Answer:
F = 8.82 N
the force in N on the bottom plate of the tank due to the water is 8.82 N
Explanation:
Force F = mass × acceleration
F = ma
In the case of weight;
F = mg .......1
Where;
m = mass
g = acceleration due to gravity
a = acceleration
Mass m = density × volume
m = dV ......2
Volume = cross sectional area × height
V = Ah ........3
Substituting equation 3 to 2;
m = dAh .....4
Substituting equation 4 to 1;
F = dAhg ........5
Given;
d = 1000 kg/m^3
A = 10cm × 10cm = 100cm^2 = 0.01 m^2
h = 9 cm = 0.09 m
g = 9.8 m/s^2
Substituting the given values into equation 5;
F = 1000×0.01×0.09×9.8 N
F = 8.82 N
the force in N on the bottom plate of the tank due to the water is 8.82 N
A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s along the negative y-axis, all in 1.60 s. Determine the x-component and the y-component of the rabbit's acceleration. (Enter your answers in m/s2. Indicate the direction with the signs of your answers.)
Answer:
aₓ = 0 , ay = -6.8125 m / s²
Explanation:
This is an exercise that we can solve with kinematics equations.
Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.
x axis
vₓ = v₀ₓ = 1.10 m / s
aₓ = 0
y axis
initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration
[tex]v_{y}[/tex]= v_{oy} -ay t
ay = (v_{oy} -v_{y}) / t
ay = (0 -10.9) / 1.6
ay = -6.8125 m / s²
the sign indicates that the acceleration goes in the negative direction of the y axis
A 432 kg merry-go-round in the shape of a horizontal disk with a radius of 2.3 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.1 rad/s in 2.1 s
Answer:
τ = 1679.68Nm
Explanation:
In order to calculate the required torque you first take into account the following formula:
[tex]\tau=I\alpha[/tex] (1)
τ: torque
I: moment of inertia of the merry-go-round
α: angular acceleration
Next, you use the following formulas for the calculation of the angular acceleration and the moment of inertia:
[tex]\omega=\omega_o+\alpha t[/tex] (2)
[tex]I=\frac{1}{2}MR^2[/tex] (3) (it is considered that the merry-go-round is a disk)
w: final angular speed = 3.1 rad/s
wo: initial angular speed = 0 rad/s
M: mass of the merry-go-round = 432 kg
R: radius of the merry-go-round = 2.3m
You solve the equation (2) for α. Furthermore you calculate the moment of inertia:
[tex]\alpha=\frac{\omega}{t}=\frac{3.1rad/s}{2.1s}=1.47\frac{rad}{s^2}\\\\I=\frac{1}{2}(432kg)(2.3)^2=1142.64kg\frac{m}{s}[/tex]
Finally, you replace the values of the moment of inertia and angular acceleration in the equation (1):
[tex]\tau=(1142.64kgm/s)(1.47rad/s^2)=1679.68Nm[/tex]
The required torque is 1679.68Nm
Underline your answer for each situation: If you advance the movie one frame, the knot at point A would be a) in the same place b) higher c) lower d) to the right e) to the left If the person generates a new pulse like the first but more quickly, the pulse would be a) same size b) wider c) narrower If the person generates another pulse like the first but he moves his hand further, the pulse would be a) same size b) taller c) shorter If the person generates another pulse like the first but the rope is tightened, the pulse will move a) at the same rate b) faster c) slower Now the person moves his hand back and forth several times to produce several waves. You freeze the movie and get this snapshot. Underline your answer for each situation: If you advance the movie one frame, the knot at point A would be a) in the same place b) higher c) lower d) to the right e) to the left If you advance the movie one frame, the pattern of the waves will be _________relative to the hand. a) in the same place b) shifted right c) shifted left d) shifted up e) shifted down If the person starts over and moves his hand more quickly, the peaks of the waves will be a) the same distance apart b) further apart c) closer together If you lower the frequency of a wave on a string you will lower its speed. b) increase its wavelength. c). lower its amplitude. d) shorten its period.
Answer:
a) correct answer is b higher , b) correct answer is b higher , c) correct answer is b faster , d) traveling wave , e)
Explanation:
A traveling wave is described by the expression
y = A sin (kx - wt)
where k is the wave vector and w is the angular velocity
let's examine every situation presented
a) a new faster pulse is generated
A faster pulse should have a higher angular velocity
equal speed is related to the period and frequency
w = 2π f = 2π / T
therefore in this case the period must decrease so that the angular velocity increases
the correct answer is c narrower
b) Generate a pulse, but move your hand more.
