Answer:
a) y = 0.98 t², t=1s y= 0.98 m,
b) he two blocks must move the same distance
c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m
Explanation:
For this exercise we can use Newton's second law
Big Block
Y axis
N-W = 0
N = M g
X axis
T- fr = Ma
the friction force has the expression
fr = μ N
fr = μ Mg
small block
w- T = m a
we write the system of equations
T - fr = M a
mg - T = m a
we add and resolved
mg- μ Mg = (M + m) a
a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]
a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]
a = 9.8 (6/30)
a = 1.96 m / s²
a) now we can use the kinematic relations
y = v₀ t + ½ a t²
the blocks come out of rest so their initial velocity is zero
y = ½ a t²
y = ½ 1.96 t²
y = 0.98 t²
for t = 1s y = 0.98 m
t = 2s y = 1.96 m
b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.
As the curda is in tension the two blocks must move the same distance
c) the velocity of the block M
v = vo + a t
v = 0 + 1.96 t
for t = 1 s v = 1.96 m / s
t = 2 s v = 3.92 m / s
d) the deceleration if the chain is cut
when removing the chain the tension becomes zero
-fr = M a
- μ M g = M a
a = - μ g
a = - 0.2 9.8
a = -1.96 m / s²
e) the distance to stop the block is
v² = vo² - 2 a x
0 = vo² - 2a x
x = vo² / 2a
x = 1.96² / 2 1.96
x = 0.98 m
the time to travel this distance is
v = vo - a t
t = vo / a
t = 1.96 /1.96
t = 1 s
You want to swim from one side of a river to another side. Assume your speed is three miles per hour in the west direction, with negligible water velocity. When you reach a certain point, you will encounter water flow with a velocity of 6.2 miles per hour in the north direction. What is your resultant speed and direction
Answer:
speed = 6.71 mph and angle is 71.2 degree.
Explanation:
speed of person, u = 3 miles per hour
speed of water, v = 6 miles per hour
Resultant speed
[tex]V =\sqrt{v^2 + u^2}\\\\V = \sqrt{3^2 + 6^2}\\\\V = 6.71 mph[/tex]
The angle from the west is
tan A = 6/2 = 3
A = 71.6 degree
What is the order of magnitude of the distance of Sun to nearest star in meters?
Answer:
Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.
Explanation:
Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.
Let's model the Milky Way galaxy as a disk.
The volume of a disk is:
V
=
π
⋅
r
2
⋅
h
Plugging in our numbers (and assuming that
π
≈
3
)
V
=
π
⋅
(
10
21
m
)
2
⋅
(
10
19
m
)
V
=
3
×
10
61
m
3
Is the approximate volume of the Milky Way.
Now, all we need to do is find how many stars per cubic meter (
ρ
) are in the Milky Way and we can find the total number of stars.
Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of
4
×
10
16
m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.
ρ
=
n
V
Using the volume of a sphere
V
=
4
3
π
r
3
ρ
=
1
4
3
π
(
4
×
10
16
m
)
3
ρ
=
1
256
10
−
48
stars /
m
3
Going back to the density equation:
ρ
=
n
V
n
=
ρ
V
Plugging in the density of the solar neighborhood and the volume of the Milky Way:
n
=
(
1
256
10
−
48
m
−
3
)
⋅
(
3
×
10
61
m
3
)
n
=
3
256
10
13
n
=
1
×
10
11
stars (or 100 billion stars)
Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.
Answer:
Mercury, 46,001,272 km from the sun at the nearest point.
Explanation:
RATIO of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is:
1) 7/29
2) 9/31
3) 5/27
4) 5/23
Answer:
[tex]5/27[/tex]
Explanation:
wavelengths for Lyman series
[tex]\lambda=\frac{1}{R(1-\frac{1}{4} })=\frac{4}{3R}[/tex]
wavelengths for Balmer series
[tex]\lambda_B=\frac{1}{R(\frac{1}{4}-\frac{1}{9}) } =\frac{1}{R(\frac{5}{36}) } =\frac{36}{5R}[/tex]
[tex]\frac{ \lambda_L}{ \lambda_B} =\frac{4}{3R} \times\frac{5R}{36} =5/27[/tex]
OAmalOHopeO
The ratio of longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is 5/27. The correct option is 3.
