Answer:
Precise and Reproducible
Explanation:
For a system or a process to be useful a measurement standard for any physical quantity, it must be:
Precise, Precision is important in any measurement.The values of the measurement obtained should be reproducibleThe measurement may be large or smallIt may or may not be in SI unitsIt should be available easily.Answer:
precise
reproducible
Explanation:
What is the weight of a 2.5-kg pumpkin? Remember to always show all work.
Final answer:
The weight of a 2.5-kg pumpkin is calculated by multiplying the mass by the acceleration due to gravity, resulting in 24.5 newtons (N) when rounded to one decimal place.
Explanation:
To find out the weight of a 2.5-kg pumpkin, you would need to multiply its mass by the acceleration due to gravity. In most physics problems, this is approximately 9.8 meters per second squared (m/s²). Therefore, the weight W can be calculated using the formula W = m×g, where m is the mass in kilograms and g is the acceleration due to gravity.
So, for a 2.5-kg pumpkin, W = 2.5 kg × 9.8 m/s² = 24.5 kg·m/s². Since weight is a force, it is measured in newtons (N) in the International System of Units (SI). Therefore, the pumpkin weighs 24.5 N.
Final answer:
The weight of a 2.5-kg pumpkin is calculated by multiplying its mass by the gravitational acceleration, which results in a weight of 24.5 Newtons (N) on Earth.
Explanation:
The weight of a pumpkin is the force with which it is pulled towards the Earth due to gravity. To calculate the weight, you can use the formula:
Weight (W) = mass (m) × gravitational acceleration (g)
Where the gravitational acceleration on Earth is typically 9.8 meters per second squared (9.8 m/s²). For a 2.5-kg pumpkin, the calculation would be:
Weight (W) = 2.5 kg × 9.8 m/s²
Weight (W) = 24.5 kg·m/s²
Since weight is measured in Newtons (N), and 1 kg·m/s² is equivalent to 1 Newton, the weight of the pumpkin is 24.5 Newtons (N). When converting measurements and rounding, it's important to consider significant figures and the least precise measurement. In this calculation, both 2.5 kg and 9.8 m/s² are exact enough, so our final answer in Newtons does not need additional rounding.
How far (in feet) could a pitcher throw a baseball on flat, level ground if he can throw it at 100mph? (Neglect wind drag and the pitcher’s height.)
Answer:
R = 668.19 ft
Explanation:
given,
speed of the ball thrown by the pitcher = 100 mph
to travel maximum distance θ = 45°
distance traveled by the ball = ?
using formula
1 mph = 0.44704 m/s
100 mph = 44.704 m/s
[tex]R = \dfrac{u^2sin2\theta}{g}[/tex]
[tex]R = \dfrac{44.704^2sin2\times 45}{9.81}[/tex]
R = 203.71 m
1 m = 3.28 ft
R = 203.71 × 3.28
R = 668.19 ft
hence, ball will go at a distance of 668.19 ft when pitcher throw it at 100 mph.
A cat jumps off a piano that is 1.3m high. The initial velocity of the cat is 3m/s at an angle of 37degrees above the horizontal. How far from the edge of the piano does the cat strike the floor.
Final answer:
To find the distance the cat strikes from the edge of the piano, we can break the motion into horizontal and vertical components, and solve for the time of flight and horizontal distance traveled. By substituting the given values into relevant equations and solving, we can determine that the cat will strike the floor approximately 1.71 meters from the edge of the piano.
Explanation:
To solve this problem, we can break the motion into horizontal and vertical components. The horizontal component of the initial velocity will not change, so the cat will travel a horizontal distance of 3m/s *cos(37°) * t, where t is the time of flight. To find t, we can use the equation h = v0y * t + (1/2) * g * [tex]t^2[/tex], where h is the vertical displacement, v0y is the vertical component of the initial velocity, and g is the acceleration due to gravity.
Substituting the given values into the equation, we get 1.3m = 3m/s * sin(37°) * t - (1/2) * 9.8m/[tex]s^2[/tex] * [tex]t^2[/tex]. Solving this quadratic equation, we find two possible values for t: approximately 0.364s and 0.203s. Since the cat jumps off the piano, we only consider the positive value of t. So the cat will strike the floor approximately 3m/s * cos(37°) * 0.364s = 1.71 meters from the edge of the piano.
An ideal, monotomic gas initially at a temperature of 450K, a pressure of 4.00 atm and a volume of 10.0L, undergoes an adiabatic compression to 1/3 its original volume. Find the final temperature of the gas. A. 72 K B. 150 K C. 216 K D. 936 K E. 1350 K
Answer: D) 936 K
Explanation:
Given:
Initial temperature of the gas, [tex]T = 450\ K[/tex]
Initial Pressure of the gas, [tex]P=4\ atm[/tex]
initial volume of the gas, [tex]V=10\ L[/tex]
It it given that the process is adiabatic, so for a adiabatic process we have
Let [tex]T_f \ \ and\ \ V_f [/tex] be the final temperature and volume of the gas.
[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}[/tex]
For monotomic gas [tex]\gamma=1.67[/tex]
[tex]450\times V_i^{1.67 -1} =T_f\left (\dfrac{V_i}{3} \right )^{1.67-1}\\T_f=936 K[/tex]
Hence the final temperature of the gas is 936 K. So option D is correct
If your car accelerates from rest at a steady rate of 4.2 m/s^2, how soon will it reach 72 km/h ( 44.7 mph or 20 m/s)?
