Answer:
The ability of a water strider to move along the surface of water without breaking the surface is due to:
SURFACE TENSION
Explanation:
this is because Water molecules are more attracted to each other than they are to other materials, so they generate a force to stay together called surface tension. Which allows the strider to move without breaking the surface
The highest mountain on mars is olympus mons, rising 22000 meters above the martian surface. If we were to throw an object horizontaly off the mountain top, how long would it take to reach the surface? (Ignore atmospheric drag forces and use gMars=3.72m/s^2
a. 2.4 minutes
b. 0.79 minutes
c. 1.8 minutes
d. 3.0 minutes
Answer:
t = 1.81 min , the correct answer is c
Explanation:
This is a missile throwing exercise
The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
the final height is y = 0 and the initial height is y₀ = 22000 m
0 = y₀ + 0 - ½ g t²
t = √y 2y₀ / g
let's calculate
t = √(2 22000 / 3.72)
t = 108.76 s
let's reduce to minutes
t = 108.76 s (1 min / 60 s)
t = 1.81 min
The correct answer is c
A car travels at 45 km/h. If the driver breaks 0.65 seconds after seeing the traffic light turn yellow, how far will the car continue to travel before it begins to slow?
Answer:
8.1 m
Explanation:
Convert km/h to m/s.
45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s
Distance = speed × time
d = (12.5 m/s) (0.65 s)
d = 8.125 m
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current
Answer:
Explanation:
From the question we are told that
The radius is [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]
The current is [tex]I = 4.5 \ A[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{J}{\sigma }[/tex]
Where [tex]\sigma[/tex] is the conductivity of aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]
J is the current density which mathematically represented as
[tex]J = \frac{I}{A}[/tex]
Here A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (1.4*10^{-3})^2[/tex]
[tex]A = 6.158*10^{-6} \ m^2[/tex]
So
[tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]
[tex]J = 730757 A/m^2[/tex]
So
[tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]
[tex]E = 0.021 \ N/C[/tex]
What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×10-34 Js.)
Answer:
9.82 × [tex]10^{-35}[/tex] Hz
Explanation:
De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:
λ = [tex]\frac{h}{mv}[/tex]
where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.
Given that: h = 6.63 ×[tex]10^{-34}[/tex] Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;
λ = [tex]\frac{h}{mv}[/tex]
= [tex]\frac{6.63*10^{-34} }{2.5*2.7}[/tex]
= [tex]\frac{6.63 * 10^{-34} }{6.75}[/tex]
= 9.8222 × [tex]10^{-35}[/tex]
The wavelength of the object is 9.82 × [tex]10^{-35}[/tex] Hz.
When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass
Answer:
This means that mercury has a higher or faster expansion rate than glass
Explanation:
This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).
3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.
Answer:
Check your router connections then restart your router.
Explanation:
Answer:
Check your router connections then restart your router.
Explanation:
Most internet access comes from routers so the problem is most likely the router.
A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?
Answer:
the child is 1.581 m far from the fence
Explanation:
The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.
From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:
[tex]x - x_o = u_xt[/tex]
[tex]\mathtt{x = u_xt \ \ \ since (x_o = 0)}[/tex] ---- (1)
the equation of the motion y is :
[tex]\mathtt{y - y_o =u_yt - 0.5 gt^2}[/tex]
[tex]\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}[/tex]
[tex]\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }[/tex]
[tex]\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}[/tex]
[tex]\mathtt{1 = 5.14t - 4.9t^2}[/tex]
[tex]\mathtt{4.9t^2 - 5.14t +1 = 0}[/tex]
By using the quadratic formula, we have;
[tex]\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }[/tex]
where;
a = 4.9, b = -5.14 c = 1
[tex]= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }[/tex]
[tex]= \mathbf{ 0.791 \ \ OR \ \ 0.258} }[/tex]
In as much as the ball is traveling upward, then we consider t= 0.258sec.