What is Lyman and Balmer series?
Lyman and Balmer series are sets of spectral lines in the emission spectrum of hydrogen, which result from the transitions of the electron from higher energy levels to lower energy levels.
The Lyman series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=1 energy level. These transitions release energy in the form of ultraviolet photons. The lowest energy level in hydrogen is the n=1 energy level, which is also called the ground state. Therefore, the Lyman series includes the transition of the electron from any energy level greater than or equal to n=2 to the ground state.
The Balmer series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=2 energy level. These transitions release energy in the form of visible photons. The lowest energy level in the Balmer series is the n=2 energy level. Therefore, the Balmer series includes the transition of the electron from any energy level greater than or equal to n=3 to the n=2 energy level.
Lyman and Balmer's series are named after the scientists who discovered them. The Lyman series is named after Theodore Lyman, an American physicist who discovered the series in 1906. The Balmer series is named after Johann Balmer, a Swiss mathematician who discovered the series in 1885.
Here in the Question,
The longest wavelength in the Lyman series of the hydrogen spectrum corresponds to the transition from the n = 2 energy level to the n = 1 energy level, while the longest wavelength in the Balmer series corresponds to the transition from the n = 3 energy level to the n = 2 energy level.
The wavelengths of these transitions can be calculated using the Rydberg formula:
1/λ = R(1/n1^2 - 1/n2^2)
where λ is the wavelength of the photon emitted, R is the Rydberg constant (1.097 × 10^7 m^-1), and n1 and n2 are the initial and final energy levels of the electron.
For the longest wavelength in the Lyman series, we have n1 = 2 and n2 = 1, so:
1/λ_lyman = R(1/2^2 - 1/1^2) = 3R/4
For the longest wavelength in the Balmer series, we have n1 = 3 and n2 = 2, so:
1/λ_balmer = R(1/3^2 - 1/2^2) = 5R/36
Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is:
λ_lyman/λ_balmer = (3R/4)/(5R/36) = 27/20
Simplifying this ratio gives:
λ_lyman/λ_balmer = 27/20
Multiplying both the numerator and denominator by 1/3R, we get:
λ_lyman/λ_balmer = (1/2)/(1/3) = 3/2
Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is 3:2, or 3/5 in fractional form. Simplifying this ratio gives:
λ_lyman/λ_balmer = 5/3
Taking the reciprocal of both sides, we get:
λ_balmer/λ_lyman = 3/5
Therefore, the correct answer is (3) 5/27.
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Cho hai mặt cầu đồng tâm O tích điện đều. Bán kính của hai mặt cầu lần lượt là R1 và R2 (R2>R1). Điện tích mặt trong là q và mặt ngoài là Q
Tính cường độ điện trường tại một điểm cách tâm O một đoạn r (biết R1 < r < R2)
Tính hiệu điện thế giữa hai mặt cầu
Answer:
you will stc ohxoyxct txxtx xigigjjgjvvixiffjz,iffzikzfjvixii. hi h ohigiogooigoh
Explanation:
k jjvhvojvovvojivivivihvi hj
Earth’s Moon has a diameter of 3,474 km and orbits at an average distance of 384,000 km. At that distance it subtends and angle just slightly larger than half a degree in Earth’s sky. Pluto’s moon Charon has a diameter of 1,186 km and orbits at a distance of 19,600 km from the dwarf planet. Compare the appearance of Charon in Pluto’s skies with the Moon in Earth’s skies. Describe where in the sky Charon would appear as seen from various locations on Pluto.