Moving the hand increases the amplitude (A) of the pulse
the correct answer is b higher
c) generates a pulse but the force is tightened
Set means that more tension force is applied to the string, so the velicate changes
v = √ (T /μ)
the correct answer is b faster
d) move your hand back and forth
in this case you would see a pulse series whose sum corresponds to a traveling wave
e) Advance a frame the movie
in this case the wave will be displaced a whole period to the right
the correct answer is b
f) move your hand faster
the waves will have a maximum fast, so they are closer
answer C
g) decrease wave frequency
Since the speed of the wave is a constant m ak, decreasing the frequency must increase the wavelength to keep the velocity constant.
the correct answer is b increases its wavelength
An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is the kinetic energy of the electron at the end of the motion
Answer:
K.E = 15.57 x 10⁻¹⁷ J
Explanation:
First, we find the acceleration of the electron by using the formula of electric field:
E = F/q
F = Eq
but, from Newton's 2nd Law:
F = ma
Comparing both equations, we get:
ma = Eq
a = Eq/m
where,
E = electric field intensity = 120 N/C
q = charge of electron = 1.6 x 10⁻¹⁹ C
m = Mass of electron = 9.1 x 10⁻³¹ kg
Therefore,
a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)
a = 2.11 x 10¹³ m/s²
Now, we need to find the final velocity of the electron. Using 3rd equation of motion:
2as = Vf² - Vi²
where,
Vf = Final Velocity = ?
Vi = Initial Velocity = 1.4 x 10⁷ m/s
s = distance = 3.5 m
Therefore,
(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²
Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)
Vf = 1.85 x 10⁷ m/s
Now, we find the kinetic energy of electron at the end of the motion:
K.E = (0.5)(m)(Vf)²
K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²
K.E = 15.57 x 10⁻¹⁷ J
High voltage power is often carried in wire bundles made up of individual strands. In your initial post to the discussion, discuss the forces on the strands of wire due to the current flowing through them. What would happen if the force acted opposite of the known behavior
Answer:
More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.
Explanation:
A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.
If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.
The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.
In 1898, the world land speed record was set by Gaston Chasseloup-Laubat driving a car named Jeantaud. His speed was 39.24 mph (63.15 km/h), much lower than the limit on our interstate highways today. Repeat the calculations of Example 2.7 (assume the car accelerates for 6 miles to get up to speed, is then timed for a one-mile distance, and accelerates for another 6 miles to come to a stop) for the Jeantaud car. (Assume the car moves in the +x direction.)
Find the acceleration for the first 6 miles.
Answer:
the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]
Explanation:
Given that:
the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.
So; the acceleration for the first 6 miles can be calculated by using the formula:
v₂² = v₁² + 2a (Δx)
Making acceleration a the subject of the formula in the above expression ; we have:
v₂² - v₁² = 2a (Δx)
[tex]a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}[/tex]
[tex]a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}[/tex]
[tex]a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}[/tex]
[tex]a =0.0159 \ m/s^2[/tex]
Thus;
Assume the car moves in the +x direction;
the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]
A circuit contains two elements, but it is not known if they are L, R, or C. The current in this circuit when connected to a 120-V 60 Hz source is 5.3 A and lags the voltage by 65∘.
Part A. What are the two elements?
Part B. What are their values?
Express your answer using two significant figures
Answer:
the two elements are resistor R and inductor L
answers in two significant figures
R = 9.6Ω
L = 54mH
Explanation:
When mapping the equipotentials on the plates with different electrode configurations you may find that some have significant areas with uniform distribution of the equipotential lines. If the distance between such lines is 0.5 cm, what is the electric field there (in units SI)
Answer:
E = V/5 x10⁻³
Explanation:
if the potential difference is V
then electric field E is given by
E = V/d
d = 0.5cm = 5 x 10⁻³m
E = V/5 x10⁻³
The energy band gap of GaAs is 1.4eV. calculate the optimum wavelength of light for photovoltaic generation in a GaAs solar cell.
Answer:
λ = 8.88 x 10⁻⁷ m = 888 nm
Explanation:
The energy band gap is given as:
Energy Gap = E = 1.4 eV
Converting this to Joules (J)
E = (1.4 eV)(1.6 x 10⁻¹⁹ J/1 eV)
E = 2.24 x 10⁻¹⁹ J
The energy required for photovoltaic generation is given as:
E = hc/λ
where,
h = Plank's Constant = 6.63 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = ?
