Answer:
n=2.053
Explanation:
We will use Snell's Law defined as:
[tex]n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}[/tex]
Where n values are indexes of refraction and [tex]\theta[/tex] values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:
[tex]1*Sin(39.9)=n_{2}*Sin(18.2)[/tex]
Solving for [tex]n_{2}[/tex] yields:
[tex]n_{2}=\frac{Sin(39.9)}{Sin(18.2)}=2.053[/tex]
Remember to use degrees for trigonometric functions instead of radians!
Measurements indicate that there is an electric field surrounding the Earth. Its magnitude is about 150 N/C at the Earth's surface and points inward toward the Earth's center. What is the magnitude of the electric charge on the Earth? Express your answer using two significant figures.
Answer:
Magnitude of electric field on Earth = Q = 6.8 × 10⁵ C
Explanation:
Electric field = E = 150 N/C
Distance from the center of the earth to the surface = Radius of the earth
Radius of the earth = R = 6.38× 10⁶ m
E = k Q / R² is the basic formula for the electric field. k = 9 × 10⁹ N m²/C²
150 = (9 × 10⁹)(Q) / (6.38× 10⁶ )²
⇒ Charge = Q = (150)(6.38× 10⁶ )²/(9 × 10⁹)
= 6.8 × 10⁵ C(2 significant figures).
Answer:
The charge of earth is[tex]-6.8\times 10^{5}Columbs[/tex]
Explanation:
Assuming earth as a spherical body we have
For a sphere of radius 'r' and charge 'q' the electric field generated at a distance 'r' form the center of sphere is given by the equation
[tex]E=\frac{1}{4\pi \epsilon _o }\cdot \frac{Q}{r^{2}}[/tex]
where
'Q' is the total charge on sphere
Now at a distance 'r' equal to radius of earth(6371 km) we have the electric field strength is 150N/C
Using the given values we obtain
[tex]150=\frac{1}{4\pi \epsilon _o}\frac{Q}{(6371\times 10^{3})^2}\\\\\therefore Q=150\times (6371\times 10^{3})^{2}\times 4\pi \epsilon _o\\\\\therefore Q=6.8\times 10^{5}Columbs[/tex]
Now since the electric field is inwards thus we conclude that this charge is negative in magnitude.
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. How much time does it take until he catches his friend (after his friend passes him)?
Final answer:
It will take approximately 1.61 seconds for the bicyclist to catch up to his friend after his friend passes him.
Explanation:
To determine the time it takes for the bicyclist to catch up to his friend, we can use the equation:
distance = initial velocity * time + 0.5 * acceleration * time^2
Since the friend is traveling at a constant speed of 3.1 m/s, the distance traveled by the bicyclist during the 2-second delay is 6.2 m. Using the equation above:
6.2 m = 0 m/s * t + 0.5 * 2.4 m/s^2 * t^2
Simplifying the equation:
2.4 m/s^2 * t^2 = 6.2 m
t^2 = 6.2 m / 2.4 m/s^2
t^2 = 2.5833 s^2
t ∼ 1.61 s
Therefore, it will take approximately 1.61 seconds for the bicyclist to catch up to his friend.
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
Two tiny conducting spheres are identical and carry charges of -23.0C and +67.2C. They are separated by a distance of 3.18 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 3.18 cm. Determine the magnitude of the force that each sphere now experiences.
Explanation:
Given that,
Charge 1, [tex]q_1=-23\ C[/tex]
Charge 2, [tex]q_1=+67\ C[/tex]
Distance between charges, r = 3.18 cm = 0.0318 m
(a) Let F is the magnitude of force that each sphere experiences. The force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{-23\times 67}{(0.0318)^2}[/tex]
[tex]F=-1.37\times 10^{16}\ N[/tex]
(b) The spheres are brought into contact and then separated to a distance of 3.18 cm. When they are in contact, both possess equal charges. Net charge is :
[tex]q=\dfrac{q_1+q_2}{2}[/tex]
[tex]q=\dfrac{-23+67}{2}=22\ C[/tex]
Electric force is given by :
[tex]F=9\times 10^9\times \dfrac{22^2}{(0.0318)^2}[/tex]
[tex]F=4.307\times 10^{15}\ N[/tex]
Hence, this is the required solution.