Answer:
The car will reach 20 m/s in 4.76 seconds
Explanation:
The car moves following an uniformly accelerated motion, therefore we know that [tex]v=v_0+ a(t-t_0)[/tex]
Where [tex]v_0 = 0[/tex] are [tex]t_0 =0[/tex]
Then [tex]t= \frac{v}{a}=\frac{20 m/s}{4.2 m/s^{2}}=4.76s[/tex]
In physics, why would an equation like y=mx+b be plotted as a straight line rather than as a parabola, as is done in math?
Answer:
Explanation:
In Both Physics and Math
y=mx+b is plotted as straight line where
m=slope of line
b=intercept on Y-axis
whereas Equation of parabola is something like this
[tex]y^2=4ax[/tex]
or
[tex]x^2=4ay[/tex]
Math is a tool to solve Physics problems so equations are same in math and physics
The height of a helicopter above the ground is given by h = 3.30t3, where h is in meters and t is in seconds. At t = 2.30 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? 9.857 Incorrect: Your answer is incorrect.
Answer:
The total time to reach ground is 24.89 seconds
Explanation:
Since the height of the helicopter is given by
[tex]h(t)=3.30t^{3}[/tex] thus at time t = 2.30 seconds the height of the helicopter is
[tex]h(2.30)=3.30\times (2.30)^{3}=40.151m[/tex]
The velocity of helicopter upwards at time t = 2.30 is given by
[tex]v=\frac{dh(t)}{dt}\\\\v=\frac{d}{dt}(3.30t^{3})\\\\v(t)=9.90t^{2}\\\\\therefore v(2.30)=9.90\times (2.30)^2=120.45m/s[/tex]
Now the time after which it becomes zero can be obtained using the equations of kinematics as
1) Time taken by the mailbag to reach highest point equals
[tex]v=u+gt\\\\0=120.45-9.81\times t\\\\\therefore t_{1}=\frac{120.45}{9.81}=12.28s[/tex]
2) Time taken by the mailbag to reach ground from a height of 40.151 meters equals
[tex]s=ut+\frac{1}{2}gt^{2}\\\\40.151=120.45t+4.9t^{2}[/tex]
Solving for t we get[tex]t_{2}=0.3289secs[/tex]
Now the total time of the journey is
[tex]\\\\2\times t_{1}+t_{2}\\\\=2\times 12.28+0.3289=24.89secs[/tex]
Use dimensional analysis to determine how the linear acceleration a in m/s2 of a particle traveling in a circle depends on some, or all, of the following properties: the radius r of the circle, the angular frequency ω in s−1 with which the particle orbits about the circle, and the mass m of the particle. There is no dimensionless constant involved in the relation.
I have to say, i love this kind of problems.
So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:
[tex][a]=[\frac{distance}{time^2}][/tex].
Now, we got the radius
[tex][r] = [distance][/tex]
the angular frequency
[tex][\omega] = \frac{1}{s}[/tex]
and the mass
[tex][m]=[mass][/tex].
Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.
The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write
[tex]a \propto r[/tex].
The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared
[tex]a \propto r \omega^2[/tex].
We are almost there. If this were any other problem, we would write:
[tex]a = A r \omega^2[/tex]
where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:
[tex]a = r \omega^2[/tex]
Final answer:
The linear acceleration a of a particle in circular motion depends on the radius r of the circle and the square of the angular frequency ω, which is expressed by the formula a = rω².
Explanation:
To determine how the linear acceleration a in m/s² of a particle traveling in a circle depends on the properties such as the radius r of the circle, the angular frequency ω in s⁻¹, and the mass m of the particle, we can use dimensional analysis and known relationships from physics.
Firstly, we recognize that mass m does not directly affect a, because a depends on the force per unit mass. By considering the relationship between linear and angular variables, we know that linear velocity v is related to ω by v = rω, and the centripetal acceleration of an object moving in a circle is ac = v²/r. Substituting v = rω into the expression for centripetal acceleration, we get ac = (rω)²/r = rω².
Therefore, the linear acceleration a of a particle undergoing circular motion is directly proportional to the radius r and the square of the angular frequency ω. This relationship can be expressed as a = rω², where a is the linear acceleration, r is the radius of the circle, and ω is the angular frequency.
An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accelerated by the presence of the electric field. 50% Part (a) What is the change in the electron’s kinetic energy, in kiloelectron volts, if it travels over a distance of 0.25 m in this field? ΔK = - 4.63 * 105|
Answer:
[tex]W = 462.5 keV[/tex]
Explanation:
As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron
So here we have to find the work done by electric field on moving electron
So we have
[tex]F = qE[/tex]
[tex]F = (1.6 \times 10^{-19})(1.85 \times 10^6)[/tex]
[tex]F = 2.96 \times 10^{-13} N[/tex]
now the distance moved by the electron is given as
[tex]d = 0.25 m[/tex]
so we have
[tex]W = F.d[/tex]
[tex]W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)[/tex]
[tex]W = 7.4 \times 10^{-14} J[/tex]
now we have to convert it into keV units
so we have
[tex]1 keV = 1.6 \times 10^{-16} J[/tex]
[tex]W = 462.5 keV[/tex]
Two ice skaters, Frank and Mary, are initially facing each other and separated by 2000 m on a smooth frozen lake. Mary begins skating towards Frank at t = 0 s with an acceleration of 0.75 m/s^2. After she reaches a speed of 12 m/s, she continues skating at a constant 12 m/s. Three seconds after Mary begins to skate, Frank starts skating. He accelerates at 1.1 m/s^2 for 9 s and then skates at a constant speed. a) When do the skaters pass each other? b) How far from Mary’s starting position is the point at which the skaters pass each other?