Therefore,
2.24 x 10⁻¹⁹ J = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ
λ = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.24 x 10⁻¹⁹ J)
λ = 8.88 x 10⁻⁷ m = 888 nm
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
Required:
a. If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
b. What does this tell you about the shape of the nearsighted eye?
1. This distance is greater than for the normal eye.
2. This distance is shorter than for the normal eye.
Answer:
a) The distance from the cornea vertex to the retina is 2.37 cm
b) This distance is shorter than for the normal eye.
Explanation:
a) Let refractive index of air,
n(air) = x = 1
Let refractive index of lens,
n(lens) = y = 1.4
Object distance, s = 36 cm
Radius of curvature, R = 0.65 cm
The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.
Image distance, s' = ?
(x/s) + (y/s') = (y-x)/R
(1/36) + (1.4/s') = (1.4 - 1)/0.65
1.4/s' = 0.62 - 0.028
1.4/s' = 0.592
s' = 1.4/0.592
s' = 2.37 cm
Distance from the cornea vertex to the retina is 2.37 cm
(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.
3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charge on the paper of the same magnitude, find the weight of the paper in newtons. Remember to convert the distance to meters and show your work here.
Answer:
[tex]\mathbf{1.51\times10^{-15}N}[/tex]
Explanation:
The computation of the weight of the paper in newtons is shown below:
On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.
Therefore the paper charge is
[tex]q_{paper}=-4.1\times10^{-15}C[/tex]
Now the distance from the charge is
[tex]r=1cm=0.01m[/tex]
Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.
[tex]mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}[/tex]
[tex]\Rightarrow W=mg[/tex]
[tex]=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}[/tex]
[tex]=\mathbf{1.51\times10^{-15}N}[/tex]
The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of friction between the block and incline is o.3, the block will:
Answer:
a. 94.54 N
b. 0.356 m/s^2
Explanation:
Given:-
- The mass of the inclined block, M = 100 kg
- The mass of the vertically hanging block, m = 10 kg
- The angle of inclination, θ = 20°
- The coefficient of friction of inclined surface, u = 0.3
Find:-
a) The magnitude of tension in the cable
b) The acceleration of the system
Solution:-
- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.
- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.
- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:
[tex]T - m*g = m*a\\\\[/tex] ... Eq 1
Where,
a: The acceleration of the system
- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.
- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.
- We will employ the static equilibrium of the inclined block in the normal direction and we have:
[tex]N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )[/tex]
- The frictional force ( F ) is proportional to the contact force ( N ) as follows:
[tex]F = u*N\\\\F = u*M*g *cos ( Q )[/tex]
- Now we will apply the Newton's second law of motion parallel to the plane as follows:
[tex]M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\[/tex] .. Eq2
- Add the two equation, Eq 1 and Eq 2:
[tex]M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}[/tex]
- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).
- Using equation 1, we determine the tension ( T ) in the cable as follows:
[tex]T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N[/tex]
Imagine a metal rod 0.4 m long with a mass of 2 kg. You attach the rod at one end by a lightweight 3.0-m-long cord and twirl the rod around your head. Which expression below gives the best estimate of the moment of inertia of the rod-cord system?
a) I=mr^2, where m = 2kg and r = 3.2m
b) I=1/3mr^2, where m = 2kg and r = 3.4
c) I=1/12mr^2, where m = 2kg and r = 1.7m
d) I=mr^2, where m= 2kg and r = 3.4m
Answer:
Explanation:
Given that
a metal rod 0.4 m long
mass of 2 kg
lightweight 3.0-m-long cord
a) I=mr^2, where m = 2kg and r = 3.2m
b) I=1/3mr^2, where m = 2kg and r = 3.4
c) I=1/12mr^2, where m = 2kg and r = 1.7m
d) I=mr^2, where m= 2kg and r = 3.4m
[tex]I=\int\limits^{3.4}_3 \lambda dx \,x^2\\\\=\lambda\frac{x^3}{3} |^{3.4}_{3}[/tex]
[tex]I=\frac{M}{L}[\frac{3.4^3}{3}-\frac{3^3}{3} ] \\\\I=\frac{2}{(0.4\times3)}[\frac{3.4^3}{3}-\frac{3^3}{3} ]\\\\I=\frac{2}{1.2}(39.304-9)\\\\I=1.67\times12.304\\\\I=20.51[/tex]
[tex]a)I_1=mr^2\\\\I_1=2\times3.2^2\\\\I_1=20.48kgm^2\\\\b)I_2=\frac{1}{3} mr^2\\\\=\frac{1}{3} \times2\times3.4^2\\\\I_2=7.71kgm^2\\\\c)I_3=\frac{1}{12} mr^2\\\\I_3=\frac{1}{12} \times2\times1.7^2\\\\I_3=0.461kg/m^2\\\\d)I_4=mr^2\\\\I_4=2\times3.462\\\\I_4=23.12kg/m^2[/tex]
You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in terms of some unknown units r and v. Rank the wires in order of decreasing electron current. Rank from most to least electron current. To rank items as equivalent, overlap them.
