Answer:
Explanation:
Given that;
horizontal circle at a rate of 2.33 revolutions per second
the magnetic field of the Earth is 0.500 gauss
the baton is 60.1 cm in length.
the magnetic field is oriented at 14.42°
we wil get the area due to rotation of radius of baton is
[tex]\Delta A = \frac{1}{2} \Delta \theta R^2[/tex]
The formula for the induced emf is
[tex]E = \frac{\Delta \phi}{\Delta t}[/tex]
[tex]\phi = \texttt {magnetic flux}[/tex]
[tex]E=\frac{\Delta (BA) }{\Delta t}[/tex]
[tex]=B\frac{\Delta A}{\Delta t}[/tex]
B is the magnetic field strength
substitute
[tex]\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A[/tex]
[tex]E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega[/tex]
The magnetic field of the earth is oriented at 14.42
[tex]\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5[/tex]
we plug in the values in the equation above
so, the induce EMF will be
[tex]E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega[/tex]
[tex]=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V[/tex]
A 35.0-kg child swings in a swing supported by two chains, each 2.96 m long. The tension in each chain at the lowest point is 436 N. (a) Find the child's speed at the lowest point. Consider all the vertical components of force acting on the swing when it is at its lowest point and relate them to the acceleration of the swing at that instant. m/s (b) Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.)
Answer:
6.69m/s,529N
Explanation:
Now we have 3 forces acting on the boy at its least point, the two tensions on the string and its weight. The tensions are acting upwards why its weight is acting downwards.
Hence the net force causes the child to swing in a circular fashion without skidding off the swing.
This net force is the centripetal force.
The weight of the child is mass × g
35×9.8=343N
The net force = 436+436-343 = 529N
The centripetal force is defined mathematically as;
F = mass × velocity square/ length of string.
Hence 529 = 35V^2/2.96
V^2 = 529×2.96/35 =44.7383
V=√44.7383 =6.69m/s
B. The force exerted by the seat on the child is the net force which keeps the boy from falling.
529N
A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?
Answer:
Explanation:
We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth
[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]
[tex]L=\frac{.5}{.24}[/tex]
L= 2.05
The length will appear to be 2.05 m . and width will appear to be .5 m to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.
An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.
Answer:
[tex]\delta = 0.385\,m[/tex] (Compression)
Explanation:
The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:
[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]
Where:
[tex]P[/tex] - Load experimented by the bar, measured in newtons.
[tex]L[/tex] - Length of the bar, measured in meters.
[tex]A[/tex] - Cross section area of the bar, measured in square meters.
[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.
The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])
[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]
Where:
[tex]D_{o}[/tex] - Outer diameter, measured in meters.
[tex]D_{i}[/tex] - Inner diameter, measured in meters.
[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]
[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]
The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)
[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]
[tex]\delta = -3.852\times 10^{-4}\,m[/tex]
[tex]\delta = -0.385\,mm[/tex]
[tex]\delta = 0.385\,m[/tex] (Compression)
Consider a situation in which you are moving two point charges such that the potential energy between them decreases. (NOTE: ignore gravity).
This means that you are moving the charges:
a) Closer to each other
b) Farther apart
c) Either A or B
Answer: Option A
Explanation:
The potential energy decreases in the case when the charges are opposite and they attract each other.
In this case there is no external energy required in order to put the charges together.
This is so because the charges are opposite and they will attract each other. Yes, the only condition should be that the charges should be alike.
Example: a negative charge and a positive charge.
А
mass exerts force of
5.6 X 10^-10N on
another mass
when
seperated 93cm apart. If
one mass is the square root of
the other
Find
value of
the two masses.
Answer:
i hope it will be useful for you
Explanation:
F=5.6×10^-10N
R=93cm=0.93m
let take m1 and m2 =m²
according to newton's law of universal gravitation
F=m1m2/r²
F=m²/r²
now we have to find masses
F×r²=m²
5.6×10^10N×0.93m=m²
5.208×10^-9=m²
taking square root on b.s
√5.208×10^-9=√m²
so the two masses are m1=7.2×10^-5
and m2=7.2×10^-5
In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 N to the ball for a time of 0.70×10−3s. Part A Assuming that this force is constant, what is the magnitude J of the impulse on the ball?
Answer:
J = 8.4 kg*m/s
Explanation:
The magnitude of the impulse, J, on the ball can be calculated using the following equation:
[tex] J = F*t [/tex] (1)
Where:
F: is the force = 12000 N
t: is the time = 0.70x10⁻³ s
So, by entering the values above into equation (1) we can find the impulse on the ball:
[tex] J = F*t = 12000 N*0.70 \cdot 10^{-3} s = 8.4 kg*m/s [/tex]
Therefore, the impulse on the ball is 8.4 kg*m/s.