Answer:
a) 99.1 s
b) 1093.2 m
Explanation:
The equation for speed with constant acceleration is
V(t) = V0 + a * t
Mary starts accelerating with a speed of zero, so
V0 = 0
V(t) = a * t
To reach a speed of 12 m/s
t = V(t) / a
t = 12 / 0.75 = 16 s
The equation for position under constant acceleration is
X(t) = X0 + V0 * t + 1/2 * a * t^2
Mary's starting position is zero
X0 = 0
X(t) = 1/2 * a * t^2
X(16) = 1/2 * 0.75 * 16^2 = 96 m
Frank starts skating 3 seconds after Mary, and he accelerates at 1.1 m/s^2 for 9 s. He will stop accelerating at second 12 (9 + 3).
His position after accelerating will be:
X(12) = X0 + V0 * (t - 3) + 1/2 * a * (t - 3)^2
His initial position is 2000, and his initial speed is zero
x(12) = 2000 - 1/2 1.1 * (12 - 3)^2 = 1955.5 m
Shi speed will be
V(12) = -1.1 * (12 - 3) = -9.9 m/s
From there they will move at constant speed from these positions. We can consider them as moving at constant speed starting at t0 = 16 and t0 = 12 respectively.
For Mary:
X(t) = X0 + V0 * (t - t0)
X(t) = 96 + 12 * (t - 16)
For Frank:
X(t) = 1955.5 - 9.9 * (t - 12)
Equating these two we can find the time when they meet:
96 + 12 * (t - 16) = 1955.5 - 9.9 * (t - 12)
96 + 12*t - 192 = 1955.5 - 9.9*t + 118.8
21.9*t = 2170.3
t = 2170.3 / 21.9 = 99.1 s
Replacing this time value on either equation we get the position:
X(99.1) = 96 + 12 * (99.1 - 16) = 1093.2 m
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.1 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.4 m/s^2 until he catches his friend. a) How much time does it take until he catches his friend (after his friend passes him)?
Express your answer using two significant figures.
Answer:
5.91 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Distance = Speed × Time
⇒Distance = 3.1t
Distance traveled by bicycle that passes through = 3.1t
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 2.4\times (t-2)^2\\\Rightarrow s=1.2(t-2)^2[/tex]
They both travel the same distance
[tex]3.1t=1.2(t-2)^2\\\Rightarrow 31t=12\left(t-2\right)^2\\\Rightarrow 12t^2-79t+48=0[/tex]
[tex]t=\frac{79+\sqrt{3937}}{24},\:t=\frac{79-\sqrt{3937}}{24}\\\Rightarrow t=5.91, 0.67[/tex]
Hence, time taken by the bicyclist to catch the other bicyclist is 5.91 seconds
A projectile is fired at t = 0 with an initial velocity ofvi at an angle of θ with respectto the horizontal towards the vertical side of a building that is adistance d away as shown in the figure.
(a) Determine an equation for the time at which the projectilestrikes the building, in terms of the quantities d, v_i, theta, andg.
(b) Determine an equation for the height h above theground at which the projectile strikes the building, in terms ofthe quantities d, v_i, theta, and g.
Answer:
[tex]t=d/(v_{o}*cos(\alpha ))[/tex]
[tex]y =d*tan(\alpha )- 1/2*g^{2}d^{2}/(v_{o}^{2}*(cos(\alpha ))^2)[/tex]
Explanation:
Kinematics equation in the axis X:
[tex]x=v_{o}*cos(\alpha )*t[/tex]
The projectile strikes the building at time t:
[tex]d=v_{o}*cos(\alpha )*t[/tex]
[tex]t=d/(v_{o}*cos(\alpha ))[/tex] (1)
Kinematics equation in the axis Y:
[tex]y =v_{o}*sin(\alpha )*t - 1/2*gt^{2}[/tex] (2)
We replace (1) in (2):
[tex]y =v_{o}*sin(\alpha )*d/(v_{o}*cos(\alpha )) - 1/2*g(d/(v_{o}*cos(\alpha )))^{2}[/tex]
[tex]y =d*tan(\alpha )- 1/2*g^{2}d^{2}/(v_{o}^{2}*(cos(\alpha ))^2)[/tex] (2)
If my weight on Earth is 140lbs, what is my mass?
Answer:
63.57 kg
Explanation:
weight = 140 lbs
Let the mass is m.
1 lbs = 4.45 N
The weight of an object is defined as the force with which our earth attracts the body towards its centre.
Weight is the product of mass of the body and the acceleration due to gravity of that planet.
W = m x g
On earth surface g = 9.8 m/s^2
Now convert lbs in newton
So, 140 lbs = 140 x 4.45 = 623 N
So, m x 9.8 = 623
m = 63.57 kg
Thus, the mass is 63.57 kg.