a) radius=3r, drift speed=1v
b) radius=4r, drift speed=0.5v
c) radius=1r, drift speed=5v
d) radius=2r, drift speed=2.5v
Answer:
d > a > b > c
Explanation:
Given that
a) radius = 3r and drift speed = 1v
b) radius = 4r and drift speed = 0.5v
c) radius = 1r and drift speed = 5v
d) radius = 2r and drift speed = 2.5v
Based on the above information, the ranking of the wires for reducing the electron current is
As we can see that the radius i.e to be less and the drift speed that is highest so it should be rank one
And, According to that, other options are ranked
Therefore, the ranking would be d > a > b > c
A wave travels at a speed of 242 m/s. If the distance between crests is 0.11
m, what is the frequency of the wave? Use |
A. 0.00045 Hz
B. 27 Hz
C. 2200 Hz
D. 190 Hz
Answer:
f = 2200 Hz
Explanation:
It is given that,
Speed of a wave is 242 m/s
The distance between crests is 0.11 m
We need to find the frequency of the wave. The distance between crests is called wavelength of a wave. So,
[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{242}{0.11}\\\\f=2200\ Hz[/tex]
So, the frequency of the wave is 2200 Hz.
Answer:2200 hz
Explanation:
A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________
Answer:
Explanation:
Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.
This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.
So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.
A charge of 4.5 × 10-9 C is located 3.2 m from a charge of -2.8 × 10-9 C. Find the
electrostatic force exerted by one charge on the other.
Answer:
[tex]F = -1.107*10^{-8} N\\|F| = 1.107*10^{-8} N[/tex]
Explanation:
[tex]q_1 = 4.5 * 10^{-9} C[/tex]
[tex]q_2 = -2.8 * 10^{-9} C[/tex]
The distance separating the two charges, r = 3.2 m
According to Coulomb's law of electrostatic attraction, the electrostatic force between the two charges can be given by the formula:
[tex]F = \frac{kq_{1} q_{2} }{r^2}[/tex]
Where [tex]k = 9.0 * 10^9 Nm^2/C^2[/tex]
[tex]F = \frac{9*10^9 * 4.5*10^{-9} * (-2.8*10^{-9}}{3.2^2} \\F = \frac{-113.4*10^{-9}}{10.24}\\F = -11.07 *10^{-9}\\F = -1.107*10^{-8}N[/tex]
Answer:
F = 1.1074 × [tex]10^{-8}[/tex] N
Explanation:
An electrostatic force is either a force or attraction or repulsion between two charges. It can be determined by:
F = [tex]\frac{kq_{1}q_{2} }{r^{2} }[/tex]
where: F is the force, k is a constant, [tex]q_{1}[/tex] is the first charge, [tex]q_{2}[/tex] is the second charge and r the distance between the charges.
Given that: k = 9 × [tex]10^{9}[/tex] N[tex]m^{2}[/tex][tex]C^{-2}[/tex], [tex]q_{1}[/tex] = 4.5 × [tex]10^{-9}[/tex]C, [tex]q_{2}[/tex] = -2.8 × [tex]10^{-9}[/tex]C and r = 3.2 m.
Then,
F = [tex]\frac{9*10^{9}*4.5*10^{-9}*2.8*10^{-9} }{3.2^{2} }[/tex]
= [tex]\frac{1.134*10^{-7} }{10.24}[/tex]
= 1.1074 × [tex]10^{-8}[/tex]
The electrostatic force exerted is 1.1074 × [tex]10^{-8}[/tex] N, and it is a force of attraction.
