Answer:
British East India Company
The battle was fought at Buxar, a "small fortified town" within the territory of Bihar, located on the banks of the Ganga river about 130 kilometres (81 mi) west of Patna; it was a decisive victory for the British East India Company.
In a 2-dimensional Cartesian coordinate system the y-component of a given vector is equal to that vector's magnitude multiplied by which trigonometric function, with respect to the angle between vector and y-axis?
a. sine
b. cosine
c. tangent
d. cotangent
Answer:
Option b, cosine.
Explanation:
Below you can see an image that illustrates this situation.
Remember that for a triangle rectangle with a given angle θ, we have:
Cos(θ) = (adjacent cathetus)/(hypotenuse)
In the image, you can see a vector of magnitude M, and the angle θ defined between the vector and the positive y-axis.
In this case, the y-component is the adjacent cathetus and the hypotenuse is the magnitude of the vector.
Then we will have:
Cos(θ) = (adjacent cathetus)/(hypotenuse) = y/M
solving that for y, we get:
y = Cos(θ)*M
Then the y-component is the vector's magnitude multiplied by the cosine of the angle between the vector and the y-axis.
The correct option is b.
Answer:
(b) cosine
Explanation:
In a 2-dimensional Cartesian coordinate system, a vector has a x-component and/or a y-component. To get these components, the magnitude of the vector is resolved with respect to the x-axis and the y-axis by multiplying it (the magnitude) by some trigonometric function with respect to the angle between the vector and the x or y axis.
For example, given a vector A of magnitude A which makes an angle α with the x-axis and an angle β with the y-axis, the x and y components of the vector A can be found as follows;
i. x-component is given by [tex]A_{x}[/tex]
[tex]A_{x}[/tex] = A cos α (with respect to the angle between A and the x-axis) or
[tex]A_{x}[/tex] = A sin β (with respect to the angle between A and the y-axis)
ii. y-component is given by [tex]A_{y}[/tex]
[tex]A_{y}[/tex] = A sin α (with respect to the angle between A and the x-axis) or
[tex]A_{y}[/tex] = A cos β (with respect to the angle between A and the y-axis)
Therefore, the y-component of a vector in a 2-dimensional Cartesian coordinate is given by the product of the magnitude of the vector and the cosine of the angle between the vector and the y-axis.
If the outermost electron in an atom is excited to a very high energy state, its orbit is far beyond that of the other electrons. To a good approximation, we can think of the electron as orbiting a compact core with a charge equal to the charge of a single proton. The outer electron in such a Rydberg atom thus has energy levels corresponding to those of hydrogen.
Sodium is a common element for such studies. How does the radius you calculated in part A compare to the approximately 0.20 nm radius of a typical sodium atom?
r100/rNa = _______.
Answer:
the calculated ratio to the radius of the sodium [tex]r_{100[/tex] / [tex]r_{Na[/tex] is 2645.0
Explanation:
Given the data in the question;
the calculated ratio to the radius of the sodium = [tex]r_{100[/tex] / [tex]r_{Na[/tex]
so from here we can write the number of energy states as 100
The number of energy states; n = 100
A;
We know that the radius of the sodium atom is;
[tex]r_n[/tex] = n²α₀
Now, the value of the Bohr radius; α₀ = 5.29 × 10⁻¹¹ m
so lets determine the radius of the sodium atom; by substituting in our values;
[tex]r_{100[/tex] = (100)² × (5.29 × 10⁻¹¹ m )
[tex]r_{100[/tex] = 5.29 × 10⁻⁷ m
B
given that, the theoretical value of the radius of the sodium is;
[tex]r_{Na[/tex] = 0.2 nm = 2 × 10⁻¹⁰ m
so we calculate the ratio of the radii of the sodium;
[tex]r_{100[/tex] / [tex]r_{Na[/tex] = ( 5.29 × 10⁻⁷ m ) / ( 2 × 10⁻¹⁰ m )
[tex]r_{100[/tex] / [tex]r_{Na[/tex] = 2645.0
Therefore, the calculated ratio to the radius of the sodium [tex]r_{100[/tex] / [tex]r_{Na[/tex] is 2645.0
A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.50 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?
Answer:
K = 0.076 J
Explanation:
The height of the target, h = 0.860 m
The mass of the steel ball, m = 0.0120 kg
Distance moved, d = 1.50 m
We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.