I hope it helps you!
The magnitude J of the impulse on the ball is 8.4 kg m/s.
Calculation of the impulse:Since it applies a force of 12,000 N to the ball for a time of 0.70×10−3s.
So,
The magnitude is
[tex]= Force \times time\\\\= 12,000 \times 0.70\times 10^{-3}[/tex]
= 8.4 kg m/s
Here we basically multiplied the force with the time to determine the magnitude.
Learn more about impulse here: https://brainly.com/question/16324614
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her feet are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor
Answer:
μ = 0.336
Explanation:
We will work on this exercise with the expressions of transactional and rotational equilibrium.
Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation
fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0
fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0
fr = cotan θ (W / 2 + 0,3 W_painter)
Now let's write the equilibrium translation equation
X axis
F1 - fr = 0
F1 = fr
the friction force has the expression
fr = μ N
Y Axis
N - W - W_painter = 0
N = W + W_painter
we substitute
fr = μ (W + W_painter)
we substitute in the endowment equilibrium equation
μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)
μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)
we substitute the values they give
μ = cotan θ (12/2 + 0.3 55) / (12 + 55)
μ = cotan θ (22.5 / 67)
μ = cotan tea (0.336)
To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45
cotan 45 = 1 / tan 45 = 1
the result is
μ = 0.336
In Fig. on the right, what is the acceleration at 1.0 s?
Answer:
10 m/s²
Explanation:
Acceleration is the slope of the velocity vs time graph.
At t = 1.0 s, the slope of the line is:
a = Δv/Δt
a = (20 m/s) / (2.0 s)
a = 10 m/s²
Answer:
10m/s2
Explanation:
Acceleration is defined as change in velocity over the time period
Now at 1second the velocity is 10m/s.
Hence the acceleration is V-U/t;
Where V is velocity at 10s and U is velocity when the object is at rest.
So we have;
Acceleration = 10-0/ 1 = 10m/s2
Four heavy elements (A, B, C, and D) will fission when bombarded by neutrons. In addition to fissioning into two smaller elements, A also gives off a beta particle, B gives off gamma rays, C gives off neutrons, and D gives off alpha particles. Which element would make a possible fuel for a nuclear reactor
Answer:
Element C will be best for a nuclear fission reaction
Explanation:
Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process. For the reaction to be continuous in a chain reaction, the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.
What is the impulse on a car (750 kg) that accelerates from rest to 5.0 m/s in 10 seconds
Explanation:
impulse J = m × (v2-v1) =750 × ( 5 - 0 ) =3750( N×s)
Answer:
3750Ns
Explanation:
Impulse is defined as Force × time
Force = mass × acceleration,
Hence impulse is;
mass × acceleration × time.
From Newton's second law
Force × time = mass × ∆velocity
750× 5 = 3750Ns
∆velocity = Vfinal-Vinitial ; the initial velocity is zero since the body starts from rest.
You have two charges; Q1 and Q2, and you move Q1 such that the potential experienced by Q2 due to Q1 increases.
Gravity should be ignored.
Then, you must be:
a) Moving Q1 further away from Q2.
b) Moving in the opposite direction to that of the field due to Q1
c) Moving Q1 closer to Q2.
d) Moving in the same direction as the field due to Q1.
e) Any of the above
Given that,
First charge = Q₁
Second charge = Q₂
The potential experienced by Q2 due to Q1 increases
We know that,
The electrostatic force between two charges is defined as
[tex]F=\dfrac{kQ_{1}Q_{2}}{r^2}[/tex]
Where,
k = electrostatic constant
[tex]Q_{1}[/tex]= first charge
[tex]Q_{2}[/tex]= second charge
r = distance
According to given data,
The potential experienced by Q₂ due to Q₁ increases.
We know that,
The potential is defined from coulomb's law
[tex]V=\dfrac{Q_{1}}{4\pi\epsilon_{0}r}[/tex]
[tex]V\propto\dfrac{1}{r}[/tex]
If r decrease then V will be increases.
If V decrease then r will be increases.
Since, V is increases then r will decreases that is moving Q₁ closer to Q₂.
Hence, Moving Q₁ closer to Q₂.
(c) is correct option.
A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Find the coordinates of the particle at t = 4.00 s. x = m y = m (d) Find the speed of the particle at t = 4.00 s.