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.4 m/s2 while Kathy maintains an acceleration of 4.49 m/s2. (a) Find the time at which Kathy overtakes Stan. 6.7 Correct: Your answer is correct. seenKey 6.7 s from the time Kathy started driving
Answer:
time at which Kathy overtakes Stan is 6.70 sec
Explanation:
given data
time = 1 sec
acceleration= 3.4 m/s²
acceleration = 4.49 m/s²
to find out
the time at which Kathy overtakes Stan
solution
we consider here travel time for kathy = t1
and travel time for stan is = t2
and we know initial velocity = 0
so
t1 = 1 + t2
and distance travel equation by kinematic is
d1 = ut + [tex]\frac{1}{2}[/tex] at²
d1 = 0+ [tex]\frac{1}{2}[/tex] 4.49 (t2)² ................1
and
d2 = 0 + [tex]\frac{1}{2}[/tex] 3.4 (1+t2)² ..................2
and when overtake distance same so from equation 1 and 2
[tex]\frac{1}{2}[/tex] 4.49 (t2)² = [tex]\frac{1}{2}[/tex] 3.4 (1+t2)²
t2 = 6.703828 sec
so time at which Kathy overtakes Stan is 6.70 sec
A 1200 kg car surrounds an 80 meter radius curve. If the coefficient of friction between the tires and the road is 0.65, what maximum speed should the car have to not leave the curve to : a. 23 m / s
b. 10 mph
c. 62 km / h
d. 6.3 m / s
Answer: a) 23 m/s
Explanation: In order to solve this problem we have to consider the circular movement where the friction force change de direcion of the velocty to keep a circular trajectory.
By using Second Newton law, we have:
F=m*acentripeta
μ*N= m*v^2/r where N ( equal to mg) is the normal force and μ is the coefficient of friction. r is the redius of the trajectory.
so
the maximun speed permited to keep a circular trajectory is calculated as:
v= (μ*r*g)^1/2=(0.65*80m*9.8 m/s^2)^1/2=22.57 m/s
Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they form a diamond shape centered on the origin. A charge of +5 μC is placed at the origin. If the force between a +2 μC and a +5 μC charge separated by 10 cm has a magnitude of 9 N, which of the following can we say about the force on the +5 μC charge at the origin in this case?
Answer:
The force will be zero
Explanation:
Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.
Two 2.0-cm-diameter disks face each other, 1.0 mm apart. They are charged to ± 8.0 nC . What is the electric field strength between the disks. A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
Answer:
E = 2.88*10^6 N/C
v = 0.742 *10^6m/s
Explanation:
The two disks form a parallel-plate capacitor, which will cause an electric field equal to:
[tex]E = \frac{Q}{e_0 A}[/tex]
Where Q is the charge of the disks, A is the area of the disks and e_0 is the vacuum permisivity equal to 8.85*10^-12 C^2/Nm^2:
[tex]E = \frac{8*10^{-9} C}{(\frac{1}{4}\pi(0.02m)^2)*8.85*10^{-12}C^2/Nm^2}= 2.88*10^6 N/C[/tex]
Now, for the second part of the problem, we can use conservation of energy. The addition of potential and kinetic energy at launch point should be equal to the addition at the positive disk. Because the proton has positive charge, the potential energy of the proton will increase as its distance to the negative disk increases too. This is because the proton will be attracted towards the negative disk. The potential energy is given by:
[tex]E_p = V*q[/tex]
Where V is the difference in potential (voltage) between the disks. In a parallel-plate capacitor:
[tex]V = E*d[/tex], where d is the difference in position with the frame of reference. Our frame of reference will be the negative disk.
q is the charge of the proton.
The kinetic energy is given by:
[tex]E_k = \frac{1}{2}mv^2[/tex]
Then:
[tex]E_p_1+E_k_1 = E_p_2+E_k_2\\V_1*q+\frac{1}{2}mv_1^2= V_2*q+\frac{1}{2}mv_2^2 |v_2=0, E_p_1 = 0\\\frac{1}{2} mv_1^2 = E*d*q \\v = \sqrt{\frac{2Edq}{m}} = \sqrt{\frac{2*2.88*10^6N/m*0.001m*1.6*10^{-19}C}{1.67*10^{-27}}} = 0.742 *10^6m/s[/tex]
(a) The electric field strength between the disks is 2.88 N/C
(b) The launch speed of the proton to reach the positive disk is 7.43 x 10⁵ m/s.
The given parameters;
diameter of the disk, d = 2 cm distance between the disks, r = 1 mmcharge on the disks, q = 8 nCradius of the disk, r = 1 cmThe electric field strength between the disks is calculated as follows;
[tex]E = \frac{Q}{\varepsilon _o A} \\\\E = \frac{Q}{\varepsilon _o \pi r^2} \\\\E = \frac{8 \times 10^{-9} }{8.85\times 10^{-12} \times \pi \times (0.01)^2} \\\\E = 2.88 \times 10^ 6 \ N/C[/tex]
The launch speed of the proton to reach the positive disk is calculated as follows;
[tex]K.E = W\\\\\frac{1}{2} mv^2 = Fd\\\\\frac{1}{2} mv^2 = Eqd\\\\mv^2 = 2Eqd\\\\v = \sqrt{\frac{2Eqd}{m} } \\\\v = \sqrt{\frac{2\times 2.88 \times 10^6 \times 1.6\times 10^{-19} \times 0.001 }{1.67 \times 10^{-27}} }\\\\v = 7.43 \times 10^5 \ m/s[/tex]
Learn more here:https://brainly.com/question/13263538
A ceiling fan has 5-inch blades (so the radius of the circular fan is inches). Suppose the fan turns at a rate of 2 revolutions per minute. (a) Find the angular speed of the fan in radians per minute. (b) Find the linear speed of the tip of a blade in inches per minute.
Answer:
a) 12.56 rad per minute
b)62.8 inches per minute
Explanation:
Given:
Radius of the blades, R=5 inchesFrequency of the blades, f=2 revolutions per minutea)
Let [tex]\omega[/tex] be the angular velocity of the blade which is given by
[tex]\omega =2\pi f\\\omega=2\times 3.14 \times 2\\\omega=12.56\ \rm rad/min[/tex]
b)
Let v be the linear speed of the tip of the blade which is given by
[tex]v=\omega \times R\\v=12.56\times 5\\v=62.8\ \rm inches/min[/tex]
Hence we have calculated the angular velocity and linear speed.