From equation (1)
[tex]\mathtt{x = u_x(0.258)}[/tex]
[tex]\mathtt{x = ucos 40^0 (0.258)}[/tex]
[tex]\mathtt{x = 8 \ cos 40^0 (0.258)}[/tex]
[tex]\mathbf{x = 1.581 \ m}[/tex]
Thus, the child is 1.581 m far from the fence
Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object
Answer:
Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.
Explanation:
A 26-g rifle bullet traveling 220 m/s embeds itself in a 3.8-kg pendulum hanging on a 2.7-m-long string, which makes the pendulum swing upward in an arc, Determine the vertical and horizontal component of the pendulum's maximum displacement
Answer:
displacements are 0.776m, 0.114m
Explanation:
We were given mass of 26-g rifle bullet , then we can convert to Kg since
Momentum is conserved here.
The initial momentum before impact = (Mi * Vi)
Where Mi= initial given mass
Vi=initial velocity given
= 0.026 * 220 = 5.72 kgm/s
The final momentum after impact is (Mf * Vf )
Mf= final mass
5.72=( 3.82* Vf )
= 5.72/ 3.82
= 1.497 m/s
the speed of the pendulum bob with bullet afterwards= 1.497 m/s
the total energy after the collision is the addition of the kinetic energy of the bob+bullet and the potential energy of the bob and bullet, potential energy can be taken as zero.
M = 3.82 kg the mass of the bob containing the bullet
E(total) = ¹/₂MV² = 1/2 * (3.82kg)*(1.497m/s)² = 4.280J
When the Bob got to highest point the kinetic energy is zero and the potential energy is due to the increase in height of the bob, and the addition of the potential and kinetic energies still equal the total energy from before
E(total) = Mgh + 0 = Mgh = 4.280J
solving for h and substituting,
h = 4.280 J/(9.8m/s^2*3.82kg) = 0.114 m
Since the height is found,we the angle of the pendulum at the top of the swing can also be determined
A = arccos[(2.7 - 0.114) / 2.7] or A = 16.71degrees
Since A is known, the displacement along the horizontal axis can be calculated as
x = 2.7* sin(A) = 0.776m
therefore, displacement is 0.776m, 0.114m
the vertical and horizontal component of the pendulum's maximum displacement are displacement is 0.776m, 0.114m
A body is thrown vertically upwards with a speed of 95m / s and after 7s it reaches its maximum height. How fast does it reach its maximum height? What was the maximum height reached?
Explanation:
u = 95 m/sec ( Initial speed)
t = 7 sec ( Time of ascent)
According to Equations of Motion :
[tex]s = ut - \frac{1}{2} g {t}^{2} [/tex]
Max. Height = 95 * 7 - 4.9 * 49 = 424. 9 = 425 m
Answer:
332.5 m
Explanation:
At the maximum height, the velocity is 0.
Given:
v₀ = 95 m/s
v = 0 m/s
t = 7 s
Find: Δy
Δy = ½ (v + v₀) t
Δy = ½ (0 m/s + 95 m/s) (7 s)
Δy = 332.5 m
The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.
Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.
Hope this helps!
Answer:
The electromagnetic waves reach Earth, while the mechanical waves do not
Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3
Answer:
v₂ = 9 v
Explanation:
For this exercise in fluid mechanics, let's use the continuity equation
v₁ A₁ = v₂ A₂
where v is the velocity of the fluid, A the area of the pipe and the subscripts correspond to two places of interest.
The area of a circle is
A = π R²
let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint
In this case v₁ = v and the area is
A₁ = π R²
in the second point
A₂= π (R / 3)²
we substitute in the continuity equation
v π R² = v₂ π R² / 9
v = v₂ / 9
v₂ = 9 v
To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 2 cm
Answer:
The self-inductance in henries for the solenoid is 0.0274 H.