The reason for arriving at the above solutions is as follows:
The given dimensions and distance from the Earth of the Moon are;
The diameter of the Moon, d = 3,474 km
The average distance of the Moon from the Earth, R = 384,000 km
Required:
The comparison between Charon's appearance in Pluto and the Moon's appearance on Earth Earth
Solution:
The distance of the Moon's travels in an orbit, C = 2·π·R
∴ C = 2 × π × 384,000 km
The angle subtended by the Moon, θ = d/C × 360°
∴ θ = 3,474/(2 × π × 384,000) × 360° ≈ 0.518°
Pluto's moon Charon, has the following parameters;
The diameter of the Charon, d₂ = 1,186 km
The average distance of the Charon from Pluto, R₂ = 19,600 km
Therefore, the distance of the Moon's travels in an orbit, C₂ = 2·π·R₂
∴ C₂ = 2 × π × 19,600 km
The angle subtended by the Moon, θ₂ = d₂/C₂ × 360°
∴ θ₂ = 1,186/(2 × π × 19,900) × 360° ≈ 3.415°
The angle subtended by Charon in Pluto's sky ≈ 3.415°
Charon therefore, appears 7 times larger in Pluto's skies than the Moon's appearance in Earth's skies
Required:
The appearance of Charon as seen from different locations on Pluto
Solution:
Charon is gravitationally locked to Pluto, therefore, the same side of Pluto is faced with the same side of Charon
Therefore;
Charon appears constantly overhead from the side of Pluto locked to CharonCharon appears constantly at the horizon from the poles on either side of the axis of rotation of Pluto and CharonLearn more about Pluto's moon Charon here:
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A 150g copper bowl contains 220g of water, both at 20.0oC, A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.00 g being converted to steam. The final temperature of the system is 100oC, Neglect energy transfers with the environment.
a) How much energy (in calories) is transfered to the water as heat?
b) How much to the bowl?
c) What is the original temperature of the cylinder?
We have that the energy (in calories) is transferred to the water as heat,to the bowl and the original temperature of the cylinder is mathematically given as
Qw=20.3 kcal Q= 1.11 kcal Ti=873°CEnergy
Generally the equation for the is mathematically given as
(a)
The heat transferred to the H20
Qw= CwMwdT+Lvms
Qw=((220g)(100°C-20.0T)+(539 caVg)(5.00 g)
Qw=20.3 kcal .
(b)
The heat transferred to the bowl is
Qb= CbmbdT
Q= (0.0923 cal/gC)(150g)(100°C-20.0°C)
Q= 1.11 kcal
(c)
original temperature of the cylinder
-Qw- Qb = CcMc(T2-T1)
[tex]T1=\frac{Qw+Qm}{CcMc}+T2[/tex]
T1=873C
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01.04 Law of Conservation of Energy
science question
Answer:
law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
krichoffs law of current questions
Answer:
Explanation:
Kirchhoff's Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.
#I AM ILLITERATE
explain why our sweat is salty?
Answer:
Sweat also contains ammonia and urea, which are produced by the body when it breaks down proteins from the foods you eat.
Hope this helps..
elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extension of 2.0cm
Answer:
40g
Explanation:
20g range > 1.0cm
Therefore,
40g range > 2.0cm
A 69.0-kg astronaut is floating in space, luckily he has his trusty 28.0-kg physics book. He throws his physics book and accelerates at 0.0130 m/s2 in the opposite direction. What is the magnitude of the acceleration of the physics book?
Answer:
0.032 [tex]m/s^2[/tex]
Explanation:
Given :
Weight of the astronaut = 69 kg
Weight of the physics book = 28 kg
Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]
The force that is applied on the astronaut :
[tex]F=ma[/tex]
[tex]$=69 \times 0.013$[/tex]
= 0.897 N
Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N
Therefore, the acceleration of the physics book is given by :
[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]
[tex]$a = \frac{0.897}{28}$[/tex]
a = 0.032 [tex]m/s^2[/tex]
Hence, the acceleration of the physics book is 0.032 [tex]m/s^2[/tex].
Answer:
The acceleration of astronaut is 5.27 x 10^-3 m/s^2.
Explanation:
mass of astronaut, M = 69 kg
Mass of book, m = 28 kg
acceleration of book, a = 0.013 m/s^2
Let the acceleration of astronaut is A.
According to the Newton's third law, for every action there is an equal and opposite reaction.
So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.
M A = m a
69 x A = 28 x 0.013
A = 5.27 x 10^-3 m/s^2
Light of a given wavelength is used to illuminate the surface of a metal, however, no photoelectrons are emitted. In order to cause electrons to be ejected from the surface of this metal you should: ___________
a. use light of the same wavelength but increase its intensity.
b. use light of a shorter wavelength.
c. use light of the same wavelength but decrease its intensity.
d. use light of a longer wavelength.
