Answer:
distance r from the uranium atom is 18.27 nm
Explanation:
given data
uranium and iron atom distance R = 44.10 nm
uranium atom = singly ionized
iron atom = doubly ionized
to find out
distance r from the uranium atom
solution
we consider here that uranium electron at distance = r
and electron between uranium and iron so here
so we can say electron and iron distance = ( 44.10 - r ) nm
and we know single ionized uranium charge q2= 1.602 × [tex]10^{-19}[/tex] C
and charge on iron will be q3 = 2 × 1.602 × [tex]10^{-19}[/tex] C
so charge on electron is q1 = - 1.602 × [tex]10^{-19}[/tex] C
and we know F = [tex]k\frac{q*q}{r^{2} }[/tex]
so now by equilibrium
Fu = Fi
[tex]k\frac{q*q}{r^{2} }[/tex] = [tex]k\frac{q*q}{r^{2} }[/tex]
put here k = [tex]9*10^{9}[/tex] and find r
[tex]9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }[/tex] = [tex]9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }[/tex]
[tex]\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}[/tex]
r = 18.27 nm
distance r from the uranium atom is 18.27 nm
Light travels at a speed of close to 3 x 10^5 km/s in vacuum. Given that it takes light 8 min and 19 s to travel the distance from the center of the Earth to the center of the Sun, how far away is the Sun from the Earth? (Astronomers use this as a "distance unit" called 1 Astronomical Unit or 1 au)
Answer:
1497×10⁵ km
Explanation:
Speed of light in vacuum = 3×10⁵ km/s
Time taken by the light of the Sun to reach the Earth = 8 min and 19 s
Converting to seconds we get
8×60+19 = 499 seconds
Distance = Speed × Time
[tex]\text{Distance}=3\times 10^5\times 499\\\Rightarrow \text{Distance}=1497\times 10^5\ km[/tex]
1 AU = 1497×10⁵ km
The Sun is 1497×10⁵ km from Earth
Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a stopwatch just as the top of the sun disappears. You then stand, elevating your eyes by a height H = 1.43 m, and stop the watch when the top of the sun again disappears. If the elapsed time is it = 11.9 s, what is the radius r of the planet to two significant figures? Notice that duration of a solar day at the far-off planet is the same that is on Earth.
Answer:
R=3818Km
Explanation:
Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.
Since you travel 2π radians in 24H, the angle can be calculated as:
[tex]\alpha =\frac{2*\pi *t}{24H}[/tex] t being expressed in hours.
[tex]\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad[/tex]
From the triangle formed by A,B and the center of the planet, we know that:
[tex]cos(\alpha )=\frac{r}{r+H}[/tex] Solving for r, we get:
[tex]r=\frac{H*cos(\alpha) }{1-cos(\alpha) } =3818Km[/tex]
which graph shows the variation with amplitude a of the intensity i for a wavelength for this spectrum
Answer: the right answer is C
Explanation:
A ball is thrown down vertically with an initial speed of 20 m/s from a height of 60 m. Find (a) its speed just before it strikes the ground and (b) how long it takes for the ball to reach the ground. Repeat (a) and (b) for the ball thrown directly up from the same height and with the same initial speed.
Answer:
Explanation:
Ball is thrown downward:
initial velocity, u = - 20 m/s (downward)
height, h = - 60 m
Acceleration due to gravity, g = - 9.8 m/s^2 (downward)
(a) Let the speed of the ball as it hits the ground is v.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]
v = 39.69 m/s
(b) Let t be the time taken
Use First equation of motion
v = u + a t
- 39.69 = - 20 - 9.8 t
t = 2 second
Now the ball is thrown upwards:
initial velocity, u = 20 m/s (upward)
height, h = - 60 m
Acceleration due to gravity, g = - 9.8 m/s^2 (downward)
(c) Let the speed of the ball as it hits the ground is v.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]
v = 39.69 m/s
(d) Let t be the time taken
Use First equation of motion
v = u + a t
- 39.69 = + 20 - 9.8 t
t = 6.09 second
The temperature of 1 m^3 of water is decreased by 10°C. If this thermal energy is used to lift the water vertically against gravity, what is the change in height of the center of mass?
Answer:
h = 4271.43 m
Explanation:
given,
Volume of the water = 1 m³
temperature decrease by = 10°C
heat removed from water
Q = m c ΔT
Q = ρ V c ΔT
= 1000 × 1 × 4186 × 10
= 4.186 × 10⁷ J
energy is used to do work to move the water against its weight
Q = force × displacement
4.186 × 10⁷ J = m g × h
4.186 × 10⁷ J = 1000 × 1 × 9.8 × h
h = 4271.43 m
hence, the change in height of is equal to h = 4271.43 m
When an object is thrown upwards and reaches its maximum height its speed is: a. Greater than the initial
b. Less than the initial
c. Same as the initial
d. Zero
Answer:
Option d
Explanation:
When we throw an object in the upward direction, we provide it with certain initial velocity due to which it covers a certain distance up to the maximum height.