A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angle of 110 with the positive xaxis. The resulta displacement has a magnitude of 131 cm and is directed at an angle of 38.0 to the positive axis. Find the magnitude and direction of the second displacement magnitude direction 1 (countercockwise from the positive x-axis)
Answer:
D₂= 167,21 cm : Magnitude of the second displacement
β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement
Explanation:
We find the x-y components for the given vectors:
i: unit vector in x direction
j:unit vector in y direction
D₁: Displacement Vector 1
D₂: Displacement Vector 2
R= resulta displacement vector
D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j
D₂= -D₂(i)-D₂(j)
R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j
We propose the vector equation for sum of vectors:
D₁+ D₂= R
-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j
-51.99i+D₂x(i)=103.23i
D₂x=103.23+51.99=155.22 cm
+142.83j-D₂y(j) =+80.65j
D₂y=142.83-80.65=62.18 cm
Magnitude and direction of the second displacement
[tex]D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }[/tex]
[tex]D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }[/tex]
D₂= 167.21 cm
Direction of the second displacement
[tex]\beta = tan^{-1} \frac{D_{y}}{D_{x} }[/tex]
[tex]\beta = tan^{-1} \frac{62.18}{155.22 }[/tex]
β= 21.8°
D₂= 167,21 cm : Magnitude of the second displacement
β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement
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.
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 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
Suppose a single electron orbits about a nucleus containing two protons (+2e), as would be the case for a helium atom from which one of the naturally occurring electrons is removed. The radius of the orbit is 2.99 × 10-11 m. Determine the magnitude of the electron's centripetal acceleration.
Answer:
[tex]a=5.66*10^{23} \frac{m}{s^2}[/tex]
Explanation:
In this case we will use the Bohr Atomic model.
We have that: [tex]F=m*a[/tex]
We can calculate the centripetal force using the coulomb formula that states:
[tex]F=k*\frac{q*q'}{r^2}[/tex]
Where K=[tex]9*10^9 \frac{Nm^2}{C}[/tex]
and r is the distance.
Now we can say:
[tex]m*a=k*\frac{q*q'}{r^2}[/tex]
The mass of the electron is = [tex]9.1*10^{-31}[/tex] Kg
The charge magnitud of an electron and proton are= [tex]1.6*10^{-19}C[/tex]
Substituting what we have:
[tex][tex]a=\frac{9*10^{9}*(1.6*10^{-19} )*(2(1.6*10^{-19} ))}{9.1*10^{-31}*(2.99*10^{-11})^2 }[/tex][/tex]
so:
[tex]a=5.66*10^{23} \frac{m}{s^2}[/tex]
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).
In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-like nature of the board to help propel him into the air. Assume that the diver’s motion is essentially vertical. He leaves the board, which is 3.0 m above the water, with a speed of 6.3 m/s. How long is the diver in the air, from the moment he leaves the board until
Answer:
The diver is in the air for [tex]1.65s[/tex].
Explanation:
Hi
Known data [tex]v_{i}=6.3m/s, y_{i}=3.0m[/tex] and [tex]g=9.8m/s^{2}[/tex].We need to find the time when the diver reaches the highest point, so we use the following equation [tex]v_{f} =v_{i}-gt[/tex] with [tex]v_{f}=0ms[/tex] so [tex]t_{1}=\frac{v_{i}-v_{f} }{g}=\frac{6.3m/s}{9.8m/s^{2} }=0.64 s[/tex]. Then we need to find the highest point, so we use [tex]y=v_{i}t-\frac{1}{2} gt^{2}=(6.3m/s)(0.64s)-\frac{1}{2} (9.8m/s^{2})(0.64s)^{2}=2.03m[/tex], this is above the springboard so the highest point is [tex]y_{max}=5.03m[/tex].Finally, we need to find the time in free fall, so we use [tex]y_{f}=y_{i}+v_{i}t-\frac{1}{2}gt^{2}[/tex], at this stage [tex]v_{i}=0m/s, y_{i}=5.03m[/tex] and [tex]y_{f}=0m[/tex], therefore we have [tex]0m=5.03-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex], and solving for [tex]t_{2}=\sqrt{\frac{5.03m}{4.9m/s^{2}}} =\sqrt{1.02s^{2}}=1.01s[/tex].
Last steep is to sum [tex]t_{1}[/tex] and [tex]t_{2}[/tex], so [tex]t_{T}=t_{1}+t_{2}=0.64s+1.01s=1.65s[/tex].
The total time spent by the diver in the air from the moment he leaves the board until he gets to the water is 1.65 s
We'll begin by calculating the time taken to get to the maximum height from the board.