Explanation:
Given;
number of turns, N = 1500 turns
length of the solenoid, L = 13 cm = 0.13 m
radius of the wire, r = 2 cm = 0.02 m
The self-inductance in henries for a solenoid is given by;
[tex]L = \frac{\mu_oN^2A}{l}[/tex]
where;
[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]
A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²
[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]
Therefore, the self-inductance in henries for the solenoid is 0.0274 H.
When light of wavelength 233 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.98 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface
Answer:
λmax = 372 nm
Explanation:
First we find the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m
c = speed of light = 3 x 10⁸ m/s
h = Planks Constant = 6.626 x 10⁻³⁴ J.s
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)
E = 8.5 x 10⁻¹⁹ J
Now, from Einstein's Photoelectric Equation:
E = Work Function + Kinetic Energy
8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)
Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J
Work Function = 5.332 x 10⁻¹⁹ J
Since, work function is the minimum amount of energy required to emit electron. Therefore:
Work Function = hc/λmax
λmax = hc/Work Function
where,
λmax = maximum wavelength of light that will produce photoelectrons = ?
Therefore,
λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)
λmax = 3.72 x 10⁻⁷ m
λmax = 372 nm
How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m
Answer:
Explanation:
volume of water being lifted
= π r² h , where r is radius of cylinder and h is height of cylinder
= 3.14 x5² x 10
= 785 m³
mass of water = 785 x 10³ kg
mass of this much of water is lifted so that its centre of mass is lifted by height
10 / 2 = 5m .
So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity
= 785 x 10³ x 9.8 x 5
= 38.465 x 10⁶ J
Object A, with heat capacity CA and initially at temperature TA, is placed in thermal contact with object B, with heat capacity CB and initially at temperature TB. The combination is thermally isolated. If the heat capacities are independent of the temperature and no phase changes occur, the final temperature of both objects is
Answer:
d) (CATA + CBTB) / (CA + CB)
Explanation:
According to the given situation, the final temperature of both objects is shown below:-
We assume T be the final temperature
while m be the mass
So it will be represent
m CA (TA - T) = m CB (T - TB)
or we can say that
CATA - CA T = CB T - CBTB
or
(CA + CB) T = CATA + CBTB
or
T = (CA TA + CBTB) ÷ (CA + CB)
Therefore the right answer is d
The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
The given parameters;
heat capacity of object A = CAinitial temperature of object A = TAheat capacity of object B = CBinitial temperature of object B = TBThe final temperature of both objects is calculated as follows;
heat lost by object A is equal to heat gained by object B
[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]
Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
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Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?
Answer:
10.75 A
The current is in opposite direction since it causes a repulsion force between the wires
Explanation:
Force per unit length on the wires = 4.30×10^−5 N/m
distance between wires = 2.6 cm = 0.026 m
current through one wire = 0.52 A
current on the other wire = ?
Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as
[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]
where [tex]F/l[/tex] is the force per unit length on the wires
[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A
[tex]I_{1}[/tex] = current on the first wire = 0.520 A
[tex]I_{2}[/tex] = current on the other wire = ?
r = the distance between the two wire = 0.026 m
substituting the value into the equation, we have
4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] = [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]
4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]
[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A
The current is in opposite direction since it causes a repulsion force between the wires.
A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?
Answer:
Pls seeattached file
Explanation:
A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.
What is Resistance?Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.
When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.
According to the question,
R = R₀ (1 + α ΔT)
(1 + 0.0135)R₀ = R₀(1 + α ΔT)
ΔT = (1 + 0.0135) / α
= 0.0135 / 0.0004
= 33.75 °C.
ΔT = [(1 - 0.0135) -1]/0.004
= -33.75 °C
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If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 120 years into the future. According to special relativity, how fast must you travel
Answer:
I must travel with a speed of 2.97 x 10^8 m/s
Explanation:
Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months
Time that elapses on the spacecraft = 1 year
On earth the people have advanced 120 yrs
According to relativity, the time contraction on the spacecraft is gotten from
[tex]t[/tex] = [tex]t_{0} /\sqrt{1 - \beta ^{2} }[/tex]
where
[tex]t[/tex] is the time that elapses on the spacecraft = 120 years
[tex]t_{0}[/tex] = time here on Earth = 1 year
[tex]\beta[/tex] is the ratio v/c
where
v is the speed of the spacecraft = ?