An ice cube of volume 15 cm^3 and density 917 kg/m^3 is placed in ethyl alcohol of density 811 kg/m^3. What is the buoyant force on the ice? Will it float?
Answer:
the buoyant force acting on the ice cube is 0.119 N
Explanation:
given,
volume of ice cube = 15 cm³ = 15 × 10⁻⁶ m³
density of ice cube = 917 kg/m³
density of ethyl alcohol = 811 kg/m³
buoyant force = ?
The density of ice is more than ethyl alcohol hence it will sink.
buoyant force acting on the ice cube = ρ V g
= 811 × 15 × 10⁻⁶ × 9.81
=0.119 N
so, the buoyant force acting on the ice cube is 0.119 N
A catfish is 2.4 m below the surface of a smooth lake. (a) What is the diameter of the circle on the surface through which the fish can see the world outside the water? (b) If the fish descends, does the diameter of the circle increase, decrease, or remain the same? Assume that the index of refraction of water is 1.333.
Answer:
(a) 5.45 m
(b) no change
Explanation:
refractive index of water, n = 1.333
h = 2.4 m
Let the radius of circle is r.
Here we use the concept of total internal reflection.
Let i be the angle of incidence.
By using Snell's law
Refractive index of water with respect to air is the ratio of Sin of angle of incidence to the Sin of angle of refraction.
here refraction takes place from denser medium to rarer medium
So, [tex]\frac{1}{n}=\frac{Sin i}{Sin r}[/tex]
Here, angle r is 90 for total internal reflection
[tex]\frac{1}{1.33}=\frac{Sin i}{Sin 90}[/tex]
Sin i = 0.75
i = 48.6°
According to the triangle ΔOAB
[tex]tan i = \frac{AB}{OA}[/tex]
[tex]tan 48.6 = \frac{r}{2.4}[/tex]
r = 2.73 m
Diameter = 2 r = 2 x 2.73 = 5.45 m
Thus, the diameter of the circle is 5.45 m.
(b) As the value of r depends on the angle of incidence i and angle of incidence depends on the value of refractive index. As refractive index is constant, so the value of diameter remains same.
Due to refraction, a fish sees the world outside through a circle on the water's surface. The diameter of this circle, around 3.6 meters for a fish 2.4 m deep, decreases as the fish descends due to the narrowing range of incident light angles.
Explanation:The question revolves around the concept of refraction, a phenomenon that occurs when light changes the medium, therefore, also changing direction. This comes into play when light travels from water to air (or vice versa), like how a fish observes the world outside the water body.
In response to the student's question part (a): When the fish is 2.4 m below the water's surface, it sees through a circle on the surface due to refraction. Knowing that the index of refraction of water is 1.333, the apparent depth of the fish to an observer on the surface would be the real depth divided by the refractive index, or around 1.8 meters. Therefore, the light from objects outside the lake would reach the fish within a circle having a diameter of about 3.6 meters (twice the apparent depth).
For part (b) of the question: If the fish were to dive deeper, the diameter of the circle through which it could observe the world outside would diminish. This is because as the actual depth increases, the range of incident angles for light rays entering the water from air, for which total internal reflection doesn’t occur, gets narrower. Therefore, the fish's field of view on the outside world decreases.
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A parallel-plate capacitor consists of two plates, each with an area of 21 cm2 separated by 3.0 mm. The charge on the capacitor is 9.8 nC . A proton is released from rest next to the positive plate. How long does it take for the proton to reach the negative plate?
Answer:
it take for the proton to reach the negative plate is 0.1089 ns
Explanation:
given data
area = 21 cm² = 21 × [tex]10^{-4}[/tex] m²
separated d = 3 mm = 3 × [tex]10^{-3}[/tex] m
charge Q = 9.8 nC = 9.8 × [tex]10^{-9}[/tex] C
to find out
How long does it take for the proton to reach the negative plate
solution
we know that electric field between the plate is express as
electric field E = [tex]\frac{Q}{A*\epsilon}[/tex] ..................1
here Q is charge and A is area and ∈ is permittivity of free space
put here value
electric field E = [tex]\frac{9.8 ×10^{-9}}{21×10^{-4}*8.85×10^{-12}}[/tex]
electric filed E = 527306.968 N/C
and
we know force on proton by electric field is
Force = electric filed × charge on proton
and force according to newton law
force = mass × acceleration
so
mass × acceleration = electric filed × charge on proton
here put all value and find acceleration
1.67 × [tex]10^{-27}[/tex] × acceleration = 527306.968 × 1.6 × [tex]10^{-19}[/tex]
acceleration = 5.05 × [tex]10^{13}[/tex] m/s²
and
final velocity of proton by equation of motion
v² - u² = 2as ..........2
here u is zero and v is final velocity and s is distance and a is acceleration
v² - 0 = 2(5.05 × [tex]10^{13}[/tex]) (3 × [tex]10^{-3}[/tex] )
velocity = 5.50 × [tex]10^{5}[/tex] m/s
so time is
time = [tex]\frac{velocity}{acceleration}[/tex]
time = [tex]\frac{5.50*10^{5}}{5.05*10^{13}}[/tex]
time = 0.1089 × [tex]10^{-9}[/tex] s
so it take for the proton to reach the negative plate is 0.1089 ns
In a parallel plate, the capacitor plate is parallel while the plates are separated by distance. The time required for the proton to reach the negative plate will be 0.1089×10⁻⁹ second.
What is a parallel plate capacitor?