Initial velocity (u) = 6.3 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Time to reach maximum height (t₁) =?
v = u – gt (since the diver is going against gravity)
0 = 6.3 – 9.8t₁
Collect like terms
0 – 6.3 = –9.8t₁
–6.3 = –9.8t₁
Divide both side by –9.8
t₁ = –6.3 / –9.8
t₁ = 0.64 sNext, we shall determine the maximum height reached by the diver from the board
Initial velocity (u) = 6.3 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Maximum Height from the board (h₁) =?
v² = u² – 2gh (since the diver is going against gravity)
0² = 6.3² – (2 × 9.8 × h₁)
0 = 39.69 – 19.6h₁
Collect like terms
0 – 39.69 = –19.6h₁
–39.69 = –19.6h₁
Divide both side by –19.6
h₁ = –39.69 / –19.6
h₁ = 2.03 mNext, we shall determine the height from the maximum height reached by the diver to the water.
Maximum height from the board (h₁) = 2.03 m
Height of board from water (h₂) = 3 m
Height of diver from maximum height to water (H) =?
H = h₁ + h₂
H = 2.03 + 3
H = 5.03 mNext, we shall determine the time taken by the diver to fall from the maximum height to the water.
Height (H) = 5.03 m
Acceleration due to gravity (g) = 9.8 m/s²
Time to fall from maximum height to water (t₂) =?
H = ½gt²
5.03 = ½ × 9.8 × t₂²
5.03 = 4.9 × t₂²
Divide both side by 4.9
t₂² = 5.03 / 4.9
Take the square root of both side
t₂ = √(5.03 / 4.9)
t₂ = 1.01 sFinally, we shall determine the total time spent by the diver in the air.
Time to reach maximum height (t₁) = 0.64 s
Time to fall from maximum height to water (t₂) = 1.01 s
Total time in air (T) =?T = t₁ + t₂
T = 0.64 + 1.01
T = 1.65 sTherefore, the total time spent by the diver in the air is 1.65 s
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A 3.00 kg steel ball strikes a massive wall at 10.0m/s at
anangle of 60.0 degree with the plane of the wall. It bouncesoff
the wall with the same speed and angle. If the ball is incontact
with the wall for 0.200s, what is the average force exertedby the
wall on the ball?
Answer:259.80 N
Explanation:
Given
mass of ball=3 kg
ball velocity =10 m/s
angle made by ball with plane of wall [tex]\theta [/tex]=60
Momentum change in Y direction remains same and there is change only in x direction
therefore
initial momentum[tex]=mvsin\theta [/tex]
=30sin60
Final momentum=-30sin60
Change in momentum is =30sin60+30sin60
=60sin60
and Impulse = change in momentum
Fdt=dP
where F=force applied
dP=change in momentum
[tex]F\times 0.2=60sin60[/tex]
[tex]F\times 0.2=51.96[/tex]
F=259.80 N
The average force exerted by the wall on the 3.00 kg steel ball after it bounces off is 150 N. We calculated this using the principle of conservation of momentum and Newton's second law.
Explanation:The subject of this question is Physics, specifically, the topic of motion and force. To answer this question, we should apply the law of conservation of momentum and Newton's second law (Force = change in Momentum / change in Time).
Firstly, the ball is projected against the wall at an angle of 60 degrees, but we are concerned with the component of velocity perpendicular to the wall. This means we will consider the velocity component of the ball towards the wall, which is 10 cos(60), giving us 5.0 m/s. The incoming momentum of the ball can then be calculated as the mass times the velocity (3.0 kg * 5.0 m/s = 15 kg*m/s).
Since the ball bounces off with the same speed at the same angle, its outgoing momentum is -15 kg*m/s. The change in momentum is therefore Outgoing momentum - Incoming momentum = -15 kg*m/s - 15 kg*m/s = -30 kg*m/s. The force exerted by the wall on the ball equals the Change in momentum divided by the time it takes for the change to occur (-30 kg*m/s / 0.200 s = -150 N). Given that force is a vector and we are asked for the magnitude of the force, the answer is 150 N.
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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 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
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 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
Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving in a circular path perpendicular to a magnetic field of magnitude B. (Use any variable or symbol stated above along with the following as necessary: q and c.)
Explanation:
Let [tex]m_p[/tex] is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.