[tex]h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}[/tex]
Put all the values,
[tex]t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s[/tex]
The velocity of the ball is :
[tex]v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s[/tex]
The kinetic energy of the ball is :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J[/tex]
So, the required kinetic energy is 0.076 J.
A certain electric stove has a 16 Ω heating element. The current going through the element is 15 A. Calculate the voltage across the element.
Answer:
V = 240V
Explanation:
V = I*R
V = 15A*16ohms
V = 240V
a soap bubble was slowly enlarged from radius 4cm to 6cm and amount of work necessary for enlargement is 1.5 *10 calculate the surface tension of soap bubble joules
Answer:
The surface tension is 190.2 N/m.
Explanation:
Initial radius, r = 4 cm
final radius, r' = 6 cm
Work doen, W = 15 J
Let the surface tension is T.
The work done is given by
W = Surface Tension x change in surface area
[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]
A conducting sphere of radius R carries an excess positive charge and is very far from any other charges. Draw the graphs that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere?
Answer:
See annex
Explanation:
By convention potential at ∞ V(∞ ) = 0
As the distance from the sphere decreases the potential increases up to the point d = R ( R is the radius of the sphere. That potential remains constant while d = R and becomes 0 inside the sphere where there is not free charges and therefore the electric field is 0 and so is the potential.
I am sorry I could not make a better graph
The graph that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere is attached as an image below
[tex]V = \frac{KQ}{R}[/tex]
for r <= R
[tex]V = \frac{KQ}{r}[/tex]
for r > R
Therefore the graph will be
For more information on potentials as function of distance
https://brainly.com/question/24146175?referrer=searchResults
Two astronauts, each having a mass of 88.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.40 m/s. Treating the astronauts as particles, calculate each of the following.
a. the magnitude of the angular momentum of the system
b. the rotational energy of the system
c. What is the new angular momentum of the system?
d. What are their new speeds?
e. What is the new rotational energy of the system
Answer:
a) L = 4.75 103 kg m² / s, b) K_total = 2.57 10³ J,
c) L₀ = L_f =4.75 103 kg m² / s, d) K = 1.03 10⁴ J, K = 1.03 10⁴ J
Explanation:
a) the angular momentum is the sum of the angular momentum of each astronaut
the distance is measured from the center of the circle r = 10/2 = 5.0 m
L = 2m v r
L = 2 88.0 5.40 5.0
L = 4.75 103 kg m² / s
b) rotational kinetic energy
K = ½ I w²
As there are two astronauts, the total energy is the sum of the energy of each no.
The moment of inertia of a point mass
I = m r²
I = 88 5²
I = 2.2 10³ kg m²
the angular velocity is given by
v = w r
w = v / r
w = 5.40 / 5
w = 1.08 rad / s
the kinetic energy of the system
K_total = 2 K
K_total = 2 (½ I w²)
K_total = 2.2 10³ 1.08²
K_total = 2.57 10³ J
c, d) as astronauts are isolated in space, these speeds do not change unless there is an interaction between them, for example they approach each other, suppose they reduce their distance by half
r = 2.5 m
I = 88 2.5²
I = 5.5 10² kg m²
for the change in angular velocity let us use the conservation of moment
L₀ = L_f
2Io wo = 2 I w
w = Io / I wo
w = 2.2 10³ / 5.5 10² 1.08
w = 4.32 rad / s
linear velocity is
v = w r
v = 4.32 2.5
K = 1.03 10⁴ J
the kinetic energy of the system is
K = 5.5 10² 4.32²
K = 1.03 10⁴ J
A bowling ball with a mass of 8 kg is moving at a speed of 5 m/s. What is its
kinetic energy?
Answer:
100J
Explanation:
Kinetic energy=1/2mv^2
Kinetic energy=(1/2 x 8)x5^2
Kinetic energy=4x25
Kinetic energy=100
100J
Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.6 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
a. How far from the wall is she?
b. What is the period of her up-and-down motion?
Answer:
a) [tex]d=11m[/tex]
b) [tex]T=4.68s[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=22m[/tex]
Velocity [tex]v=4.6m/s[/tex]
a)
Generally the equation for distance between her and the wall d is mathematically given by
Since
The First Anti node distance is [tex]\frac{\lambda}{2}[/tex]
Therefore
[tex]d= \frac{\22}{2}[/tex]
[tex]d=11m[/tex]
b)
Generally the equation for her up-and-down motion is mathematically given by
[tex]T=\frac{22}{4.7}[/tex]
[tex]T=4.68s[/tex]
6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if its surface normal isin the XY plane and along a line that isinclined at 60 degrees to the positive Y axisand 30 degrees to the positive X axis
Answer:
Flux is 21 Nm^2/C.