It is a type of capacitor is in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.
A dielectric medium is a must in between these plates helps to stop the flow of electric current through it due to its non-conductive nature.
The following date given in the problem,
A id the area of plates = 21 cm² = 21 × 10m²
d is the distance separated= 3 mm = 3 ×10⁻³m
q is the charge on capicitor = 9.8 nC = 9.8 ×10⁻⁹C
The formula of the electric field is as
[tex]\rm E = \frac{q}{A\varepsilon } \\\\ \rm E = \frac{9.8\times10^{-9}}{ 21\times10^{-4}\times8.85\times10^{-12}} \\\\\rm E = 527306\; N/C}[/tex]
The electric force is equal to the product of the charge and the electric field. Here the electric force is balanced by mechanical force.
[tex]\rm F= qE\\\\\rm{ F= ma}\\\\qE =ma\\\\ \rm{ a= \frac{qE}{m} }\\\\\rm{ a= \frac{1.6\times10^{-9}\times527306}{1.2\times10^{-10}} }\\\\ \rm a = 5.05\times10^{-31}\;m/sec^2[/tex]
From newton's third equation of motion
As the initial velocity is zero u=0
[tex]\rm{v^2= u^2-2as} \\\\ \rm{v^2= 2\times5.05\times10^-3\times10^{-3}} \\\\ \rm{v= \sqrt{2\times5.05\times10^-3\times10^{-3}}\\\\[/tex]
[tex]\rm v = 5.50\times10^5 \;m/sec.[/tex]
Time is defined as the ratio of velocity and acceleration.
[tex]\rm t= \frac{v}{a} \\\\ \rm t= \frac{5.50\times10^5}{5.05\times10^{-3}} \\\\ \rm t= 0.1089\times 10^{-9}}\;sec[/tex]
Hence the time required for the proton to reach the negative plate will be 0.1089×10⁻⁹ second.
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The acceleration due to gravity on the Moon's surface is
one-sixth that on Earth. An astronaut's life support backpack
weighs300lb on Earth. What does it weigh on the moon?
The weight of the astronaut's life support backpack on the moon would be one-sixth of its weight on Earth, which equates to 50lb.
Explanation:The acceleration due to gravity on the moon is one-sixth that on Earth. Therefore, the weight of any object on the moon will be one-sixth of its weight on Earth. Since weight is a product of mass and gravity, if gravity is reduced, weight is also reduced accordingly.
To calculate the weight of an astronaut's life support backpack on the moon, you need to divide its Earth weight (300lb) by 6 (the relative gravity of the moon compared to Earth). So, on the moon, the backpack would weigh 300/6 = 50lb.
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Express Planck's radiation law in terms of wavelength (a) as opposed to frequency. Hint: Start by performing a change of variable from v to 2 in equation 1.7 using c = va.
Answer:
[tex]I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}[/tex]
Explanation:
The intensity of the radiation emitted by a black body with a certain temperature T and frequency [tex]\nu[/tex], is given by Planck's law:
[tex]I(\nu,T)=\frac{2h\nu^3}{c^2}(\frac{1}{e^{\frac{h\nu}{kT}}-1})[/tex]
Considering the frequency range between [tex]\nu[/tex] and [tex]\nu + \delta \nu[/tex] and [tex]dI[/tex] the amount of energy emitted in this frequency range. Since an increase in frequency means a decrease in wavelength. Then:
[tex]I(a,T)da=-I(\nu,T)d\nu\\I(a,T)=-\frac{d\nu}{da}I(\nu,T)[/tex]
Now recall that [tex]\nu=\frac{c}{a}[/tex], differentiate both sides:
[tex]d\nu=-\frac{c}{a^2}da\\\frac{d\nu}{da}=-\frac{c}{a^2}[/tex]
Replacing this in previous equation:
[tex]I(a,T)=\frac{c}{a^2}I(\nu,T)\\I(a,T)=\frac{c}{a^2}(\frac{2h\nu^3}{c^2}\frac{1}{e^{\frac{h\nu}{kT}}-1})[/tex]
Rewriting [tex]\nu^3[/tex] as [tex]\frac{c^3}{a^3}[/tex] and [tex]\nu[/tex] as [tex]\frac{c}{a}[/tex]
[tex]I(a,T)=\frac{c}{a^2}(\frac{2hc^3}{a^3c^2}\frac{1}{e^{\frac{hc}{akT}}-1})\\I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}[/tex]
Finally, we obtain Planck's radiation law in terms of wavelength
Suppose a person pushes thumbtack that is 1/5 centimeter long into a bulletin board, and the force (in dynes) exerted when the depth of the thumbtack in the bulletin board is x centimeters is given by F(x) = 1000 (1 + 2x)2 for 0 ≤ x ≤ 1 5 . Find the work W done by pushing the thumbtack all the way into the board.
Answer:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate
[tex]W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx[/tex]
W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm
In this exercise, we have to use work knowledge to calculate the total, in this way:
[tex]W = 290.7[/tex]
Given some information about the exercise we find that:
d = 1/5 cm = 0.2 cm[tex]F(x) = 1000 * (1 + 2*x)^2[/tex]Using the working definition we find that:
[tex]W = F * d\\W = [1000*x + 2000*x^2 + 1333*X^3]\\W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3\\W = 200 + 80 + 10.7 = 290.7[/tex]
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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height in feet after t seconds is given by y = 27 t − 20 t 2 y=27t-20t2. A. Find the average velocity for the time period beginning when t=3 and lasting .01 s: Preview .005 s: Preview .002 s: Preview .001 s: Preview NOTE: For the above answers, you may have to enter 6 or 7 significant digits if you are using a calculator. Estimate the instanteneous velocity when t=3. Preview Get help: Video
Answer:
[tex]V_{3.01}=-93.2m/s[/tex]
[tex]V_{3.005}=-93.1m/s[/tex]
[tex]V_{3.002}=-93.04m/s[/tex]
[tex]V_{3.001}=-93.02m/s[/tex]
[tex]V_{3}=-93m/s[/tex]
Explanation:
To calculate average velocity we need the position for both instants t0 and t1.