Answer:
use light of the same wavelength but decrease it's intensity
If the horizontal range and the max height of a body projected at an angle titre to the horizontal is K and Q respectively. Show that the muzzle velocity Vo is given by: Vo=√[2gQ + K²/8Q]
Explanation:
Recall that
[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]
and
[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]
From Eqn(2), we can write
[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]
Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as
[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]
Squaring the above equation, we get
[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]
[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]
Use Eqn(3) on Eqn(4) and we will get the following:
[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]
This simplifies to
[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]
Rearranging this further, we get
[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]
Putting [tex]v_0^2[/tex] to the left side, we get
[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]
Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as
[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]
A 1.50 kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s , and at point B it has slowed to 1.25 m/s .
Part A
How much work was done on the book between A and B ?
Part B
If -0.750J of work is done on the book from B to C , how fast is it moving at point C ?
Part C
How fast would it be moving at C if 0.750J of work were done on it from B to C ?
I assume friction is the only force acting on the book as it slides.
(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²
W ≈ -6.56 J
(B) Using the work-energy theorem again, the speed v of the book at point C is such that
-0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²
==> v = 0.750 m/s
(C) Take the left side to be positive, then solve again for v.
0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²
==> v ≈ 1.60 m/s
which watch is more preferable for the measurement of time among pendulum, quartz and atomic watch
Answer:
pendulum, quartz
Explanation:
Do you believe in ghost
Answer:
well its about our thinking but i do believe in ghost a little
a standard bathroom scale is placed on an elevator. A 34 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.4 m/s2
Answer:F = 255 N
Explanation:
It is given that,
Mass of the boy, m = 25 kg
Acceleration of the elevator,
The elevator is accelerating in upward direction. The net force acting on the boy is given by :
g is the acceleration due to gravity
F = 255 N
The scale reading is 255 N as it begins to accelerate upward. hence, this is the required solution.
In order to test an intentionally weak adhesive, the bottom of the small 0.15-lb block is coated with adhesive and then the block is pressed onto the turntable with a known force. The turntable starts from rest at t = 0 and uniformly accelerates with a = 2 rad/s^2. If the adhesive fails at exactly t = 3 s, then determine:
a. the magnitude of the ultimate shear force that the adhesive supports
b. the angular displacement of the turntable at the time of failure
Answer:
answer
Explanation:
it is the answer which was presented in the year
potential diffetence
Answer:
6v
Explanation:
V=IR
V= 2* 3
V= 6 volts
A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude . What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards
The question is incomplete. The complete question is :
A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.
Solution :
Given :
Length of the wire, L = 0.6 m
Mass of the wire length, m = 11 g
= [tex]11 \times 10^{-3}[/tex] kg
Magnetic field , B = 0.4 T
Know we know that :
ILB = mg
or [tex]$I=\frac{mg}{BL}$[/tex]
[tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]
[tex]I=0.44963\ A[/tex]
[tex]I = 449.63 \ mA[/tex]
A student of mass 50kg takes 15seconds to run up a flight of 50 steps. If each step is 20cm, calculate the potential energy of the student at the maximum height
Answer:
the answer is 49000 joules at the maximum height
Explanation:
we know the mass (50kg)
we know the acceleration due to gravity(9.8m/s²)
we know the height too(maximum height meaning the 50th step so we multiply 50 with 20cm as each step is 20 cm and we get 1000 cm, convert to m it is 100 m
the formula is potential energy=mgh
m for mass
g for acceleration due to gravity
h for height
multiply them
50x9.8x100
we get 49000
the unit of potential energy is joules so the answer is
49000 joules
Answer:
49000 joules
Explanation:
hope it helpss
Two masses of 3 kg and 5 kg are connected by a light string that passes over a smooth polley as shown in the Figure.