Answer:
a) d = (6.00 t i ^ + 0.500 t²) m , b) v = (6.00 i ^ + 1.00 t j ^) m / s
c) d = (24.00 i ^ + 8.00 j^ ) m , d) v = (6.00 i ^ + 5 j^ ) m/s
Explanation:
This exercise is about kinematics in two dimensions
a) find the position of the particle on each axis
X axis
Since there is no acceleration on this axis, we can use the relation of uniform motion
v = x / t
x = v t
we substitute
x = 6.00 t
Y Axis
on this axis there is an acceleration and there is no initial speed
y = v₀ t + ½ a t²
y = ½ at t²
we substitute
y = ½ 1.00 t²
y = 0.500 t²
in vector position is
d = x i ^ + y j ^
d = (6.00 t i ^ + 0.500 t²) m
b) x axis
as there is no relate speed is concatenating
vₓ = v₀
vₓ = 6.00 m / s
y Axis
there is an acceleration and the initial speed is zero
[tex]v_{y}[/tex] = v₀ + a t
v_{y} = a t
v_{y} = 1.00 t
the velocity vector is
v = vₓ i ^ + v_{y} j ^
v = (6.00 i ^ + 1.00 t j ^) m / s
c) the coordinates for t = 4 s
d = (6.00 4 i ^ + 0.50 4 2 j⁾
d = (24.00 i ^ + 8.00 j^ ) m
x = 24.0 m
y = 8.00 m
d) the velocity of for t = 4 s
v = (6 i ^ + 1 5 j ^)
v = (6.00 i ^ + 5 j^ ) m/s
Please help in the 2nd question
Answer:
[tex]q=4\times 10^{-16}\ C[/tex]
Explanation:
It is given that,
The charge on an object is 2500 e.
We need to find how many coulombs in the object. The charge remains quantized. It says that :
q = ne
[tex]q=2500\times 1.6\times 10^{-19}\ C\\\\q=4\times 10^{-16}\ C[/tex]
So, the charge on the object is [tex]4\times 10^{-16}\ C[/tex].
What is the gravitational potential energy of a ball of mass 2.00 kg which is tossed to a height of 13.0 m above the ground? Answer in J, taking the potential energy to be 0.00 J at the ground.
Answer:
I believe the answer is 254.8 J, or rounded 255 J.
Explanation:
The formula for potential energy is:
PE=m(h)g
This means the mass (m) times height (h) times gravity (m). Gravity is 9.8 m/s (meters per second). Putting all of the numbers into it would equal:
PE=2(13)9.8
This equals 254.8 exactly, or if the assignment calls for you to round, 255.
0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distance of 2 m
Answer:
1.6 J
Explanation:
Work = change in energy
W = ΔKE
Fd = KE
(0.8 N) (2 m) = KE
KE = 1.6 J
You hold block A with a mass of 1 kg in one hand and block B with a mass of 2 kg in the other. You release them both from the same height above the ground at the same time. (Air resistance can safely be ignored as they fall.) Which of the following describes and explains the motion of the two objects after you release them?
A) Before A hits the ground before the block B. The same force of gravity acts on both objects, but because block A has a lower mass, it will have a larger acceleration
B) Block B hits the ground before the block A. The larger force on block B causes it to have a greater acceleration than block A
C) The two blocks hit the ground at the same time, because they both ackelerate at the same rate while falling. The greater gravitational force on block B is compersated giving it the same acceleration
D) The two blocks hit the ground at the same time because they move at the same constant speed while falling.The force on the blocks are irrelevent once they start falling
ok so im not that good when it comes to physics but i think its C)
How far does a roller coaster travel if it accelerates at 2.83 m/s2 from an initial
velocity of 3.19 m/s for 12.0 s?
Answer:
b
Explanation:
A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s. Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop
Answer:
The velocity is [tex]v = 8.85 m/s[/tex]
Explanation:
From the question we are told that
The mass of the skier is [tex]m_s = 60 \ kg[/tex]
The initial speed is [tex]u = 4.0 \ m/s[/tex]
The height is [tex]h = 10 \ m[/tex]
According to the law of energy conservation
[tex]PE_t + KE_t = KE_b + PE_b[/tex]
Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as
[tex]PE_t = mg h[/tex]
substituting values
[tex]PE_t = 60 * 4*9.8[/tex]
[tex]PE_t = 2352 \ J[/tex]
And [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill
And [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as
[tex]KE_b = 0.5 * m * v^2[/tex]
substituting values
[tex]KE_b = 0.5 * 60 * v^2[/tex]
=> [tex]KE_b = 30 v^2[/tex]
Where v is the velocity at the bottom
And [tex]PE_b[/tex] is the potential energy at the bottom which equal to zero due to the fact that height is zero at the bottom of the hill
So
[tex]30 v^2 = 2352[/tex]
=> [tex]v^2 = \frac{2352}{30}[/tex]
=> [tex]v = \sqrt{ \frac{2352}{30}}[/tex]
[tex]v = 8.85 m/s[/tex]
Answer:
The Skier's velocity at the bottom of the hill will be 18m/s
Explanation:
This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.