A clarinetist, setting out for a performance, grabs his 3.010 kg clarinet case (including the clarinet) from the top of the piano and carries it through the air with an upward force of 28.05 N. Find the case's vertical acceleration. Indicate an upward acceleration as positive and a downward one as negative. Answer needs to be in m/s2.
Answer:
-0.481 m/s^2
Explanation:
The force equation of this problem is given as:
F - W = ma
where F = upward force holding the clarinet bag
W = downward force (weight of the clarinet)
The mass of the clarinet bag is 3.010 kg, therefore, its weight is:
W = mg
W = 3.010 * 9.8 = 29.498
F = 28.05 N
Therefore:
28.05 - 29.498 = 3.010 * a
-1.448 = 3.010a
=> a = -1.448 / 3.010
a = -0.481 m/s^2
The acceleration of the bag is downward.
The Sun's energy comes from which nuclear reaction?
A. Nuclear fission
B. Gamma decay
C. Positron emission
D. Nuclear fusion
SUBMIT
Sun's energy comes from the nuclear fusion taking place inside. In nuclear fusion two light nuclei fuses together to form a heavy nuclei with the release of greater amount of energy.
What is nuclear fusion :Nuclear fusion is the process of combining two light nuclei to form a heavy nuclei. In this nuclear process, tremendous energy is released. This is the source of heat and light in stars.
On the other hand, nuclear fission is the process of breaking of a heavy nuclei into two lighter nuclei. Fission also produces massive energy. But in comparison, more energy is produced by nuclear fusion.
Nuclear fission is used in nuclear power generators. The light energy and heat energy comes form the nuclear fusion of hydrogens to form helium nuclei. Hence, option D is correct.
Find more on nuclear fusion:
https://brainly.com/question/12701636
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A brass ring of diameter 10.00 cm at 19.0°C is heated and slipped over an aluminum rod with a diameter of 10.01 cm at 19.0°C. Assuming the average coefficients of linear expansion are constant. What if the aluminum rod were 10.02 cm in diameter?
Answer:
the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]
Explanation:
The change in length of a bar can be expressed with the relation;
[tex]\Delta L = L_f - L_i[/tex] ---- (1)
Also ; the relative or fractional increase in length is proportional to the change in temperature.
Mathematically;
ΔL/L_i ∝ kΔT
where;
k is replaced with ∝ (the proportionality constant )
[tex]\dfrac{ \Delta L}{L_i}=\alpha \Delta T[/tex] ---- (2)
From (1) ;
[tex]L_f = \Delta L + L_i[/tex] --- (3)
from (2)
[tex]{ \Delta L}=\alpha \Delta T*{L_i}[/tex] ---- (4)
replacing (4) into (3);we have;
[tex]L_f =(\alpha \Delta T*{L_i} ) + L_i[/tex]
On re-arrangement; we have
[tex]L_f = L_i + \alpha L_i (\Delta T )[/tex]
from the given question; we can say that :
[tex](L_f)_{brass}}} = (L_f)_{Al}[/tex]
So;
[tex]L_{brass} + \alpha _{brass} L_{brass}(\Delta T) = L_{Al} + \alpha _{Al} L_{Al}(\Delta T)[/tex]
Making the change in temperature the subject of the formula; we have:
[tex]\Delta T = \dfrac{L_{Al}-L_{brass}}{\alpha _ {brass} L_{brass}-\alpha _{Al}L_{Al}}[/tex]
where;
[tex]L_{Al}[/tex] = 10.02 cm
[tex]L_{brass}[/tex] = 10.00 cm
[tex]\alpha _{brass}[/tex] = 19 × 10⁻⁶ °C ⁻¹
[tex]\alpha_{Al}[/tex] = 24 × 10⁻⁶ °C ⁻¹
[tex]\Delta T = \dfrac{10.02-10.00}{19*10^{-6} \ \ {^0}C^{-1} *10.00 -24*10^{-6} \ \ {^0}C^{-1} *10.02}[/tex]
[tex]\Delta T[/tex] = −396.1965135 ° C
[tex]\Delta T[/tex] ≅ −396.20 °C
Given that the initial temperature [tex]T_i = 19^0 C[/tex]
Then ;
[tex]\Delta T = T_f - T_i[/tex]
[tex]T_f = \Delta T + T_I[/tex]
Thus;
[tex]T_f =(-396.20 + 19.0)^0 C[/tex]
[tex]\mathbf{T_f = -377.2^0 C}[/tex]
Thus; the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]
A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-the loop which has a radius 5 m. The acceleration of gravity is 9.8 m/s^2.