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
120 = 1/[tex]\sqrt{1 - \beta ^{2} }[/tex]
squaring both sides of the equation, we have
14400 = 1/[tex](1 - \beta ^{2} )[/tex]
14400 - 14400[tex]\beta ^{2}[/tex] = 1
14400 - 1 = 14400[tex]\beta ^{2}[/tex]
14399 = 14400[tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 14399/14400 = 0.99
[tex]\beta = \sqrt{0.99}[/tex] = 0.99
substitute β = v/c
v/c = 0.99
but c = 3 x 10^8 m/s
v = 0.99c = 0.99 x 3 x 10^8 = 2.97 x 10^8 m/s
The following equation is an example of
decay.
181
185
79
Au →
4
2
He+
Answer:
Alp decay.
Explanation:
From the above equation, the parent nucleus 185 79Au produces a daughter nuclei 181 77 Ir.
A careful observation of the atomic mass of the parent nucleus (185) and the atomic mass of the daughter nuclei (181) shows that the atomic mass of the daughter nuclei decreased by a factor of 4. Also, the atomic number of the daughter nuclei also decreased by a factor of 2 when compared with the parent nucleus as shown in the equation given above.
This simply means that the parent nucleus has undergone alpha decay which is represented with a helium atom as 4 2He.
Therefore, the equation is an example of alpha decay.
Four charges each of magnitude 15 µC are arranged on the corners of a square of side 5 cm. What is the total potential energy of the system?
Answer:
-105J
Explanation:
See attached file
A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The induced current through resistor R is:
Answer:
The induced current in the resistor is I = BLv/R
Explanation:
The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by
ε = BLv.
Now, the current I in the resistor is given by
I = ε/R where ε = induced emf in circuit and R = resistance of resistor.
So, the current I = ε/R.
substituting the value of ε the induced emf, we have
I = ε/R
I = BLv/R
So, the induced current through the resistor is given by I = BLv/R
A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Complete Question
A 590-turn solenoid is 12 cm long. The current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Answer:
The force is [tex]F = 0.1602 \ N[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 590 \ turns[/tex]
The length of the solenoid is [tex]L = 12 \ cm = 0.12 \ m[/tex]
The current is [tex]I = 36 \ A[/tex]
The diameter is [tex]D = 4.5 \ cm = 0.045 \ m[/tex]
The current carried by the wire is [tex]I = 27 \ A[/tex]
The length of the wire is [tex]l = 2 cm = 0.02 \ m[/tex]
Generally the magnitude of the force on this wire assuming the solenoid's field points due east is mathematically represented as
[tex]F = B * I * l[/tex]
Here B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{L}[/tex]
Here [tex]\mu _o[/tex] is permeability of free space with value [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]
substituting values
[tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]
[tex]B = 0.2225 \ T[/tex]
So
[tex]F = 0.2225 * 36 * 0.02[/tex]
[tex]F = 0.1602 \ N[/tex]
A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.
Answer: Ф = 17.2657 ≈ 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
∴ tanФ = a/g
Ф = tan⁻¹ a / g
Ф = tan⁻¹ ( 10 / 32.17405)
Ф = tan⁻¹ 0.31080948777
Ф = 17.2657 ≈ 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.630 cm thick and heat conduction occurs through an area of 1.45 ✕ 10−2 m2 at a rate of 500 J/s, what is the temperature difference across it (in °C)? Ceramic has the same thermal conductivity as glass and concrete brick.