While the object is moving in the upward direction, its velocity keeps on reducing due to the acceleration due to gravity which acts vertically downwards in the opposite direction thus reducing its velocity.
So, the maximum height attained by the object is the point where this upward velocity of the body becomes zero and after that the object starts to fall down.
A ball is dropped from rest at the top of a 6.10 m
tallbuilding, falls straight downward and collides inelastically
withthe ground, and bounces back. The ball loses 10% of itskinetic
energy every time it collides with the ground. Howmany bounces can
the ball make and still reach a windowsill that is2.38 m above the
ground?
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
[tex]\frac{v_1}{v}[/tex] = e ( coefficient of restitution ) = [tex]\frac{1}{\sqrt{10} }[/tex]
and
[tex]\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }[/tex]
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
[tex]e = \sqrt{\frac{h_1}{6.1} }[/tex]
[tex]e = \sqrt{\frac{h_2}{h_1} }[/tex]
So on
[tex]e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }[/tex]
[tex](\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}[/tex]= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
The ball can bounce approximately 8 times before it reaches the height of 2.38 meters after accounting for a 10% energy loss each collision.
This problem involves a ball undergoing inelastic collisions, losing 10% of its kinetic energy with each bounce, and determining how many times it can bounce to still reach a windowsill 2.38 m high.
First, let's calculate the initial potential energy (PEinitial) of the ball when it is dropped from a height (hinitial) of 6.10 m:
PEinitial = mghinitial
where, g = 9.8 m/s² (acceleration due to gravity)
As it falls, this entire potential energy converts into kinetic energy (KEinitial) at the ground:
KEinitial = PEinitial = mghinitial
Upon each bounce, the ball loses 10% of its kinetic energy. Therefore, it retains 90% of its kinetic energy:
KEnew = 0.9 × KEprevious
To find out how high it can bounce after each loss of energy, convert kinetic energy back into potential energy:
PE = KE = mgh
After each bounce, the height the ball can reach is calculated by applying the 90% retention factor:
hnew = 0.9 × hprevious
Starting with h0 = 6.10 m:
Next height (h1) = 0.9 × 6.10 m = 5.49 mNext height (h2) = 0.9 × 5.49 m = 4.94 mNext height (h3) = 0.9 × 4.94 m = 4.45 mContinue this process until hn is less than 2.38 m:The number of bounces can be calculated using the formula: hn = 6.10 × (0.9)n, where n is the number of bounces. Set hn = 2.38 m and solve for n:
2.38 = 6.10 × (0.9)n
Divide both sides by 6.10:
0.39 ≈ (0.9)n
Taking the natural log (ln) of both sides:
ln(0.39) = n × ln(0.9)
Finally, solving for n:
n ≈ ln(0.39) / ln(0.9) ≈ 8
Therefore, the ball can bounce approximately 8 times and still reach the windowsill that is 2.38 m above the ground.
An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1 microsecond is 1x10 65). (Express your answers in vector form.) Part 1 (a) What is the average velocity of the electron? Vavg = < > m/s Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER Part 2 (b) If the electron continues to travel at this average velocity, where will it be in another 9 us? 7 = < > m
Final answer:
The electron's average velocity is found to be (0 m/s, 316,000 m/s, -146,000 m/s), and after another 9 microseconds, it will be at the position (0.02 m, 4.464 m, -2.104 m).
Explanation:
To calculate the average velocity of an electron, we use the formula:
Vavg = (rf - ri) / Δt, where rf is the final position, ri is the initial position, and Δt is the time interval between the positions.
Given the initial position (0.02 m, 0.04 m, -0.06 m) and the final position (0.02 m, 1.62 m, -0.79 m), with a time difference of 5 microseconds (μs), which is 5 x 10-6 seconds:
The position change in vector form is
Δr = (0.02 m - 0.02 m, 1.62 m - 0.04 m, -0.79 m - (-0.06 m))
= (0 m, 1.58 m, -0.73 m).
Thus, the average velocity is
Vavg = Δr / Δt
= (0 m, 1.58 m, -0.73 m) / (5 x 10-6 s)
= (0 m/s, 316,000 m/s, -146,000 m/s)
The electron's new position after another 9 μs, moving with the same average velocity, is calculated by:
rnew = rf + Vavg × Δtnew
Here, Δtnew is 9 μs, which is 9 x 10-6 seconds, so:
rnew = (0.02 m, 1.62 m, -0.79 m) + (0 m/s, 316,000 m/s, -146,000 m/s) × (9 x 10-6 s)
= (0.02 m, 1.62 m + (2.844 m), -0.79 m - (1.314 m))
= (0.02 m, 4.464 m, -2.104 m).