Required:
How large is the normal force on it if its mass is 5g?
Answer:
N₁ = 393.96 N and N = 197.96 N
Explanation:
In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop
Lowest point, we write Newton's second law n for the y-axis
N -W = m a
where the acceleration is ccentripeta
a = v² / r
N = W + m v² / r
N = mg + mv² / r
we can use energy to find the speed at the bottom of the circle
starting point. Highest point where the ball is released
Em₀ = U = m g h
lowest point. Stop curl down
[tex]Em_{f}[/tex] = K = ½ m v²
Emo = Em_{f}
m g h = ½ m v²
v² = 2 gh
we substitute
N = m (g + 2gh / r)
N = mg (1 + 2h / r)
let's calculate
N₁ = 5 9.8 (1 + 2 17.6 / 5)
N₁ = 393.96 N
headed up
we repeat the calculation in the longest part of the loop
-N -W = - m v₂² / r
N = m v₂² / r - W
N = m (v₂²/r - g)
we seek speed with the conservation of energy
Em₀ = U = m g h
final point. Top of circle with height 2r
[tex]Em_{f}[/tex] = K + U = ½ m v₂² + mg (2r)
Em₀ = Em_{f}
mgh = ½ m v₂² + 2mgr
v₂² = 2 g (h-2r)
we substitute
N = m (2g (h-2r) / r - g)
N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)
N = mg (2h/r - 3)
N = 5 9.8 (2 17.6 / 5 -3)
N = 197.96 N
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Ludwig Boltzmann performed a simple, but powerful experiment to gather evidence concerning the velocity distribution of a sample of gas particles. His experiment revealed that the velocities of gases:
Answer:
Ludwig Boltzmann was an Austrian Physicist and he performed a simple but powerful experiment to gather evidence concerning the velocity distribution of a sample of gas particles.
His experiment revealed that the way Gases are distributed in its normal form across a range of temperature is dependent on the molar mass amen temperature of the gas. Gases with high temperature move faster due to the high number of colliding particles when compared to those with low temperature. Gases with lower molar mass move faster than those with higher molar mass.
A uniform, 4.5 kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.5 m/s in the opposite direction..What is the angular speed of the gate just after it is struck by the unfortunate raven?
Answer:
ω = 1.82 rad/sec
Explanation:
Given
mass of the gate = 4.5 kg
distance between the hinge 1.5 m
mass of the bird = 1.2kg
velocity of the bird = 5 m/s
The gate is rotates in circular motion
Inertia ot the gate = [tex]\frac{1}{3}[/tex]×m×r²
= [tex]\frac{1}{3}[/tex]×4.5×1.5²
= 3.3 kg/m²---------(1)
momentum of the the rave = mv
= 1.2×5
= 6.0 kg
angular momentum of the gate = Iω
By conservation of energy
6. 0= 3.3×ω
ω= 6.0/3.3
= 1.82 rad/sec
from equation 1
I = 3.3 kg/m^2
ω = 1.82 rad/sec
In 1949, an automobile manufacturing company introduced a sports car (the "Model A") which could accelerate from 0 to speed v in a time interval of Δt. In order to boost sales, a year later they introduced a more powerful engine (the "Model B") which could accelerate the car from 0 to speed 2.92v in the same time interval. Introducing the new engine did not change the mass of the car. Compare the power of the two cars, if we assume all the energy coming from the engine appears as kinetic energy of the car.
Answer: [tex]\frac{P_B}{P_A}[/tex] = 8.5264
Explanation: Power is the rate of energy transferred per unit of time: P = [tex]\frac{E}{t}[/tex]
The energy from the engine is converted into kinetic energy, which is calculated as: [tex]KE = \frac{1}{2}.m.v^{2}[/tex]
To compare the power of the two cars, first find the Kinetic Energy each one has:
K.E. for Model A
[tex]KE_A = \frac{1}{2}.m.v^{2}[/tex]
K.E. for model B
[tex]KE_B = \frac{1}{2}.m.(2.92v)^{2}[/tex]
[tex]KE_B = \frac{1}{2}.m.8.5264v^{2}[/tex]
Now, determine Power for each model:
Power for model A
[tex]P_{A}[/tex] = [tex]\frac{m.v^{2} }{2.t}[/tex]
Power for model B
[tex]P_B = \frac{m.8.5264.v^{2} }{2.t}[/tex]
Comparing power of model B to power of model A:
[tex]\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }[/tex]
[tex]\frac{P_B}{P_A} =[/tex] 8.5264
Comparing power for each model, power for model B is 8.5264 better than model A.
