A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms⁻¹ , the magnitude of the induced emf in the wire of aerial is:
Answer: Induced emf is given by:
ε= Bvl
On putting the values we get
=5×10
−5
×1.50×2
=0.15mV
Explanation:
Hope these helped have a wonderful Christmas break ❄️
Which is not true about waves?
O Compression and rarefaction are found in longitudinal waves.
O Sound waves are transverse waves.
O Amplitude, wavelength, and troughs all describe parts of a wave.
O Electromagnetic waves are transverse waves.
Answer:
sound waves are transverse waves
What is true of the bottom layer of the ocean
Answer:
well i tough it was infinte but its not an the layer is a d i c k
Explanation:
Pls Help physics 8th class question
Answer:
angle of reflection and angle of incident is always equal
what is the acceleration of a softball if it has a mass of 0.50 kg and hits the catcher’s glove with a force of 25 newtons
Answer:
a = 50 [m/s²]
Explanation:
This type of problem can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force =25 [N] (units of Newtons)
m = mass = 0.5 [kg]
a = acceleration [m/s²]
[tex]a=F/m\\a=25/0.5\\a= 50 [m/s^{2}][/tex]
Determine the weight of a 5.1 kg scooter that moves with a constant acceleration of 3.0 ms2. (Make sure you use the weight equation: W or Fg = mg)
Answer:
15.3 sorry if i got it wrong
Explanation:
Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 17600 N for 55 ms.
Answer:
0.8 m/s
Explanation:
F(avg) * Δt = I, where
F(avg) = Average force
Δt = change in time
I = impulse
From the question, we know that the
average force is given as -17600 N
time change is given as 55 milliseconds
speed of the player, v = 8 m/s
Mass of the player is given as 110 kg
Impulse, I = F(avg) * Δt
I = -17600 * 0.055
I = -968
Now, using the Impulse-Momentum theory, we have that
Δp = I and thus,
Δp = m [v(f) - v(i)], where
Δp = change in the momentum
v(f) = final speed of the player
v(i) = initial speed of the player
Substituting the values, we have
I = m [v(f) - v(i)]
-968 = m.v(f) - m.v(i)
m.v(f) = m.v(i) - 968
110v(f) = 110 * 8 - 968
110v(f) = 880 - 968
110v(f) = -88
v(f) = -88 / 110
v(f) = -0.8 m/s
Strontium chloride, SrCl2
Is ionic or covalent
Answer:
it is a ionic bond because electrons are transferred from chlorine to strontium
Strontium chloride, SrCl₂, is an ionic compound because Ionic compounds are formed through the transfer of electrons from one atom to another, resulting in the formation of positively charged ions (cations) and negatively charged ions (anions).
In the case of strontium chloride, strontium (Sr) is a metal, and chlorine (Cl) is a non-metal. Strontium readily donates two valence electrons to chlorine, which has a strong affinity for gaining one electron to achieve a stable electron configuration. The electron transfer between strontium and chlorine leads to the formation of S[tex]r^2^+[/tex] cations and C[tex]l^-[/tex] anions. The resulting ionic bond holds the cations and anions together in a crystal lattice structure. The attraction between the oppositely charged ions creates the ionic bond, making strontium chloride an ionic compound.
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A typical donation of whole blood is equal to a pint (473 ml). If the average density of blood rhoBLOOD = 1060 kg/m^3, what is the mass of donated blood?
a. 5.01 kg
b. 0.501 lb
c. 0.501 kg
d. 0.501 g
e. 5.01 g
Answer:
c. 0.501 kg
Explanation:
Given;
volume of the denoted blood, V = 473 ml = 0.000473 m³
density of the denoted blood, ρ = 1060 kg/m³
The mass of donated blood is given as;
mass of the donated blood = density of the denoted blood x volume of the denoted blood
m = 1060 x 0.000473
m = 0.501 kg
Therefore, the mass of donated blood is 0.501 kg
a girl weighing 350N runs up hill upto 5m in 20sec. calculate her work done and power
Work done against gravity becomes potential energy of the object-Earth system. Its given by:
W = Ep
W = mgh but weigh(w) = mg
W = wh
W = 350 N × 5 m
W = 1750 JPower is the amount of energy transferred or converted per unit time. This is:
P = W/t
P = 1750 J / 20 s
P = 87.5 Win what condition do acceleration and deaccelration
Im not sure what are you asking about. But in physic/ mechanic usually when the object against gravity, its deceleration but when the object follow the gravity its acceleration
Also usually when the object move to the left side the speed will be negative (-) this is probably because of deceleration? since deceleration = -acceleration.