That is
[tex]mgh = 0.5 mv^{2}[/tex]
solving for velocity, we will have
[tex]v= \sqrt{2gh}[/tex]
hence her velocity will be
[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]
This is the velocity she gains from the slope.
Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s
The Skier's velocity at the bottom of the hill will be 18m/s
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? Giancoli, Douglas C.. Physics (p. 45). Pearson Education. Kindle Edition.
Answer:
Assuming that the vertical speed of the ball is 14 m/s we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 14 m/s
[tex]V_{0}[/tex]: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]
b) The maximum height is:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]
c) The time can be found using the following equation:
[tex] V_{f} = V_{0} - gt [/tex]
[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]
d) The flight time is given by:
[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]
I hope it helps you!
Which of the following is NOT a type of electromagnetic wave?
Seismic waves
Visible light
Radio waves
Microwave
Answer:
Seismic waves
Explanation:
seismic waves are not represented by electromagnetic graphs, nor can they be reflected on an electromagnetic spectrum. Visible light, radio waves, and microwaves are all electromagnetic waves, which are represented by graphs and electromagnetic spectrums.
Answer:
Siesmic waves
Explanation:
A body moves due north with velocity 40 m/s. A force is applied
on it and the body continues to move due north with velocity 35 m/s. W. .What is the direction of rate of change of momentum,if it takes
some time for that change and what is the direction of applied
external force?
Answer:
the direction of rate of change of the momentum is against the motion of the body, that is, downward.
The applied force is also against the direction of motion of the body, downward.
Explanation:
The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:
[tex]\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v[/tex]
p: momentum
m: mass
[tex]\Delta v[/tex]: change in the velocity
The sign of the change in the velocity determines the direction of rate of change. Then you have:
[tex]\Delta v=v_2-v_1[/tex]
v2: final velocity = 35m/s
v1: initial velocity = 40m/s
[tex]\Delta v =35m/s-40m/s=-5m/s[/tex]
Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.
The applied force is also against the direction of motion of the body, downward.
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Answer:
185 N
Explanation:
Sum of forces in the x direction:
Fₓ = -(80 N cos 75°) + (120 N cos 60°)
Fₓ = 39.3 N
Sum of forces in the y direction:
Fᵧ = (80 N sin 75°) + (120 N sin 60°)
Fᵧ = 181.2 N
The magnitude of the net force is:
F = √(Fₓ² + Fᵧ²)
F = √((39.3 N)² + (181.2 N)²)
F = 185 N
We have that for the Question "Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?" it can be said that the magnitude of the resultant force is
R=200N
From the question we are told
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Generally the equation for the Resultant force is mathematically given as
For x axis resolution
[tex]Fx=80cos75+120cos60\\\\Fx=80.7N[/tex]
For y axis resolution
[tex]Fx=80sin75+120sin60\\\\Fx=181.2N[/tex]
Therefore
[tex]R=\sqrt{80.7^2+181.2N^2}\\\\R=200N[/tex]
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11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle of 40° clockwise from the x axis. Find out resultant vector with its magnitude and direction using components method.
Answer:
The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°
Explanation:
To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.
For M vector you obtain:
[tex]M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}[/tex]
For N vector:
[tex]N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}[/tex]
The resultant vector is the sum of the components of M and N:
[tex]F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}[/tex]
The magnitude of the resultant vector is:
[tex]|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm[/tex]
And the direction of the vector is:
[tex]\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°[/tex]
hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°
Light from a fluorescent lamp is observed through a cloud of cool nitrogen gas. Again, two students are having a discussion about the kind of spectra that they would see.
Student 1: We would see absorption line spectra and the missing lines would correspond to the light from the fluorescent lamp.
Student 2: I disagree. We would see an emission line spectrum corresponding to nitrogen. This would happen because the nitrogen gas would absorb some energy from the fluorescent lamp and would reemit this energy which would result in an emission line spectrum.
Which student, if any, do you agree with and why?