The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could double the charge to 2Q and at the same time reduce the distance to R/2. reduce the distance to R/4. double the charge to 2Q. reduce the distance to R/2. double the distance to 2R.
Answer:
Double the charge to 2Q
Explanation:
The magnitude of a electric field caused by a charge Q at a distance R can be find by the following expression:
[tex]E = K\frac{Q}{R^2}[/tex]
Where K is the Coulomb constant.
As the relationship between the charge and the electric fiel is proportional, by simply doubling the charge, you would double the magnitude of the electric field.
Notice that the distance affects the magnitude of the electric field by the inverse square. So if you half the distance, the magnitude of the field will quadruple.
To double the electric field's magnitude at point P from a point charge Q at distance R, double the charge to 2Q while keeping the distance R unchanged. Option B is the correct option.
To double the magnitude of the electric field at point P due to a point charge Q a distance R away from P, you can follow Coulomb's law. Electric field (E) is directly proportional to the charge (Q) and inversely proportional to the square of the distance (R) between the point charge and the observation point.
If you double the charge to 2Q while keeping the distance R the same, the new electric field magnitude at point P will be double the original E. This is because E ∝ Q. Other options like reducing the distance to R/2 or R/4, or doubling the distance to 2R, would not result in a doubling of the field magnitude but rather lead to different field strengths according to the inverse square law, and altering both the charge and distance simultaneously is not necessary to achieve this specific goal. Option B is the correct option.
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The appropriate question is :
The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could double the charge to 2Q and at the same time reduce the distance to R/2 OR
A) reduce the distance to R/4.
B) double the charge to 2Q.
C)reduce the distance to R/2.
D) double the distance to 2R.
Which of the following does not have the appropriate SI unit? work - Joule
acceleration - m/s2
power - Watt
momentum - kg.m/s
force - Pound
Answer:
Explanation:
The SI system of units is the system which is Standard International system. It is used internationally.
The SI units of the fundamental quantities are given below
Mass - Kilogram
Length - metre
Time - second
temperature - Kelvin
Amount of substance - mole
Electric current - Ampere
Luminous Intensity - Candela
So, The SI unit of work is Joule
SI unit of acceleration is m/s^2
SI unit of power is watt
SI unit of momentum is kg m /s
SI unit of force is newton
Thus, the last option is incorrect.
Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write R in vector form. (b) Write R showing the magnitude and direction in degrees.
Answer:
R = ( 4.831 m , 1.469 m ) Magnitude of R = 5.049 mDirection of R relative to the x axis= 16°54'33'Explanation:
Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula
[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
where [tex]| \vec{A} |[/tex] its the magnitude and θ.
So, for our vectors, we will have:
[tex]\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )[/tex]
[tex]\vec{D}= ( 2.121 m , -2.121 m )[/tex]
and
[tex]\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )[/tex]
[tex]\vec{E}= ( 2.71 m , 3.59 m )[/tex]
Now, we can take the sum of the vectors
[tex]\vec{R} = \vec{D} + \vec{E}[/tex]
[tex]\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )[/tex]
[tex]\vec{R} = ( 2.121 \ m + 2.71 \ m , -2.121 \ m + 3.59 \ m ) [/tex]
[tex]\vec{R} = ( 4.831 \ m , 1.469 \ m ) [/tex]
This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem
[tex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/tex]
[tex]|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}[/tex]
[tex]|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}[/tex]
[tex]|\vec{R}| = \sqrt{25.496 m^2}[/tex]
[tex]|\vec{R}| = 5.049 m[/tex]
To find the direction, we can use
[tex]\theta = arctan(\frac{R_y}{R_x})[/tex]
[tex]\theta = arctan(\frac{1.469 \ m}{4.831 \ m})[/tex]
[tex]\theta = arctan(0.304)[/tex]
[tex]\theta = 16\°54'33''[/tex]
As we are in the first quadrant, this is relative to the x axis.
Find the volume of a cylinder of height 10 cm if its base is a square of side 4 cm
Answer:
[tex]Volume=160 cm^3[/tex]
Explanation:
The volume of the cylinder is given by the area of the base multiplied by the height.
In this case:
Height:
[tex]h = 10 cm\\[/tex]
Base area = area of the square ([tex]area=side*side=side^2[/tex])
the side of the square is:
[tex]side=4cm[/tex]
thus, the area of the base:
[tex]area=(4cm)^2 = 16 cm^2[/tex]
Now we multiply this quantities, to find the volume:
[tex]Volume= 16 cm^2*10cm=160 cm^3[/tex]
Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in this process?
