Answer:
4900 N
Explanation:
Given :
[tex]m\ = \ 0.140\ kg\\V2= 40\ m/s\\v1\ = 30\ m/s\\t=\ 2.0 ms.[/tex]
We know that the newton second law The force * time is directly proportional to the change in the momentum .It means force *time is equal to change in ,momentum
[tex]F* t = m * ( v2 -v1 )[/tex]
it can be written as
[tex]F = \frac{m*(v2-v1)}{t}[/tex]
Putting the value of m ,v2 ,v1 and t in the previous equation we get
[tex]= \frac{0.14*(40 - (-30)}{2*10^{-3} }[/tex]
On solving
[tex]= 4900\ N[/tex]
The magnitude of the average force exerted by the bat on the ball is 4900 N.
From the question given above, the following data were obtained:
Mass (m) = 0.140 KgInitial velocity (u) = 30 m/sFinal velocity (v) = 40 m/sTime (t) = 2 ms = 2×10¯³ sForce (F) =?The magnitude of the force can be obtained as follow:
F = m(v + u) / t
F = 0.140 (40 + 30) / 2×10¯³
F = (0.140 × 70) / 2×10¯³
F = 9.8 / 2×10¯³
F = 4900 N
Thus, the magnitude of the force exerted is 4900 N
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A ball is thrown from the ground with nonzero horizontal and vertical initial velocities. While the ball is in the air, assuming that only the force of gravity acts on it, which of the following statements is true? a. Both the horizontal and vertical components of momentum are constant. b. Only the horizontal component of momentum is constant. c. Neither the horizontal nor the vertical components of momentum are constant.
Answer:
B
Explanation:
Solution:-
- Take a coordinate system as follows:
x: Directed along the ground
y: Directed vertical to ground
- We will assume that the initial vertical and horizontal non-zeroes velocities are given as follows:
[tex]v_x_i = v_o_x\\\\v_y_i = v_o_y\\[/tex]
- After a ball is thrown it continues a path of parabolic projectile. The motion of the ball can be analyzed in each coordinate system. We will assume that effects of air-resistance are negligible.
- Therefore, only gravity acts on the ball in the vertical direction. We can use kinematic equation of motions to determine the velocity of ball in either ( x-y ) direction at any instant of time ( t ).
- Use first kinematic equation of motion in both x and y directions.
[tex]v_x_f = v_o_x + a_x*t\\\\v_y_f = v_o_y + a_y*t\\[/tex]
- The accelerations ( ax and ay ) in the direction of each axis are to be determined. We know that the gravity acceleration ( g ) acts in vertical direction or along y-axis ( ay ) and always directed downwards while velocity is directed up. Since, we neglected the effects of air-resistance there is no acceleration in the x-direction ( ax = 0 ) .
[tex]v_x_f = v_o_x\\\\v_y_f = v_o_y - gt\\[/tex]
- We see that the horizontal velocity of the ball ( vxf ) at any point in time remains equal to the initial horizontal velocity; hence, it is constant throughout the journey.
- However, the velocity of the ball in vertical direction( vyf ) is changing for every unit of time ( t ) under the influence of gravitational acceleration. Hence, it is not constant throughout the journey
A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp
Answer:
sphere, disk, hoop
Explanation:
See attached file
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Complete question:
A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Answer:
The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.
Explanation:
Given;
magnitude of applied force, F = 1.5 N
Apply Newton's second law of motion;
F = ma
[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;
[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]
Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.