Now we will proceed to calculate all the positions we need:
[tex]Y_{3}=-99m/s[/tex]
[tex]Y_{3.01}=-99.932m/s[/tex]
[tex]Y_{3.005}=-99.4655m/s[/tex]
[tex]Y_{3.002}=-99.18608m/s[/tex]
[tex]Y_{3.001}=-99.09302m/s[/tex]
Replacing these values into the formula for average velocity:
[tex]V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s[/tex]
[tex]V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s[/tex]
[tex]V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s[/tex]
[tex]V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s[/tex]
To know the actual velocity, we derive the position and we get:
[tex]V=27-40t = -93m/s[/tex]
The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle between the direction of and the positive direction of x?
Answer:51.44 units
Explanation:
Given
x component of vector is [tex]-27.3\hat{i}[/tex]
y component of vector is [tex]43.6\hat{j}[/tex]
so position vector is
[tex]r=-27.3\hat{i}+43.6\hat{j}[/tex]
Magnitude of vector is
[tex]|r|=\sqrt{27.3^2+43.6^2}[/tex]
[tex]|r|=\sqrt{2646.25}[/tex]
|r|=51.44 units
Direction
[tex]tan\theta =\frac{43.6}{-27.3}=-1.597[/tex]
vector is in 2nd quadrant thus
[tex]180-\theta =57.94[/tex]
[tex]\theta =122.06^{\circ}[/tex]
Two narrow slits are separated by a distance of 0.10 mm. When illuminated with 550 nm light, a diffraction pattern is observed on a screen 2.2 m from the slits. What is the lateral separation of the first and second order maxima seen on the screen (in cm)?
Answer:
1.21 cm.
Explanation:
Given that the wavelength of light is 550 nm.
And the distance between the slits is 0.10 mm.
And the distance to the screen is 2.2 m.
As we know that the lateral separation can be defined by the formula.
[tex]LS=\dfrac{m\lambda D}{d}[/tex]
Here, m is the order of maxima, [tex]\lambda[/tex] is the wavelength , d is the distance between the slits and D is the distance to the screen.
For first maxima m=1, and for second maxima m=2.
Difference in the lateral maxima will be,
[tex]\Delta LS=\dfrac{2\lambda D}{d}-\dfrac{1\lambda D}{d}\\\Delta LS=\dfrac{\lambda D}{d}[/tex]
Substitute all the variables in the above equation.
[tex]\Delta LS=\dfrac{550 nm\times 2.2m}{0.10mm}\\\Delta LS=\dfrac{550\times 10^{-9} nm\times 2.2m}{0.10\times 10^{-3} }\\ \Delta LS=121\times 10^{-4}m\\ \Delta LS=1.21 cm[/tex]
Therefore, the lateral separation difference between between first order and second order maxima is 1.21 cm.
Two boats start together and race across a 48-km-wide lake and back. Boat A goes across at 48 km/h and returns at 48 km/h. Boat B goes across at 24 km/h, and its crew, realizing how far behind it is getting, returns at 72 km/h. Turnaround times are negligible, and the boat that completes the round trip first wins.
(1) Which boat wins? (Or is it a tie?)
(A) boat A
(B) boat B
(C) it's a tie
.
(2) By how much? answer in km
(3) What is the average velocity of the winning boat? answer needs to be km/h
Answer:
Part 1)
Boat A will win the race
Part 2)
Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line
Part 3)
average velocity must be zero
Explanation:
As we know that the distance moved by the boat is given as
[tex]d = 48 km[/tex]
now the time taken by the boat to move to and fro is given as
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{48 + 48}{48}[/tex]
[tex]t = 2 hrs[/tex]
Time taken by Boat B to cover the distance
[tex]t = \frac{48}{24} + \frac{48}{72}[/tex]
[tex]t = 2.66 h[/tex]
Part 1)
Boat A will win the race
Part 2)
Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line
Part 3)
Since the displacement of Boat A is zero
so average velocity must be zero
Final answer:
Boat A wins the race since it takes a total of 2 hours for the round trip, which is faster than Boat B's 2.6667 hours. Boat A wins by approximately 48 km when Boat B finishes. The average velocity of the winning boat (Boat A) is 48 km/h.
Explanation:
To determine which boat wins and by how much, we need to calculate the time each boat takes to make the round trip across the 48-km-wide lake. For Boat A, since it travels at 48 km/h both ways, the time to go across and back is simply the total distance (48 km there and 48 km back, for a total of 96 km) divided by the speed (48 km/h):
Time for Boat A = Total Distance / Speed = 96 km / 48 km/h = 2 hours.
For Boat B, the time to go across at 24 km/h and come back at 72 km/h must be calculated separately and then summed:
Time to go across for Boat B = Distance / Speed = 48 km / 24 km/h = 2 hours.
Time to come back for Boat B = Distance / Speed = 48 km / 72 km/h = 0.6667 hours.
Total time for Boat B = 2 hours + 0.6667 hours = 2.6667 hours.
Boat A wins since it takes less time to complete the round trip.