Answer:
a) m = 993 g
b) E = 6.50 × 10¹⁴ J
Explanation:
atomic mass of hydrogen = 1.00794
4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176
we know atomic mass of helium = 4.002602
difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158
fraction of mass lost = [tex]\dfrac{0.029158}{4.03176}[/tex]= 0.00723
loss of mass for 1000 g = 1000 × 0.00723 = 7.23
a) mass of helium produced = 1000-7.23 = 993 g (approx.)
b) energy released in the process
E = m c²
E = 0.00723 × (3× 10⁸)²
E = 6.50 × 10¹⁴ J
Answer:
(a) 992.87 g
(b) [tex]6.419\times 10^{14} J[/tex]
Solution:
As per the question:
Mass of Hydrogen converted to Helium, M = 1 kg = 1000 g
(a) To calculate mass of He produced:
We know that:
Atomic mass of hydrogen is 1.00784 u
Also,
4 Hydrogen atoms constitutes 1 Helium atom
Mass of Helium formed after conversion:
[tex]4\times 1.00784 = 4.03136 u[/tex]
Also, we know that:
Atomic mass of Helium is 4.002602 u
The loss of mass during conversion is:
4.03136 - 4.002602 = 0.028758 u
Now,
Fraction of lost mass, M' = [tex]\frac{0.028758}{4.03136} = 0.007133 u[/tex]
Now,
For the loss of mass of 1000g = [tex]0.007133\times 1000[/tex] = 7.133 g
Mass of He produced in the process:
[tex]M_{He} = 1000 - 7.133 = 992.87 g[/tex]
(b) To calculate the amount of energy released:
We use Eintein' relation of mass-enegy equivalence:
[tex]E = M'c^{2}[/tex]
[tex]E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J[/tex]
A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including its uncertainty
Answer:
(177.94 ± 3.81) cm^2
Explanation:
l + Δl = 21.7 ± 0.2 cm
b + Δb = 8.2 ± 0.1 cm
Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2
Now use error propagation
[tex]\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}[/tex]
[tex]\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}[/tex]
[tex]\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81[/tex]
So, the area with the error limits is written as
A + ΔA = (177.94 ± 3.81) cm^2
The gauge pressure in your car tires is 2.40 x 10^5 N/m^2 at a temperature of 35.0°C when you drive it onto a ferry boat to Alaska. What is their gauge pressure (in atm) later, when their temperature has dropped to −42.0°C? (Assume that their volume has not changed.)
Answer:
The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]
Explanation:
Given that,
Gauge pressure of car tires [tex]P_{1}=2.40\times10^{5}\ N/m^2[/tex]
Temperature [tex]T_{1}=35.0^{\circ}C = 35.0+273=308 K[/tex]
Dropped temperature [tex]T_{2}= -42.0^{\circ}C=273-42=231 K[/tex]
We need to calculate the gauge pressure P₂
Using relation pressure and temperature
[tex]\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{2.40\times10^{5}}{308}=\dfrac{P_{2}}{231}[/tex]
[tex]P_{2}=\dfrac{2.40\times10^{5}\times231}{308}[/tex]
[tex]P_{2}=180000 = 1.8\times10^{5}\ N/m^2[/tex]
Hence, The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]
f the electric field is zero at a particular point, must the electric potential be zero at the same point? Explain
Answer:
No
Explanation:
As we know that the electric field nullity does not define that the electric potential will be zero at that point.
For example consider the two positive charge at the mid point of these charge electric field is zero but potential is finite.The electric potential has two contribution means it is positive if charges are positive and it is negative if charges are negative.A closed system consisting of 2 lb of a gas undergoes a process during which the relation between pressure and volume is pVn = constant. The process begins with p1 = 15 lbf/in.2, ν1 = 1.25 ft3/lb and ends with p2 = 60 lbf/in.2, ν2 = 0.5 ft3/lb. Determine (a) the volume, in ft3, occupied by the gas at states 1 and 2 and (b) the value of n.
Answer:
V1=2.5ft3
V2=1ft3
n=1.51
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=2lb*1.25ft3/lb=2.5ft3
state 2
V2=m.v2
V2=2lb*0.5ft3/lb= 1ft3
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/60)/ ln (1/2.5)
n=1.51
Water has a mass per mole of 18.0 g/mol, and each water molecule (H20) has 10 electrons. (a) How many electrons are there in one liter (1.00 x 10 m ) of water?