The magnetic force is balanced by the centripetal force acting on the proton as :
[tex]\dfrac{mv^2}{r}=qvB[/tex]
r is the radius of path,
[tex]r=\dfrac{mv}{qB}[/tex]
Time period is given by :
[tex]T=\dfrac{2\pi r}{v}[/tex]
[tex]T=\dfrac{2\pi m_p}{qB}[/tex]
Frequency of proton is given by :
[tex]f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}[/tex]
The wavelength of radiation is given by :
[tex]\lambda=\dfrac{c}{f}[/tex]
[tex]\lambda=\dfrac{2\pi m_pc}{qB}[/tex]
So, the wavelength of radiation produced by a proton is [tex]\dfrac{2\pi m_pc}{qB}[/tex]. Hence, this is the required solution.
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.
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]
An aeroplane flies in a loop (a circular path in a vertical plane) of radius 200 m. The pilot's head always points toward the centre of the loop. The speed of the aeroplane is not constant; the aeroplane goes slowest at the top of the loop and fastest at the bottom. At the top of the loop, the pilot feels weightless. What is the speed of the aeroplane at this point?
At the bottom of the loop, the speed of the aeroplane is 280 km/h . What is the apparent weight of the pilot at this point? His true weight is 710 N .
Answer:
2899.24 N
Explanation:
W = Weight of pilot
r = Radius
v = Velocity
g = Acceleration due to gravity = 9.81 m/s²
[tex]mg=m\frac{v^2}{r}\\\Rightarrow v=\sqrt{gr}\\\Rightarrow v=\sqrt{9.81\times 200}\\\Rightarrow v=44.3\ m/s[/tex]
Speed of the aeroplane at the top of the loop is 44.3 m/s
Now, v = 280 km/h = 280/3.6 = 77.78 m/s
Apparent weight
[tex]A=W+\frac{W}{g}\frac{v^2}{r}\\\Rightarrow A=710+\frac{710}{9.81}\times \frac{77.78^2}{200}\\\Rightarrow A=2899.24\ N[/tex]
Apparent weight at the bottom of the loop is 2899.24 N
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.
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.
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 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. a. What is the solubility (in g salt/100 g of water) of the salt? b. How much water would it take to dissolve 25 g of this salt? c. If 10.0 g of this salt is mixed with 15.0 g of water, what percentage of the salt dissolves?
Answer:
The solubility of the salt is 35.16 (g/100 g of water).It would take 71.09 grams of water to dissolve 25 grams of salt.The percentage of salt that dissolves is 52.7 %Explanation:
a.We know that 3.20 grams of salt in 9.10 grams of water gives us a saturated solution at 25°C. To find how many grams of salt will gives us a saturated solution in 100 grams of water at the same temperature, we can use the rule of three.
[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{100 \ g \ water}[/tex]
Working it a little this gives us :
[tex] x = 100 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]
[tex] x = 35.16 \ g \ salt [/tex]
So, the solubility of the salt is 35.16 (g/100 g of water).
b.Using the rule of three, we got:
[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{25 \ g \ salt}{x \ g \ water}[/tex]
Working it a little this gives us :
[tex] x = \frac{25 \ g \ salt}{ \frac{3.20 \g \ salt}{9.10 \ g \ water}} [/tex]
[tex] x = 71.09 g \ water [/tex]
So, it would take 71.09 grams of water to dissolve 25 grams of salt.
C.Using the rule of three, we got that for 15.0 grams of water the salt dissolved will be:
[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{15.0 \ g \ water}[/tex]
Working it a little this gives us :
[tex] x = 15.0 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]
[tex] x = 5.27\ g \ salt [/tex]
This is the salt dissolved
The percentage of salt dissolved is:
[tex]percentage \ salt \ dissolved = 100 \% * \frac{g \ salt \ dissolved}{g \ salt}[/tex]
[tex]percentage \ salt \ dissolved = 100 \% * \frac{ 5.27\ g \ salt }{ 10.0 \ g \ salt}[/tex]
[tex]percentage \ salt \ dissolved = 52.7 \% [/tex]
a. The solubility (in g salt/100 g of water) of the salt is 35.16 g/100 g water. b. Amount of water it would take to dissolve 25 g of this salt is 71.10 g. c. If 10.0 g of the salt is mixed with 15.0 g of water, 52.7% of the salt will dissolve.