To find out by how much Boat A wins, we need to convert the time difference into distance. Since Boat A has already finished when Boat B has 0.6667 hours to go, we calculate how far Boat B would travel in that time at their average speed on the way back (72 km/h):
Difference in distance = Time difference x Speed of Boat B on return = 0.6667 hours x 72 km/h = 48 km.
For the average velocity of the winning boat (Boat A), considering the round trip without any rest or waiting time, it remains constant at 48 km/h because it travels at that speed in both the outward and return journeys.
Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x axis at x = 7.00 cm. And point charge C with a charge of +3.00 μC is located on the y axis at y = 6.00 cm. What is the direction of the net force exerted on charge A by the others?
Answer:
210.3 degrees
Explanation:
The net force exerted on charge A = 59.5 N
Use the x and y coordinates of net force to get the direction
arctan (y/x)
Answer:
F = -51.357i -29.958j
abs(F) = 59.45 N
Explanation:
To solve the problem we use coulomb's law with vectorial notation, F = q1*q2/(4*pi*eo*r^2) where q1 and q2 are the charges and r is the distance between them:
Point B exerts a force on A in '-i' direction
Point C exerts a force on A in '-j' direction
Fba = 4*7/(4*pi*eo*0.07^2) = 51.357N
Fca = 4*3/(4*pi*eo*0.06^2) = 29.958N
F = -51.357i -29.958j
abd(F) = 59.45 N
Carl is eating lunch at his favorite cafe when his friend Isaac calls and says he wants to meet him. Isaac is calling from a city 155 miles155 miles away, and wants to meet Carl somewhere between the two locations. Isaac says he will start driving right away, but Carl needs 31.0 min31.0 min to finish his lunch before he can begin driving. Isaac plans to drive at 70.0 mph70.0 mph , and Carl plans to drive at 50.0 mph50.0 mph . Ignore acceleration and assume the highway forms a straight line. How long will Isaac be driving before he meets Carl?
Answer:
1.507 h
Explanation:
We can think that they move at constant speed, so their positions will follow
X(t) = Xo + v * t
However Carl will start moving 31 minutes later, so we can adjust the equation like this
X(t) = Xo + v * (t - t1)
Where t1 is the time at which each of them will start moving, 0 for Isaac since he starts driving right away, and 31 minutes (0.5167 h) for Carl
Also, we can considre the initial position of Carl to be 0 and his speed to be 50, while Isaac will have an initial position of 155 and a speed of -70 (negative because he is driving towards the origin of coordinates).
Then we have these two equations:
x(t) = 0 + 50 * (t - 0.5167)
x(t) = 155 - 70 * t
These are equations of lines, the point where they intersect determines the place and time they meet.
50 * (t -0.5167) = 155 - 70 * t
50 * t - 25.83 = 155 - 70 * t
120 * t = 180.83
t = 180.83 / 120 = 1.507 h
This is the time Isaac will be driving
Issac will drive for [tex]\boxed{1.507\text{ hr}}[/tex] before he meets carl.
Further Explanation:
The acceleration of the driving is ignored. It means that the motion of the cars is purely linear on the highways.
Given:
The distance between Carl and Issac is [tex]155\text{ miles}[/tex].
The speed of Carl is [tex]50\text{ mph}[/tex].
The speed of Issac is [tex]70\text{ mph}[/tex].
The time taken by Carl to finish lunch before starting to drive is [tex]31\text{ min}[/tex].
Concept:
Since Carl is eating lunch and he will start after 31 minutes, Issac will cover some distance within this time.
The distance covered by Issac before Carl starts to drive is,
[tex]\begin{aligned}d_1&=\text{speed}\times\text{time}\\&=50\text{ mph}\times\left(\dfrac{31}{60}\right)\text{hr}\\&=36.16\text{ mile}\end{aligned}[/tex]
Now after Issac has covered [tex]36.16\text{ miles}[/tex], Carl starts to drive. So, the distance left between Carl and Issac to be covered after carl starts to drive is,
[tex]\begin{aligned}d&=155\text{ miles}-36.16\text{ miles}\\&=118.84\text{ miles}\end{aligned}[/tex]
So, now the sum of the distances covered by Issac and Carl in time [tex]t[/tex] should be equal to the [tex]118.84\text{ miles}[/tex].
[tex]\begin{aligned}D_{\text{Issac }}+D_{\text{Carl}}&=118.84\text{ miles}\\(V_{Issac}\times t)+(V_{Carl}\times t)&=118.84\text{ miles}\end{aligned}[/tex]
Substituting the values of speed of Issac and Carl,
[tex]\begin{aligned}(70\times t)+(50\times t)&=118.84\text{ miles}\\120t&=118.84\text{ miles}\\t&=\frac{118.84}{120}\text{ hr}\\t&=0.990\text{ hr}\end{aligned}[/tex]
So, Issac and Carl meet after [tex]0.990\text{ hr}[/tex] after Carl starts driving.
So, the time for which Issac has been driving is,
[tex]\begin{aligned}T_{\text{ Issac}}&=0.990\text{ hr}+\dfrac{31}{60}\text{ hr}\\&=(0.990+0.516)\text{ hr}\\&=1.507\text{ hr}\end{aligned}[/tex]
Thus, Issac will drive for [tex]\boxed{1.507\text{ hr}}[/tex] before he meets carl.
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Answer Details:
Grade: College
Subject: Physics
Chapter: Linear motion in one dimension
Keywords:
Carl, Issac, Speed, meet, 155 miles, eating lunch, 70 mph, 50 mph, 31 min, between two locations, highway forms a straight line.