Explanation:
Electric field, E = 6 N/C along X axis
Electric filed vector, E = 6 i N/C
Area, A = 4 square meter
Area vector
[tex]\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\[/tex]
The flux is given by
[tex]\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} . \left (3.5 \widehat{i} + 2 \widehat{j} \right )\\\\\phi = 21 Nm^2/C[/tex]
A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?
Explanation:
kinetic energy was converted to potential energy in the spring.
the answer is in the above image
The magnetic force exerted on a 1.2-m segment of straight wire is 1.6 N. The wire carries a current of 3.0 A in a region with a constant magnetic field of 0.50 T. What is the angle between the wire and the magnetic field
Answer:
The angle between the wire and the magnetic field is 62.74⁰
Explanation:
Given;
length of the wire, L = 1.2 m
force exerted on the wire, F = 1.6 N
current carried by the wire, I = 3.0 A
magnetic field strength, B = 0.5 T
The magnitude of a magnetic force on a current-carrying conductor is given as;
F = BIL(sinθ)
[tex]sin(\theta) = \frac{F}{BIL} = \frac{1.6}{0.5 \times 3 \times 1.2} = 0.8889 \\\\sin(\theta) =0.8889\\\\\theta = sin^{-1} (0.8889)\\\\\theta = 62.74^0[/tex]
Therefore, the angle between the wire and the magnetic field is 62.74⁰
A 50-kg copper block initially at 140°C is dropped into an insulated tank that contains 90 L of water at 10°C. Determine the final equilibrium tempera
Answer:
16.33°C
Explanation:
Applying,
Heat lost by copper = heat gained by water
cm(t₁-t₃) = c'm'(t₃-t₂).............. Equation 1
Where c = specific heat capacity of copper, m = mass of copper, c' = specific heat capacity of water, m' = mass of water, t₁ = initial temperature of copper, t₂ = initial temperature of water, t₃ = final equilibrium temperature.
From the question,
Given: m = 50 kg, t₁ = 140°C, m' = 90 L = 90 kg, t₂ = 10°C
Constant: c = 385 J/kg°C, c' = 4200J/kg°C
Substitute these values into equation 1
50(385)(140-t₃) = 90(4200)(t₃-10)
(140-t₃) = 378000(t₃-10)/19250
(140-t₃) = 19.64(t₃-10)
140-t₃ = 19.64t₃-196.6
19.64t₃+t₃ = 196.4+140
20.64t₃ = 336,4
t₃ = 336.4/20.6
t₃ = 16.33°C
Two parallel circular plates with radius carrying equal-magnitude surface charge densities of are separated by a distance of How much stored energy do the plates have? A. 120 B. 360 C. 12 D. 37
Answer:
I guess it is A. I am not sure
can someone help plz
Answer:
29.15 N
Explanation:
Applying,
Pythagoras theorem,
a² = b²+c²................ Equation 1
Where a = resultant of the two forces, b = first force, c = second force.
From the diagram,
Given: b = 15 N, c = 25 N
Substitute these values into equation 1
a² = 15²+25²
a² = 225+625
a² = 850
a = √850
a = 29.15 N
Hence the resultant of the two forces is 29.15 N
Convert Rev/min to rad/s x 2pie/60?
Anyone knows this please?
Answer:
Thus, [tex]\frac{1 rev}{min} =\frac{2\pi}{60} rad/s[/tex]
Explanation:
The angular speed is defined as the rate of change of angular velocity.
Its SI unit is rad/s and other units are rev/min or rev/s.
[tex]\frac{1 rev}{min } = \frac{1 rev}{60 sec}\\\\1 rev = 2\pi rad\\\\So\\\\\frac{1 rev}{min} = \frac{2\pi}{60} rad/s[/tex]
If you warm up the volume of a balloon but keep the pressure the same, you would be using which gas law?
Answer:
Charles law
Explanation:
Charle's law states that the volume (V) of a given gas is directly proportional to the absolute temperature (T) at a constant pressure.
That is;
: V ∝ T
: V/T = K
According to this question, the volume of a balloon is warmed up but the pressure is kept the same. Charles law will be used because it shows the relationship between the volume (V) and the temperature (heat) at a constant pressure (P).