In 1940, it was discovered that uranium could be artificially changed to plutonium, a process called transmutation. What combination of decay processes would convert U-239 (92 protons) into Pu-239 (94 protons)?
A. Beta-decay followed by beta decay.
B. Positron emission followed by electron capture.
C. Electron capture followed by alpha decay.
D. Alpha decay followed by beta decay.
The combination of decay processes that would convert U-239 into PU- 239 is beta decay followed by another beta decay. There are two different types of beta decay (positive and negative beta decays) they both occur simultaneously in order to move the atoms towards stability.
A school bus's velocity decreases from 15 m/s to 5 m/s forwards over a period of 10 s. What is the bus's average acceleration? [Vf=v,+at]
Answer:-1 m/s/s
Explanation:
The acceleration of the bus when its velocity changes from 15 m/s to 5 m/s will be a=1 m/s^2
What is acceleration?Acceleration is defined as the change of the velocity with the time. Acceleration is a vector quantity and is defined by both the magnitude and the direction.
Now from the first equation of motion we have,
[tex]a= \dfrac{v-u}{t}[/tex]
we have V=15 m/s, u=5 m/s , t=10 seconds
[tex]a=\dfrac{15-5}{10}=1\ \frac{m}{s^2}[/tex]
Hence the acceleration of the bus will be a=1 m\s^2
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If you are holding a stack of books at your waist and then you walk across the room your hands did?
Answer: 3) work
Explanation:
Work is doing something that requires you to expend energy. In walking across the room while your hands were carrying the stack of books, you expended energy through them which means that they did work.
We can neither tell the efficiency nor the power taken to do the work above from the given information. And by moving and holding books, there was definitely work done which means that all options apart from 3 are false.
calculate the aceleration of a vehicle wich start with a zero meter per second, and acelerates to 34 m/s in 21 s.
HELP PLSSS
Answer:
For the aceleration we have:
Vf = Vo + a * t
Clearing "a":
a = (Vf - Vo) / t
Replacing and resolving:
a = (34 m/s - 0 m/s) / 21 s
a = 34 m/s / 21 s
a = 1,61 m/s^2
The aceleration of the vehicle is 1,61 meters per second squared
During the pitching motion, a baseball pitcher exerted an average horizontal force of 90 N against the 0.1 kg baseball while moving it through a horizontal displacement of 2.0 m before release. (1) what was the amount of work performed by the pitcher on the baseball (2) If the velocity at the start of the pitching motion was zero, at release the ball was traveling horizontally at which velocity
Answer:
(1) 180 J
(2) 60 m/s
Explanation:
(1) From the question,
Amount of work performed by the pitcher on the baseball = Force × distance.
W = F×d............... Equation 1
Given: F = 90 N, and d = 2.0 m.
Substitute into equation 1
W = 90×2
W = 180 Joules.
(2)
F = ma........... Equation 2
Where a = acceleration of the ball.
a = F/m
Given: F = 90 N, m = 0.1 kg.
therefore,
a = 90/0.1
a = 900 m/s².
Using,
v² = u² + 2as ............ Equation 3
Where u = 0 m/s, a = 900 m/s², s = 2 m.
substitute into equation 3
v² = 0² + 2×900×2
v² = 3600
v = √3600
v = 60 m/s
A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy, what is the ratio of the brick's mass to the rock's mass
Answer:
let M be the mass of brick
let m be the mass of the pebble
.5M(3)^2 =KE
.5m(5)^2= KE
.5M9 = .5m25
9M = 25m
(9/25)M = m
Answer:
A
Explanation:
e=1/2M[tex]V^{2}[/tex]
1/2[tex]M_{1}[/tex]9=1/2[tex]M_{2}[/tex]25
[tex]\frac{M_{1} }{M_{2} }[/tex] =[tex]\frac{25}{9}[/tex]
An ambulance is traveling east at 62.4 m/s. Behind it a car travels along the same direction at 34.5 m/s. The ambulance driver hears his siren with a wavelength of 0.47 m. What wavelength would a stationary observer behind the ambulance measure for the sound? The velocity of sound in air is 343 m/s.
Answer:
The answer is "0.5555 m"
Explanation:
Where the reference leaves the list and the viewer is at rest:
[tex]\lambda'=\frac{v-v_s}{v} \times \lambda\\\\[/tex]
[tex]=\frac{343 \frac{m}{s} - (-62.4 \frac{m}{s})}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{343 \frac{m}{s} + 62.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m[/tex]
[tex]=0.5555 \ m[/tex]
If your parasympathetic nervous system was activated what might you be doing?