QL
Determine:
i. the tension in the string,
ii. the acceleration of each mass, and
iii. the distance each mass moves in the first second of motion if they start from rest
i. [tex]T = 36.8\:\text{N}[/tex]
ii. [tex]a = 2.45\:\text{m/s}^2[/tex]
iii. [tex]x = 1.23\:\text{m}[/tex]
Explanation:
Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]
Forces acting on m1:
[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]
Forces acting on m2:
[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]
Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us
[tex]m_2g - m_1g = m_2a + m_1a[/tex]
or
[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]
Solving for a,
[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]
[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]
We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).
[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]
Assuming that the two objects start from rest, the distance that they travel after one second is given by
[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]
A race car goes from a complete stop at the start line to 150 miles per hour in 5 seconds. What is its acceleration? Show your work.
Answer:
Explanation:
150/5 = 30
30mph per 1 second
Question 3 of 10
What has the same value no matter where it is located in the universe?
A. Volume
B. Weight
C. Mass
D. Density
Reset Selection
Answer:
C. Mass
Explanation:
A rabbit is moving in the positive x-direction at 2.70 m/s when it spots a predator and accelerates to a velocity of 13.3 m/s along the positive y-axis, all in 2.40 s. Determine the x-component and the y-component of the rabbit's acceleration.
Answer:
the answer is nearly 5.655 [tex]ms^{-2}[/tex]
Explanation:
Given,
[tex]v_{x}=2.7 ms^{-1}[/tex]
[tex]v_{y}=13.3 ms^{-1}[/tex]
[tex]t=2.4 s[/tex]
[tex]a_{x}=\frac{2.7}{2.4}=1.125 ms^{-2}[/tex] (as [tex]a=\frac{v-u}{t}[/tex])
[tex]a_{y}=\frac{13.3}{2.4}=5.542 ms^{-2}[/tex]
[tex]a=\sqrt{a_{x}^{2}+a_{y}^{2} }[/tex]
[tex]=\sqrt{1.125^{2}+5.542^{2} }[/tex]
[tex]=5.655 ms^{-2}[/tex]
hope you have understood this...
pls mark my answer as the brainliest
When you have a straight horizontal line on a velocity time graph, what does this tell you about the object’s motion in terms of velocity and acceleration?
Answer:
It tell you that the velocity is constant, what means that there's no acceleration
Which of the following measures is equal to 700 km?
Answer:
1km=1000m
700km=
700×1000=700000
=700000metres
hope this helps
Two different galvanometers G1 and G2, have internal resistances r1and r2. The galvanometers G1 and G2 require the same current IC1=IC2 for a full-scale deflection of their pointers. These galvanometers G1 and G2 are used to build lab-made ammeters A1 and A2 . Both ammeters A1 and A2 have the same maximum scale reading Imax1=Imax2=Imax. To build A1 ,shunt resistor of resistance Rsh1is used and to build A2 , shunt resistor of resistance Rsh2 is used. The value of these shunt resistor resistances are such that: Rsh1=3Rsh2. What is the ratio oftheir internal resistances: r1:r2?
Answer:
there are 3 photos attached. so check
Explanation:
The car has a mass of 0·50 kg. The boy
now increases the speed of the car to 6·0
ms-1 . The total radial friction between
the car and the track has a maximum
value of 7.0 N. Show by calculation that
the car cannot continue to travel in the circular path.
Answer:
A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
What is meant by centripetal force ?Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.
Here,
Mass of the car, m = 0.5 kg
Velocity of the car, v = 6 m/s
Radial friction between the car and the track, f = 7 N
The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.
So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.
So,
mv²/r > f
0.5 x 6²/r > 7
Therefore, the radius of the circular track,
r < 18/7
r < 2.57 m
Hence,
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
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Which of the following describes an electric conductor
The following describes an electric conductor : A material that has low resistance and allows the charges to move freely. The correct option is D.
What is conductor?A conductor is a material or metal which allows the electrons to flow through it. In other words, a conductor allows the current to pass through them.
A battery also called as the voltage source, provides sufficient voltage or energy to excite electrons in the conductor.
Opposition offered to the flow of current is called as the resistance. The electrical element used in the circuit is the resistor.
So, an electric conductor is a material that has low resistance and allows the charges to move freely.
Thus, the correct option is D.
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