A puck moves 2.35 m/s in a -22.0 direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50 degree direction. What was the direction of the acceleration?
Answer:
48.9 is the answer I think !
Answer:
28.4
Explanation:
A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used?
Answer:
2.86×10⁻¹⁸ seconds
Explanation:
Applying,
P = VI................ Equation 1
Where P = Power, V = Voltage, I = Current.
make I the subject of the equation
I = P/V................ Equation 2
From the question,
Given: P = 0.414 W, V = 1.50 V
Substitute into equation 2
I = 0.414/1.50
I = 0.276 A
Also,
Q = It............... Equation 3
Where Q = amount of charge, t = time
make t the subject of the equation
t = Q/I.................. Equation 4
From the question,
4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs
Q = 7.899×10⁻¹⁹ C
Substitute these value into equation 4
t = 7.899×10⁻¹⁹/0.276
t = 2.86×10⁻¹⁸ seconds
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate
Answer:
The correct answer is "4000 J".
Explanation:
Given that,
Force,
= 200 N
Displacement,
= 20 m
Now,
The work done will be:
⇒ [tex]Work=Force\times displacement[/tex]
By putting the values, we get
[tex]=200\times 20[/tex]
[tex]=4000 \ J[/tex]
Calculate the forces that the supports \rm A and \rm B exert on the diving board shown in when a 58-\rm kg person stands at its tip.
6. Which of these contain muscles that are not under the mind's control? (Select all that apply.)
Answer:
a c
Explanation:
edu 2021
Điện tích Q = 8. 10-6C đặt cố định trong
không khí , điện tích q = - 10. 10-6C di
chuyển trên đường thẳng xuyên qua Q,
từ M cách Q một khoảng 100cm, lại
gần Q thêm 50cm. Tính công của lực
điện trường trong dịch chuyển đó?
Answer:
0.72J
Explanation:
What is 1 second….???? Give a meaningful answer…..
Explanation:
.....................................
Answer:
1 second is defined as 1/86400th part of a mean solar day.
Một chất điểm chuyển động tròn đều, sau 5 giây nó quay được 20 vòng. Chu kì quay của chất điểm là:
Answer:
T=0,25s
Explanation
5s>20vong 1s>4vong
omega= 8pi
omega=2pi/T
A crucible (container) of molten metal has an open top with an area of 5.000 m^2. The molten metal acts as a blackbody radiator. The intensity spectrum of its radiation peaks at a wavelength of 320 nm. What is the temperature of that blackbody?
Answer:
T = 9056 K
Explanation:
In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation
λ T = 2,898 10⁻³
where lam is the wavelength of the maximum emission
T = 2,898 10⁻³ /λ
let's calculate
T = 2,898 10⁻³ / 320 10⁻⁹
T = 9.056 10³ K
T = 9056 K
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and speed (in m/s) when t = 5. f(t) = 18 + 48/t + 1
Answer:
The velocity of the particle = -1.92 m/s
The speed of the particle = 5.72 m/s
Explanation:
Given equation of motion;
[tex]f(t) = 18 \ + \ \frac{48}{t} \ + \ 1[/tex]
Velocity is defined as the change in displacement with time.
[tex]V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s[/tex]
The distance traveled by the particle in 5 s:
[tex]s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m[/tex]
The speed of the particle when t = 5s
[tex]Speed = \frac{28.6}{5} = 5.72 \ m/s[/tex]
A car of mass M traveling with velocity v strikes a car of mass M that is at rest. The two cars’ bodies mesh in the collision. The loss of the kinetic energy the moving car undergo in the collision is
a) a quarter of the initial kinetic energy.
b) half of the initial kinetic energy.
c) all the initial kinetic energy.
d) zero.
Answer:
the correct answer is B
Explanation:
Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved
instantly starts. Before the crash
p₀ = M v +0
final instant. After the crash
m_f = (M + M) v_f
the moment is preserved
p₀ = p_f
M v = 2 M v_f
v_f = v / 2
let's look for kinetic energy
before the crash
K₀ = ½ M v²
after the crash
K_f = ½ 2M (v_f)²
K_f = ½ 2M (v/2)²
K_f = (½ M v²) ½
K_f = K₀ / 2
therefore the correct answer is B
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground?
Explanation:
the answer is in the above image