A. Relaxing under a tree
B. Running a marathon
C. Skiing
D. Driving home in traffic
A model airplane with mass 1.0 kg is held by a wire so that it flies in a horizontal circle with radius 20.0 m. The airplane engine provides a net thrust of 1.0 N perpendicular to the wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in this horizontal flight.
Answer:
330
Explanation:
(a) The torque the net thrust produces about the center of the circle is of 20 N-m.
(b) The angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².
Given data:
The mass of model airplane is, m = 1.0 kg.
The radius of horizontal circle is, r = 20.0 m.
The magnitude of net thrust by engine is, F = 1.0 N.
(a)
The effort made to turn any object is known as the torque. The mathematical expression for the torque is given as,
T = F × r
Solving as,
T = 1.0 × 20.0
T = 20 N-m
Thus, we can conclude that the torque the net thrust produces about the center of the circle is of 20 N-m.
(b)
The expression for the angular acceleration of airplane during the horizontal flight is given as,
[tex]T = I \times \alpha[/tex]
Here, I is the moment of inertia of airplane and its value is,
[tex]I = \dfrac{1}{2}mr^{2}\\\\\\I = \dfrac{1}{2} \times 1.0 \times 20^{2}\\\\\\I =200 \;\rm kg.m^{2}[/tex]
So, the angular acceleration is,
20 = 200 × α
α = 20/200
α = 0.1 rad/s²
Thus, we can conclude that the angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².
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A
B
с
4 teaspoons sugar
2 cups water
6 teaspoons sugar
2 cups water
4 teaspoons sugar
3 cups water
Which of these lemonade glasses is the most concentrated for sugar?
Ос
Oь
Oa
charge of one of electron is 1.6022×10*-19 coulomb. what is the total charge on 1 mole of electrons
There are
[tex]6.023 \times {10}^{23} [/tex]
electrons in a mole. So charge of 1 mole electron is
[tex]6 \times {10}^{23} \times ( - 1.6022) \times {10 }^{ - 19} [/tex]
______ Is the distance and direction of an object's change in position from its starting point.
(5 Points)
Distance
Displacement
Velocity
Rate
Answer:
Displacement
Explanation:
I took the test.
Which water would you use to make salt dissolve the slowest?
Boiling water
Cold water
Hot water
Room temperature water
Answer:
boiling water because salt dissolves quicker in hot Water and the hottest is boiling
A block has mass of 20 gm and volume of 4 m3 , calculate the density *
Answer:
Density = 5 g/m³
Explanation:
Given:
Mass of block = 20 gm
Volume of block = 4 m³
Find:
Density
Computation:
Density = Mass / Volume
Density = 20 / 4
Density = 5 g/m³
5 kg/m3
Explanation:
The formula to find density is (d) = mass (m) ÷ volume (v)
6. a. find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm. B. the electron makes a transition from the n
Complete Question
(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .
(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.
Answer:
A
[tex]\Delta E = 337 \ eV[/tex]
B
[tex]\lambda = 3.439 *10^{-9} \ m[/tex]
Explanation:
Considering question a
From the question we are told that
The width of the box is [tex]w = 0.125 \ nm = 0.125 *10^{-9} \ m[/tex]
Generally the energy level of a particle confined to a box is mathematically represented as
[tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]
Generally the excitation energy is mathematically represented as
[tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]
From the question [tex]n_2 = 3\ (Third \ excited \ level ) \ \ and \ \ n_1 = 1[/tex]
Here h is the Planck's constant with a value of [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
m is the mass of electron with value [tex]m = 9.11 *10^{-31} \ kg[/tex]
So
[tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [3^2 - 1^2 ][/tex]
=> [tex]\Delta E = 539 *10^{-19} \ J[/tex]
=> [tex]\Delta E = \frac{539 *10^{-19}}{1.60 *10^{-19}} \ J[/tex]
=> [tex]\Delta E = 337 \ eV[/tex]
Considering question b
Generally the energy level of a particle confined to a box is mathematically represented as
[tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]
Generally the excitation energy is mathematically represented as
[tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]
From the question [tex]n_2 = 4 \ \ and \ \ n_1 = 1[/tex]
Here h is the Planck's constant with a value of [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
m is the mass of electron with value [tex]m = 9.11 *10^{-31} \ kg[/tex]
So
[tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [4^2 - 1^2 ][/tex]
=> [tex]\Delta E = 578 *10^{-19} \ J[/tex]
=> [tex]\Delta E = \frac{ 578 *10^{-19}}{1.60 *10^{-19 }}[/tex]
=> [tex]\Delta E = 361.45 \ eV[/tex]
Gnerally the wavelength is mathematically represented as
[tex]\lambda = \frac{hc}{\Delta E }[/tex]
=> [tex]\lambda = \frac{ 6.626 *10^{-34} * (3.0 *10^{8})}{578 *10^{-19} }[/tex]
=> [tex]\lambda = 3.439 *10^{-9} \ m[/tex]
A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block is released from rest when the
spring is stretched + 0.15 m as shown in figure below. At the instant the block is released the force vector on the block is.
