects that are moving at a consistent speed in a straight line.
Consider a circular region of a plane. If there is a changing electric field perpendicular to that plane, what can you say about any magnetic field in the circular region of the plane?
Complete Question
Consider a circular region of a plane. If there is a changing electric field perpendicular to that plane, what can you say about any magnetic field in the circular region of the plane?
A :
The magnetic field in the plane will exist in the form of closed concentric loops in the circular region.
B :
Nothing. An electric field says nothing about a magnetic field.
C :
The magnetic field will be perpendicular to the plane.
D :
There will be no magnetic field.
Answer:
Option A is the correct answer
Explanation:
Generally when there is a change in electric field, magnetic field is produced and this magnetic field is perpendicular to the changing electric field
Now from the question we are told that the changing electric field is moving in the direction perpendicular to the plane so the generated magnetic field will exist on the plane in the form of closed concentric loops in the circular region.
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale. What is the magnitude of the velocity of the shell as it hits the ground?
Answer:
[tex]V = 331.6946\ m/s[/tex]
Explanation:
First let's find the time that takes the shell to hit the ground (height zero).
To find this time, we can use the equation:
[tex]S = So + Vo*t + at^2/2[/tex]
Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time. Then, for the vertical movement of the shell, we have that:
[tex]0 = 80 + 0*t - 9.81*t^2/2[/tex]
[tex]9.81*t^2/2 = 80[/tex]
[tex]t^2 = 160/9.81 = 16.31[/tex]
[tex]t =4.0386\ seconds[/tex]
Now, to find the horizontal speed, we use the equation:
[tex]S = So + V*t[/tex]
Then, for the horizontal movement, we have:
[tex]1330 = 0 + V_h * 4.0386[/tex]
[tex]V_h = 1330/4.0386 =329.32\ m/s[/tex]
Now we need to find the vertical speed, using:
[tex]V = Vo + a*t[/tex]
[tex]V_v = 0 - 9.81*4.0386[/tex]
[tex]V_v = -39.6187\ m/s[/tex]
Finally, to find the magnitude of the velocity, we have:
[tex]V = \sqrt{V_h^2 + V_v^2}[/tex]
[tex]V = \sqrt{329.32^2 + (-39.6187)^2}[/tex]
[tex]V = 331.6946\ m/s[/tex]