Answer:
The temperature difference [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
Explanation:
From the question we are told that
The thickness is [tex]\Delta x = 0.630 cm = 0.0063 m[/tex]
The area is [tex]A = 1.45 *10^{-2 } \ m^2[/tex]
The rate is [tex]P = 500 J/s[/tex]
The thermal conductivity is [tex]\sigma = 0.84J[\cdot s \cdot m \cdot ^oC ][/tex]
Generally the rate heat conduction mathematically represented as
[tex]P = \sigma * A * \frac{\Delta T}{\Delta x }[/tex]
=> [tex]\Delta T = \frac{P * \Delta x }{\sigma * A }[/tex]
=> [tex]\Delta T = \frac{ 500 * 0.00630 }{ 0.84 * 1.45 *10^{-2} }[/tex]
=> [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment
Answer: Speed = [tex]3.10^{-31}[/tex] m/s
Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:
[tex]p_{f} = p_{i}[/tex]
Relativistic momentum is calculated as:
p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]
where:
m is rest mass
u is velocity relative to an observer
c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)
Initial momentum is zero, then:
[tex]p_{f}[/tex] = 0
[tex]p_{1}-p_{2}[/tex] = 0
[tex]p_{1} = p_{2}[/tex]
To find speed of the heavier fragment:
[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]
[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]
[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]
[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]
[tex]u_{1} = 3.10^{-31}[/tex]
The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.
A 750 gram grinding wheel 25.0 cm in diameter is in the shape of a uniform solid disk. (we can ignore the small hole at the center). when it is in use, it turns at a consant 220 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 45.0 s with constant angular acceleration due to friction at the axle. What torque does friction exert while this wheel is slowing down?
Answer:
Torque = 0.012 N.m
Explanation:
We are given;
Mass of wheel;m = 750 g = 0.75 kg
Radius of wheel;r = 25 cm = 0.25 m
Final angular velocity; ω_f = 0
Initial angular velocity; ω_i = 220 rpm
Time taken;t = 45 seconds
Converting 220 rpm to rad/s we have;
220 × 2π/60 = 22π/3 rad/s
Equation of rotational motion is;
ω_f = ω_i + αt
Where α is angular acceleration
Making α the subject, we have;
α = (ω_f - ω_i)/t
α = (0 - 22π/3)/45
α = -0.512 rad/s²
The formula for the Moment of inertia is given as;
I = ½mr²
I = (1/2) × 0.75 × 0.25²
I = 0.0234375 kg.m²
Formula for torque is;
Torque = Iα
For α, we will take the absolute value as the negative sign denotes decrease in acceleration.
Thus;
Torque = 0.0234375 × 0.512
Torque = 0.012 N.m
The positron has the same mass as an electron, with an electric charge of +e. A positron follows a uniform circular motion of radius 5.03 mm due to the force of a uniform magnetic field of 0.85 T. How many complete revolutions does the positron perform If it spends 2.30 s inside the field? (electron mass = 9.11 x 10-31 kg, electron charge = -1.6 x 10-19 C)
Answer:
5.465 × 10^10 revolutions
Explanation:
Formula for Magnetic Field = m. v/ q . r
M = mass of electron = mass of positron = 9.11 x 10^-31 kg,
radius of the positron = 5.03 mm
We convert to meters.
1000mm = 1m
5.03mm = xm
Cross multiply
x = 5.03/1000mm
x = 0.00503m
q = Electric charge = -1.6 x 10^-19 C
Magnetic field (B) = 0.85 T
Speed of the positron is unknown
0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503
0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v
v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg
v = 6.8408 ×10-22/ 9.11 x 10^-31 kg
v = 750911086.72m/s
Formula for complete revolutions =
Speed × time / Circumference
Time = 2.30s
Circumference of the circular path = 2πr
r =0.00503
Circumference = 2 × π × 0.00503
= 0.0316044221
Revolution = 750911086.72 × 2.30/0.0316044221
= 1727095499.5/0.0316044221
= 546541562294 revolutions
Approximately = 5.465 × 10^10 revolutions
If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?
Answer:
[tex]v=1.24\times 10^8\ m/s[/tex]
Explanation:
Given that,
The refractive index of benzene is 2.419
We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,
[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]
So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].