A box on a ramp is connected by a rope to a winch. The winch is turned so that the box moves down the ramp at a constant speed. The box experiences kinetic friction with the ramp. Which forces on the box do zero work as the box moves down the ramp?
a. Weight (gravitational force)
b. Normal force
c. Kinetic friction force
d. Tension force
e. None
Answer:
Option B:
The normal force
Explanation:
The normal force does no work as the box slides down the ramp.
Work can only be done when the force succeeds in moving the object in the direction of the force.
All the other forces involved have a component that is moving the box in their direction.
However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)
A power of 100 W is delivered to a certain resistor when the applied voltage is 100 V. Find the resistance. Suppose that the voltage is reduced by 10% to 90 V. By what percentage is the power reduced
Answer:
The resistance is 100 ohmsThe power has reduced 19%Explanation:
The formula for power is given as
[tex]P=\frac{V^2}{R}[/tex]
Given data
power= 100 W
voltage= 100 V
Substituting into the power formula the resistance is
[tex]100=\frac{100^2}{R} \\\\100*R=10000\\\\R= \frac{10000}{100} \\\\R= 100 ohms[/tex]
The resistance is 100 ohms
Now the power has been reduced by 10% (from 100 to 90V)
the power is
[tex]P= \frac{90^2}{100} \\\\P= \frac{8100}{100} \\\\P= 81 watts[/tex]
The power has reduced 19% from 100 watt to 81 watt
When a charge q is placed at a certain point in an electric field, it experiences a force toward the west of magnitude F. If instead a change 2q were placed at that same point what force would it experience?
Answer:
If instead a charge 2q were placed at that same point the force will be 2F.
Explanation:
The electric force is equal to:
[tex] F = q*E [/tex] (1)
Where:
F: is the electric force
q: is the charge
E: is the electric field
We can see that in equation (1) the electric force (F) is proportional to the charge q, thus, if now the charge it's the double (2q) then the force will be the double too:
Initially:
[tex] F_{1} = q_{1}*E [/tex]
Now,
[tex](2q_{1}})*E = 2(q_{1}*E) = 2F_{1}[/tex]
Therefore, if instead a charge 2q were placed at that same point the force will be 2F.
I hope it helps you!
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0°C , a volume of 8.50 cm^3 of mercury overflows the flask.Required:If the coefficient of volume expansion of mercury is βHg = 1.80×10^−4 /K , compute βglass, the coefficient of volume expansion of the glass. Express your answer in inverse kelvins.
Answer:
the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
Explanation:
Given that:
Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³
temperature of the glass flask and mercury= 1.00° C
After heat is applied ; the final temperature = 52.00° C
Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C
Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³
the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K
The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴
The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]
Increase in volume of the glass = 10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]
Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask
the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]
[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]
[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 7.05 cm from the zero volt plate (and 2.95 cm from the other) is 393 V?
Answer:
-18896.49 V/m
Explanation:
Distance between the two plates = 10 cm = 10 x [tex]10^{-2}[/tex] m = 0.1 m
Also, one of the plates is taken as zero volt.
a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V
b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V
Potential field strength = -dV/dx
where dV is voltage difference between these points,
dx is the difference in distance between these points
For the first case above,
potential field strength = -393/0.0705 = -5574.46 V/m
For the second case ,
potential field strength = -393/0.0295 = -13322.03 V/m
Magnitude of the field strength across the plates will be
-5574.46 + (-13322.03) = -5574.46 + 13322.03 = -18896.49 V/m
How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of electrons from an initial point where the electric potential is 9.20 V to a point where the electric potential is -6.90 V
Answer:
-1486 KJ
Explanation:
The work done by an electric field on a charged body is:
W = ΔV * q
where ΔV = change in voltage
q = total charge
The total charge of Avogadro's number of electrons is:
6.0221409 * 10^(23) * -1.6023 * 10^(-19) = -9.65 * 10^(4)
The change in voltage, ΔV, is:
9.20 - (6.90) = 15.4
Therefore, the work done is:
W = -9.65 * 10^(4) * 15.4 = -1.486 * 10^6 J = -1486 KJ
The negative sign means that the motion of the electrons is opposite the electrostatic force.