Answer:
total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron
Explanation:
given data
mass per mole = 18 g/mol
no of electron = 10
to find out
how many electron in 1 liter of water
solution
we know molecules per gram mole is 6.02 ×[tex]10^{23}[/tex] molecules
no of moles is 1
so
total number of electron in water is = no of electron ×molecules per gram mole × no of moles
total number of electron in water is = 10 × 6.02 ×[tex]10^{23}[/tex] × 1
total number of electron in water is = 6.02×[tex]10^{24}[/tex] electron
and
we know
mass = density × volume ..........1
here we know density of water is 1000 kg/m
and volume = 1 litter = 1 × [tex]10^{-3}[/tex] m³
mass of 1 litter = 1000 × 1 × [tex]10^{-3}[/tex]
mass = 1000 g
so
total number of electron in 1 litter = mass of 1 litter × [tex]\frac{molecules per gram mole}{mass per mole}[/tex]
total number of electron in 1 litter = 1000 × [tex]\frac{6.02*10{24}}{18}[/tex]
total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron
The fastest server in women's tennis is Sabine Lisicki, who recorded a serve of 131 mi/h (211 km/h) in 2014. Suppose that the acceleration of the ball was constant during the contact with the racket. Part A If her racket pushed on the ball for a distance of 0.15 m, what was the acceleration of the ball during her serve?
Using the formula for acceleration, a = (v^2) / 2d, and the given velocity and acceleration distance, the acceleration of the tennis ball during Lisicki's serve was approximately 11368 m/s². The force exerted by the racket is generally higher than the force due to gravity during this action.
Explanation:To calculate the acceleration of the tennis ball during Sabine Lisicki's serve, we can use the formula for acceleration: a = (v^2) / 2d, where 'v' corresponds to the final velocity and 'd' represents the distance over which the ball accelerated.
In this case, the final velocity 'v' is 211 km/h (converted to m/s gives us approximately 58.6 m/s), and Lisicki's racquet was in contact with the ball, causing it to accelerate over a distance 'd' of 0.15 m. Plugging these values into the formula gives us: a = (58.6 m/s)^2 / 2(0.15 m), which equals about 11368 m/s².
The average force exerted by the racket can be understood through the equation F = ma, a case of Newton's second law of motion. However, in this scenario, the force due to gravity is negligible as it's much smaller than the force exerted by the racket. The main focus here is the force exerted by the racket which made such a high acceleration possible.
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The car starts from rest and accelerates with an acceleration of 2 m/s^2 for 5 s. It then travels at a constant speed for 20 s, before decelerating at -5 m/s^2 until it reaches zero speed. How much distance did the car cover during this journey.
Answer:
Total distance covered during the journey is 235 m
Solution:
As per the question:
Initial velocity, v = 0 m/s
Acceleration, a = [tex]2 m/s^{2}[/tex]
Time, t = 5 s
Now,
For this, we use eqn 2 of motion:
[tex]d = vt + \farc{1}{2}at^{2}[/tex]
[tex]d = 0.t + \farc{1}{2}\times 2\times 5^{2} = 25 m[/tex]
The final speed of car after t = 5 s is given by:
v' = v + at
v' = 0 + 2(5) = 10 m/s
Now, the car travels at constant speed of 10 m/s for t' = 20 s with a = 0:
[tex]d' = vt + \farc{1}{2}at^{2}[/tex]
[tex]d' = 10\times 20 + \farc{1}{2}\times 0\times 20^{2} = 25 m[/tex]
d' = 200 m
Now, the car accelerates at a= - 5 [tex]m/s^{2}[/tex] until its final speed, v" = 0 m/s:
[tex]v"^{2} = v'^{2} + 2ad"[/tex]
[tex]0 = {10}^{2} + 2\times (- 5)d"[/tex]
[tex]100 = 10d"[/tex]
d" = 10 m
Total distance covered = d + d' + d" = 25 + 200 + 10 = 235 m
Transverse waves travel with a speed of 20 m/s on a string under a tension 0f 6.00 N. What tension is required for a wave speed of 30.0 m/s on the same string?
Answer:
[tex]T_2=13.5\ N[/tex]
Explanation:
Given that,
Speed of transverse wave, v₁ = 20 m/s
Tension in the string, T₁ = 6 N
Let T₂ is the tension required for a wave speed of 30 m/s on the same string. The speed of a transverse wave in a string is given by :
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]........(1)
T is the tension in the string
[tex]\mu[/tex] is mass per unit length
It is clear from equation (1) that :
[tex]v\propto\sqrt{T}[/tex]
[tex]\dfrac{v_1}{v_2}=\sqrt{\dfrac{T_1}{T_2}}[/tex]
[tex]T_2=T_1\times (\dfrac{v_2}{v_1})^2[/tex]
[tex]T_2=6\times (\dfrac{30}{20})^2[/tex]
[tex]T_2=13.5\ N[/tex]
So, the tension of 13.5 N is required for a wave speed of 30 m/s. Hence, this is the required solution.