Answer:
From the previous explanation Student No. 1 has the correct explanation
Explanation:
When the fluorescent lamp emits a light it has the shape of its emission spectrum, this light collides with the atoms of Nitrogen and excites it, so these wavelengths disappear, lacking in the spectrum seen by the observed, for which we would see an absorption spectrum
The nitrogen that was exited after a short time is given away in its emission lines, in general there are many lines, so the excitation energy is divided between the different emission lines, which must be weak
From the previous explanation Student No. 1 has the correct explanation
wich of the following are commonly distributed by veterinary assistants in typical veterinary practice?
Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall. Part A If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return
Answer:
36s
Explanation:
Let the objects be A and B.
Let the initial velocity of A be U and the initial velocity of B be 3U
The height sustain by A will be;
The final velocity would be zero
V2 = U2-2gH
Hence
0^2= U2 -2gH
H = U^2/2g
Similarly for object B, the height sustain is;
V2 = (3U)^2-2gH
Hence
0^2= 3U^2 -2gH
U2-2gH
Hence
0^2= U2 -2gH
H = 3U^2/2g
By comparism. The object with higher velocity sustains more height and so should fall longer than object A.
Now object A would take;
From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;
V=10×12=120m/s let g be 10m/S2
Similarly for object B,
The final velocity for B when it's falling it should be 3×that of A
Meaning
3V= gt
t =3V/g = 3× 120/10 = 36s
HELP ASAP!
There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
What real-world examples show not work being done? Can you think of examples other than resisting the force of gravity?
Answer:
Work is defined as a force doing a movement, for example, if with a force F we move an object a distance D, the work done is:
W = F*D
(note, the force is causing the movement, the product here is a dot product, this means that if the force and the displacement are perpendiculars, the product is zero)
So the examples where there are not work being done may be:
We do not have any movement:
For example, you can go to a wall in your house and push it really hard.
There is a force, but the wall will not move, so we have D = 0
and W = F*D = F*0 = 0
Because we have no motion.
Another case is where the force and the direction of motion are perpendiculars.
If we have for example a car, moving at a constant speed, and you push it sidewise (perpendicular to the direction of movement) we have a force being applied and movement, but those are in different directions (so the force does not cause the movement) so we dont have work being done.
Consider the binding energy of two stable nuclei, one with 60 nucleons and one with 200 nucleons. a. Is the total binding energy of the nucleus with 200 nucleons more than, less than, or equal to the total binding energy of the nucleus with 60 nucleons. Justify your reasoning.
Answer:
The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons
Explanation:
Binding energy can be given by the formula:
Binding energy = Binding energy per nucleon * Number of nucleons
This means that if the binding energy per nucleon = x MeV
Where x is a positive real number
The nucleus with 60 nucleons will have Binding energy = 60x MeV
The nucleus with 200 nucleons will have binding energy = 200x MeV
For a +ve x, 200x > 60x
Binding energy is proportional (directly) to nucleon volume. A further explanation is provided below.
Binding energy
Binding energy involving 200 nucleons would've been greater than 60 nucleons because so many more nucleons result in what seems like a stronger reaction.
It makes absolutely no difference out whether nucleon seems to be a proton as well as a neutron because they just have a similar strong coupling relatively steady.
Thus the response above is correct.
Find out more information about binding energy here:
https://brainly.com/question/23942204
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 64.7 N, Jill pulls with 61.5 N in the northeast direction, and Jane pulls to the southeast with 171 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey. What is the direction of the net force? Express this as the angle from the east direction between 0.
Answer:
F = 241.8 N θ = -18.7
Explanation:
To find the net force, we use Newton's second law, for this we decompose the force using trigonomer
Force 2
F2 = 61.5 N
cos 45 = F₂ₓ / F2 F₂ₓ = F2 cos 45
sin 45 = F₂_y / F2 F₂_y = F2 sin 45
force 3
F3 = 171 N
cos 45 = F₃ₓ / F3
sin 45 = F₃_y / F3
the total force can be found with its components
axis x direction (East-West)
Fₓ = 64.7 +61.5 cos 45 + 171 cos 45
Fₓ = 229.1 N
Y axis (direction North -Sur)
F_y = 61.5 sin 45 - 171 sin 45
F_y = - 77,428 N
the resulting force is
F = Fx i ^ + Fy j ^
F = (229.1 i⁻ 77,428 j⁾ N
we can use the Pythagorean theorem to find the module
F = √ (229.1 2 + 77,428 2)
F = 241.8 N
let's use trigonometry to find the direction
tan θ = F_y / Fₓ
θ = tan⁻¹ (F_y / Fₓ)
θ = tan⁻¹ (-77,428 / 229.1)
θ = -18.7