(Consider Right direction is positive direction and Left direction is negative direction)
- 34.29 N
+ 34.29 N
+ 24 N
Оe
h Oa-241
Answer:
+ 24 N
Explanation:
the computation is shown below:
Given that
Mass of the block = m = 0.7 kg
Sprint constant = k = 160 N / m
x = 0.15m
Now the force on the block is
F = kx
= (160) (0.15)
= 24 N
As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction
Therefore the third option is correct
with a mass if 74 kg, is a moving with a velocity of 11m/s, what is the kinetic energy ?
Answer:
4477 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]k = \frac{1}{2} \times 74 \times {11}^{2} \\ = 37 \times 121[/tex]
We have the final answer as
4477 JHope this helps you
from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa at point B. Friction between the water and the pipe walls is negligible. What is the rate of discharge at point B
Answer:
1.8 m/s
Explanation:
Here is the complete question
The diameters of a water pipe gradually changes from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa and point B. Friction between the water and the pipe walls is negligible.
What is the rate of discharge at point B?
Solution
Using Bernoulli's equation,
P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where P₁ = pressure at point A = 700 kPa, h₁ = height at point A, v₁ = speed at point A and P₂ = pressure at point B = 664 kPa, h₂ = height at point B, v₂ = speed at point B, ρ = density of water = 1000 kg/m³
P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²
P₁ - P₂ - ρg(h₂ - h₁) = 1/2ρ(v₂² - v₁²)
(h₂ - h₁) = 5 m
Substituting the values of the variables, we have
700000 Pa - 664000 Pa - (1000 kg/m³ × 9.8 m/s² × 5 m) = 1/2 × 1000 kg/m³(v₂² - v₁²)
36000 Pa - 49000 Pa = 500 kg/m³(v₂² - v₁²)
- 13000 Pa = 500 kg/m³(v₂² - v₁²)
(v₂² - v₁²) = - 13000 Pa/500 kg/m³
(v₂² - v₁²) = -26 m²/s²
By mass flow rate, A₁v₁ = A₂v₂ where A₁ = cross-sectional area at point A and A₂ = cross-sectional area at point B
πd₁²v₁/4 = d₂²v₂/4 where d₁ = diameter at point A = 5 cm = 0.05 m and d₂ = diameter at point B = 15 cm = 0.15 m
d₁²v₁ = d₂²v₂
v₁ = v₂(d₂/d₁)²
= v₂(0.15/0.05)²
= v₂(3)²
= 9v₂
So, substituting v₁ = 9v₂ into the above equation, we have
(v₂² - v₁²) = -26 m²/s²
v₂² - 9v₂² = -26 m²/s²
- 8v₂² = -26 m²/s²
v₂² = -26 m²/s² ÷ (-8)
v₂² = 3.25 m²/s²
taking square root of both sides,
v₂ = √(3.25 m²/s²)
= 1.8 m/s
So, the rate of discharge at point B is 1.8 m/s
Find the resultant of the following displacement:
A = 20 Km 30° south of east;
B = 50 Km due west;
C = 40 Km north east;
D = 30 Km 60° south of west.
Answer:
Explanation:
Basically you just have to find the left vectors. To do so divide A, C and D into horizontal and vertical vector. A: 10km to south and 10root3 to east. Just sine and cosine of 30 at 20km. D: 15 km to west and 15root3 to south. Again sine and cosine of 60 at 30 km. C: 45 degrees so 20root2 to north and east. Add all these up with B. Then you have 7.696 km due south and 19.395 km due west. Resultant displacement magnitude = root(7.696^2+19.395^2)=20.866 south of west with angle=arctan(7.696/19.395)=21.644 degrees