Final answer:
The tension required for a wave speed of 30.0 m/s on a string with an initial tension of 6.00 N is 900 N.
Explanation:
To find the tension required for a wave speed of 30.0 m/s, we can use the equation: wave speed = square root of (tension/linear mass density).
Given that the initial wave speed is 20.0 m/s and tension is 6.00 N, we can rearrange the equation to solve for tension using the new wave speed.
Substituting the values, we have: 30.0 m/s = sqrt(tension/linear mass density). After squaring both sides of the equation, we get: 900 = tension/linear mass density
Since the linear mass density remains constant, the tension required for a wave speed of 30.0 m/s would be 900 N.
Multiple Concept Example 9 deals with the concepts that are important in this problem. A grasshopper makes four jumps. The displacement vectors are (1) 40.0 cm, due west; (2) 26.0 cm, 32.0 ° south of west; (3) 19.0 cm, 50.0 ° south of east; and (4) 18.0 cm, 60.0 ° north of east. Find (a) the magnitude and (b) direction of the resultant displacement. Express the direction as a positive angle with respect to due west.
Answer:
Explanation:
We shall convert the movement of grasshopper in vector form. Suppose the grass hopper is initially sitting at the origin or (00) position .
It went 40 cm due west so
D₁ = -40 i
It then moves 26 cm 32 ° south of west so
D₂ = - 26 Cos32i - 26 Sin32 j = - 22 i -13.77 j
Then it moves 19 cm 50° south of east
D₃ = 19 Cos 50 i - 19 Sin 50 j = 12.2 i - 14.55 j
Then it moves 18 cm 60° north of east
D₄ = 18 Cos 60 i + 18 Sin 60 j = 9 i + 15.58 j
Total displacement = D₁ +D₂+D₃+D₄
= - 40i -22 i - 13.77 j + 12.2 i - 14.55 j + 9 i + 15.58 j
= - 40.8 i - 12.74 j
Magnitude of displacement D
D² = ( 40.8 )² + ( 12.74)²
D = 42 .74 cm
If ∅ be the required angle
Tan∅ = 12.74 / 40.80 = .31
∅ = 17 ° positive angle with respect to due west.
If you can read the bottom row of your doctor's eye chart, your eye has a resolving power of one arcminute, equal to 1.67×10^−2 degrees . If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume that the wavelength of the light is 540 nm . Express your answer in millimeters to three significant figures.
Answer:
2.26 mm.
Explanation:
According to Rayleigh criterion , angular rosolution of eye is given by the expression
Angular resolution ( in radian ) = 1.22 λ / D
λ wave length of light, D is diameter of the eye
Given
angular resolution in degree = 1.67 x 10⁻²
= 1.67 x 10⁻² x π / 180 radian ( 180 degree = π radian )
= 29.13 x 10⁻⁵ radian
λ = 540 x 10⁻⁹ m
Put these values in the expression
29.13 x 10⁻⁵ = 1.22 x 540 x 10⁻⁹ / D
D = [tex]\frac{1.22\times540\times10^{-9}}{29.13\times10^{-5}}[/tex]
D = 2.26 mm.
The effective diameter of an eye's optical system that corresponds to a resolving power of one arcminute, calculated using Rayleigh's criterion and the given values of resolving power and wavelength of light, is approximately 2.27 mm.
Explanation:The effective diameter of an eye's optical system that corresponds to a resolving power of one arcminute can be determined using Rayleigh's criterion for diffraction limit and the given values of the resolving power and wavelength of light.
Rayleigh's criterion states that the minimum resolvable angle for a diffraction-limited system is θmin = 1.22*λ/D where λ is the wavelength of light and D is the diameter of the optical system. Given that the resolving power of the eye is 1.67×10^−2 degrees and the wavelength of light is 540 nm, we can rearrange Rayleigh's formula to solve for D.
Converting 1.67×10^−2 degrees to radians gives us 0.00029 rad. Plugging in the values into Rayleigh's formula and solving for D gives us D = 1.22*λ/θmin. Substituting λ=540*10^−9 m and θmin=0.00029 into the equation, we get D = 2.27 mm to three significant figures.