To solve the problem, we need to determine the solubility of the salt at 25°C using the given data:
Part a: Solubility
The solubility of the salt is calculated as follows:A 3.20 g sample dissolves in 9.10 g of water to form a saturated solution.Solubility (g/100 g water) = (3.20 g salt / 9.10 g water) * 100 = 35.16 g/100 g water.Part b: Amount of Water Needed to Dissolve 25 g of Salt
First, use the solubility obtained in part a:Solubility = 35.16 g salt / 100 g water.To find how much water is needed to dissolve 25 g of salt: (100 g water / 35.16 g salt) * 25 g salt = 71.10 g water.Part c: Percentage of Salt Dissolved
Given 10.0 g of salt mixed with 15.0 g of water:Because we know the solubility is 35.16 g/100 g water, we find the amount that will dissolve in 15.0 g water: (35.16 g salt / 100 g water) * 15.0 g water = 5.27 g salt.Percentage dissolved = (5.27 g dissolved / 10.0 g total) * 100% = 52.7%.Therefore, 52.7% of the 10.0 g of salt will dissolve in 15.0 g of water.
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
You throw a ball straight down from an apartment balcony to the ground below. The ball has an initial velocity of 5.10 m/s, directed downward, and it hits the ground 1.92 s after it is released. Find the height of the balcony.
Answer:
the height of the balcony from where the ball is thrown is 27.874 m.
Explanation:
given,
initial velocity (u) = 5.1 m/s
time (t) = 1.92 s
height of balcony = ?
using equation;
[tex]s = u t + \dfrac{1}{2} at^2[/tex]
[tex]h =5.1 \times 1.92 + \dfrac{1}{2}\times 9.81\times 1.92^2[/tex]
h= 9.792 + 18.082
h = 27.874 m
hence, the height of the balcony from where the ball is thrown is 27.874 m.
Final answer:
Using the kinematic equations of motion, we find that the ball was thrown from a height of approximately 8.211 meters.
Explanation:
To calculate the height from which the ball was thrown, we use the kinematic equations of motion for an object under constant acceleration due to gravity. The specific equation that relates the initial velocity (Vi), time (t), acceleration due to gravity (g), and the displacement (height h in this case) is:
h = Vi * t + 0.5 * g * t2
Where:
Vi is the initial velocity = 5.10 m/s (downward, so we take it as negative)t is the time = 1.92 sg is the acceleration due to gravity = 9.81 m/s2 (downward, so we take it as positive)Plugging in these values, we get:
h = -5.10 m/s * 1.92 s + 0.5 * 9.81 m/s2 * (1.92 s)2
h = -9.792 m + 18.003 m
h = 8.211 m
Therefore, the height of the balcony is approximately 8.211 meters above the ground.
To charge an electroscope negatively by induction you need: a) a positively charged rod and a ground b) a negatively charged rod c) a negatively charged rod and a ground d) two objects of the same charge
Answer:
Option a)
Explanation:
In the process of charging anything by the method of induction, a charged body is brought near to the body which is neutral or uncharged without any physical contact and the ground must be provided to the uncharged body.
The charge is induced and the nature of the induced charge is opposite to that of the charge present on the charged body.
So when a positively charged rod is used to charge an electroscope, the rod which is positive attracts the negative charge in the electroscope and the grounding of the electroscope ensures the removal of the positive charge and renders the electroscope negatively charged.
Final answer:
To charge an electroscope negatively by induction, you need a negatively charged rod and a ground. When a positively charged rod is brought near a neutral metal sphere, it polarizes the sphere and attracts electrons from the earth's ample supply. By breaking the ground connection and removing the positive rod, the sphere is left with an induced negative charge.
Explanation:
To charge an electroscope negatively by induction, you need a negatively charged rod and a ground. When a positively charged rod is brought near a neutral metal sphere (the electroscope), it polarizes the sphere. By connecting the sphere to a ground, electrons are attracted from the earth's ample supply, resulting in an induced negative charge on the sphere. The ground connection is then broken, and the positive rod is removed, leaving the sphere with the induced negative charge.
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]
A 0.07-kg lead bullet traveling 257 m/s strikes an armor plate and comes to a stop. If all of the bullet's energy is converted to heat that it alone absorbs, what is its temperature change?
Answer:
ΔT = 258°C
Explanation:
mass of bullet, m = 0.07 kg
velocity of bullet, v = 257 m/s
According to the energy conservation law, the kinetic energy of bullet is totally converted into form of heat energy.
let ΔT be the rise in temperature of the bullet, c be the specific heat of lead.
c = 0.128 J / g°C = 128 J/kg°C
[tex]\frac{1}{2}mv^{2}=mc\Delta T[/tex]
By substituting the values
0.5 x 0.07 x 257 x 257 = 0.07 x 128 x ΔT
ΔT = 258°C
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