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A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^2, where k=0.25N*s^2/m^2. Determine the maximum speed of free fall for the sky diver and the speed reached after 100m of fall. Plot the speed of the sky diver as a function of time and as a function of distance fallen
Answer:
[tex]v_{max}=52.38\frac{m}{s}[/tex]
[tex]v_{100}=33.81[/tex]
Explanation:
the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:
[tex]\sum{F}=0=F_d-W[/tex]
[tex]F_d=W[/tex]
[tex]kv_{max}^2=m*g[/tex]
[tex]v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}[/tex]
To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:
[tex]\sum{F}=ma=W-F_d[/tex]
[tex]ma=W-F_d[/tex]
[tex]ma=mg-kv_{100}^2[/tex]
[tex]a=g-\frac{kv_{100}^2}{m}[/tex] (1)
consider the next equation of motion:
[tex]a = \frac{(v_{x}-v_0)^2}{2x}[/tex]
If assuming initial velocity=0:
[tex]a = \frac{v_{100}^2}{2x}[/tex] (2)
joining (1) and (2):
[tex]\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}[/tex]
[tex]\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g[/tex]
[tex]v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g[/tex]
[tex]v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}[/tex]
[tex]v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}[/tex] (3)
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}[/tex]
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}[/tex]
[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}[/tex]
[tex]v_{100}=\sqrt{1,143.3}[/tex]
[tex]v_{100}=33.81[/tex]
To plot velocity as a function of distance, just plot equation (3).
To plot velocity as a function of time, you have to consider the next equation of motion:
[tex]v = v_0 +at[/tex]
as stated before, the initial velocity is 0:
[tex]v =at[/tex] (4)
joining (1) and (4) and reducing you will get:
[tex]\frac{kt}{m}v^2+v-gt=0[/tex]
solving for v:
[tex]v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }[/tex]
Plots:
A student is walking with a constant speed of ????????1meters per second along High Street and sees a puddle ????????1 meters ahead of her. A bus driver is driving parallel to the student along High Street as well. At the moment the bus is ????????2meters behind the student, the bus driver decides he wants to splash he student with water by driving over the puddle as the student walks past. Determine the expression for the speed that the bus must have in terms of the given variables such that the bus and the student reach the puddle at the same time to splash the student.
Answer:
The expression is[tex]v= v_{student} ( 1 + \frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} )[/tex]
The speed must be 3 m/sExplanation:
We know that speed is:
[tex]v= \frac{distance}{time}[/tex].
So, to find the speed for the bus, we need to know:
a. How far the bus is from the puddle.b. In how much time will the student reach the puddle.Lets call [tex]v_{student}[/tex] the speed of the student, and [tex]d_{student \ to \ puddle}[/tex] the distance from the student to the puddle.
We can obtain the time taking
[tex]v_{student}= \frac {d_student \ to \ puddle}{t}[/tex]
as t must be the time that the student will take to reach the puddle:
[tex]t= \frac {d_{student \ to \ puddle}}{v_{student}}[/tex]
The bus is at a distance [tex]d_{bus \ to \ the \ student}[/tex] behind the student, so, the total distance that the bus must travel to the puddle is:
[tex]d_{bus \ to \ the \ puddle} = d_{bus \ to \ the \ student} + d_{student \ to \ puddle}[/tex]
Taking all this togethes, the formula must be:
[tex]v= \frac{d_{bus \ to \ the \ puddle}}{t}[/tex].
[tex]v= \frac{d_{bus \ to \ the \ student} + d_{student \ to \ puddle}}{\frac {d_{student \ to \ puddle}}{v_{student}}}[/tex].
[tex]v= v_{student} \frac{d_{bus \ to \ the \ student} + d_{student \ to \ puddle}}{d_{student \ to \ puddle}}[/tex].
[tex]v= v_{student} (\frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} + \frac{d_{student \ to \ puddle}}{d_{student \ to \ puddle}} )[/tex].
[tex]v= v_{student} ( 1 + \frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} )[/tex].
And this is the formula we are looking for.
Taking the values from the problem, we find
[tex]v= 1 \frac{m}{s} ( 1 + \frac{2 m}{1 m})[/tex]
[tex]v= 1 \frac{m}{s} ( 1 + 2)[/tex]
[tex]v= 3 \frac{m}{s}[/tex]
A 0.09 g honeybee acquires a charge of +23 pc while flying. The earth's electric field near the surface is typically (100 N/C, downward). What is the ratio of the electric force on the bee to the bee's weight? Multiply your answer by 10 before entering it below.
Answer:
Explanation:
Charge on honeybee q = 23 x 10⁻¹²
Force due to electric field E = E x q
= 100 x 23 x 10⁻¹²
= 23 x 10⁻¹⁰ N
Gravitational force on the honeybee
= m g = .09x 10⁻³ x 9.8
= .882x 10⁻³ N
Ratio of electric field and gravitational field
23 x 10⁻¹⁰ / .882x 10⁻³
26.07 x 10⁻⁷
= 26.07
You throw a ball upward with an initial speed of 4.3 m/s. When it returns to your hand 0.88 s later, it has the same speed in the downward direction (assuming air resistance can be ignored). What was the average acceleration vector of the ball? Express your answer using two significant figures.
Answer:
The acceleration is -9.8 m/s²
Explanation:
Hi there!!
When you throw a ball upward, there is a downward acceleration that makes the ball return to your hand. This acceleration is produced by gravity.
The average acceleration is calculated as the variation of the speed over time. In this case, we know the time and the initial and final speed. Then:
acceleration = final speed - initial speed/ elapsed time
acceleration = -4.3 m/s - 4.3 m/s / 0.88 s
acceleration = -9.8 m/s²
The average acceleration vector would be 0 m/s² due to the upward (negative) and downward (positive) accelerations cancelling. However, the average magnitude of acceleration regardless of direction is 9.8 m/s², which is simply the acceleration due to gravity.
Explanation:To answer your question about the average acceleration vector of a ball that was thrown upward, we can use the concept of free fall in physics. When you throw a ball upward, it initially slows down due to earth's gravitational pull until it stops at its highest point, then it starts accelerating downward due to gravity, until it reaches your hand again.
The acceleration when the ball is going upward will be the opposite of the acceleration when it's coming downward because they are in opposite directions but magnitude will be the same. We can assume the acceleration due to gravity as -9.8 m/s² (negative indicating upward direction) and when it's coming down it will be 9.8 m/s² (positive indicating downward direction).
So over the course of its flight (0.88s), the average acceleration would be ((-9.8)+(9.8))/2 = 0 m/s². However, this may not be what you're looking for as this is the averaged vector sum of the accelerations upward and downward.
If you are asking about the average magnitude of acceleration regardless of direction (in magnitude) it would be the average of the absolute values of the accelerations, which would be the gravitational acceleration g = 9.8 m/s².
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In your first lab, you will measure the diameter and height of a cylinder. The diameter will be measured with a micrometer and the height will be measured with a vernier caliper. Say you measure a diameter of 5.1±0.0005 cm and a height of 37.6±0.005 cm. What will be the uncertainty in your volume?
Answer:
ΔV = ±0.175 cm
Explanation:
The equation for volume is
V = π/4 * d^2 * h
All the measurements are multiplied. To propagate uncertainties in multiplication we add the relative uncertainties together.
The relative uncertainty of the diameter is:
εd = Δd/d
εd = 0.0005/5.1 = 0.000098
The relative uncertainty of the height is:
εh = Δh/h
εh = 0.005/37.6 = 0.00013
Then, the relative uncertainty of the volume is:
εV = 2 * εd + εh
εV = 2 * 0.000098 + 0.00013 = 0.000228
Then we get the absolute uncertainty of the volume, for that we need the volume:
V = π/4 * 5.1^2 * 37.6 = 768.1 cm^3
So:
ΔV = ±εV * V
ΔV = ±0.000228 * 768.1 = ±0.175 cm
A centrifuge in a medical labarotary rotates at an
angularspeed of 3600 rev/min. When switch off, it rotates
through50.0 revolutions before coming to rest. Find the
constantangular acceleration of the centrifuge.
Answer:
angular acceleration = - 217.5 rad/s²
Explanation:
given data
angular speed = 3600 rev/min
rotate = 50 revolution
to find out
angular acceleration
solution
we know here no of rotation n = 3600 rev/min i.e 60 rev/s
so initial angular velocity will be ω(i) = 2π× n
ω(i) = 2π× 60 = 376.9 rad/s
and
final angular velocity will be ω(f) = 0
so
angular displacement will be = 2π × 52 = 326.56 rad
and angular acceleration calculated as
angular acceleration = [tex]\frac{\omega(f)^2-\omega(i)^2}{2*angular displacement}[/tex]
put here value
angular acceleration = [tex]\frac{-376.9^2}{326.56}[/tex]
angular acceleration = - 217.5 rad/s²
Question #1: Consider Eratosthenes's experiment to measure the size of the Earth. Suppose the Earth were a smaller planet -- but the sun were still directly overhead in Syene at noon on the Summer Solstice, and it was still 500 miles from Syene to Alexandria. Would the shadow of the stick in Alexandria at noon on the Summer Solstice have been longer, shorter, or the same as it was on our Earth? Briefly explain your reasoning.
Answer:
It would have been longer.
Explanation:
Lets assume the Sun angle = θ
Distance between Syene and Alexandria = D
Circumference of Earth = C
As per Eratosthenes' calculations,
[tex]\frac{\theta}{360} =\frac{D}{C}[/tex]
From the above equation it is evident that if the circumference decreases value of θ will increase which implies that the shadow length would be longer as compared to that on the Earth.