Which kind of image can never be projected and forms where light rays appear to originate?
virtual
real
inverted
enlarned
Answer:
a) virtual
Explanation:
This is a lens problem, in this case the equation that describes the process is the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, weights the distance to the object and qq the distance to the image
In a projection system the image that we see is q, which can have several characteristics
a) virtual It can never be projected, since this image is formed by the extensions of the light rays, therefore it is not real but a construction of the brain that interprets where the rays must come from.
b) Real. This image can be projected since light rays pass through the image
c) Inverted. Inverted images are real so they can be projected, so rays pass through the image
d) expanded. In this case the image is greater than the object, this occurs when the object the distance to the image is greater than the distance to the object, therefore the distance q is negative, therefore this image is straight and is formed by the extensions from the rays and you can't project
Answer:
option A) virtual
Explanation:
hope this helps :)
A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in the figure below. The stone reaches point A at the top of the wall, t=3.5s after being thrown. Determine (a) the height h of the wall, (b) the maximum height H of the path of the stone, (c) the horizontal distance between the launching point and point A and (d) the horizontal reach of the stone if the wall did not exist
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
The specific heat of water is about 4200 J/(kg·oC). To heat a cup of water of 0.25 kg from 10 oC to 50 oC, at least how much heat is required?
Answer:
heat required = mass × specific heat × change in temperature
Explanation:
0.25 × 4200 × 40
=42000 J
For the last part of the lab, you should have found the mass of the meter stick. So if a mass of 85 g was placed at the 2 cm MARK and the pivot point was moved to the 38.6 cm MARK, what would have been the mass of the meter stick
Answer:
272.89g
Explanation:
Find the diagram to the question in the attachment below;.
Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moment.
Moment = Force * Perpendicular distance
Taking the moment of force about the pivot.
Anticlockwise moment:
The 85g mass will move in the anticlockwise
Moment of 85g mass = 85×36.6
= 3111gcm
Clockwise moment.
The mass of the metre stick M situated at the centre (50cm from each end) will move in the clockwise direction towards the pivot.
CW moment = 11.4×M = 11.4M
Equating CW moment to the ACW moment we will have;
11.4M = 3111
M = 3111/11.4
M = 272.89g
The mass of the metre stick is 272.89g
When moving to a new apartment, you rent a truck and create a ramp with a 244 cm long piece of plywood. The top of the moving ramp lies on the edge of the truck bed at a height of 115 cm. You load your textbooks into a wooden box at the bottom of the ramp (the coefficient of kinetic friction between the box and ramp is = 0.2). Then you and a few friends give the box a quick push and it starts to slide up the ramp. A) What angle is made by the ramp and the ground?B) Unfortunately, after letting go, the box only tables 80cm up the ramp before it starts coming back down! What speed was the box initially traveling with just after you stopped pushing it?
Answer:
A) θ = 28.1º , B) v = 2.47 m / s
Explanation:
A) The angle of the ramp can be found using trigonometry
sin θ = y / L
Φ = sin⁻¹ y / L
θ = sin⁻¹ (115/244)
θ = 28.1º
B) For this pate we can use the relationship between work and kinetic energy
W =ΔK
where the work is
W = -fr x
the negative sign is due to the fact that the friction force closes against the movement
Lavariacion of energy cineta is
ΔEm = ½ m v² - mgh
-fr x = ½ m v² - m gh
the friction force has the equation
fr = very N
at the highest part there is no speed and we take the origin from the lowest part of the ramp
To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular
Axis y . perpendicular
N- Wy = 0
cos tea = Wy / W
Wy = W cos treaa
N = mg cos tea
we substitute
- (very mg cos tea) x = ½ m v²2 - mgh
v2 = m (gh- very g cos tea x)
let's calculate
v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)
v = RA (7.84 -1.729)
v = 2.47 m / s
C.
(11) in parallel
A potentiometer circuit consists of a
battery of e.m.f. 5 V and internal
resistance 1.0 12 connected in series with a
3.0 12 resistor and a potentiometer wire
AB of length 1.0 m and resistance 2.0 12.
Calculate:
(i) The total resistance of the circuit
The current flowing in the circuit
(iii) The lost volt from the internal
resistance of battery across the
battery terminals
(iv) The p.d. across the wire AB
(v) The e.m.f. of a dry cell which can be
balanced across 60 cm of the wire
AB.
Assume the wire has a uniform cross-
sectional area.
Answer:
fggdfddvdghyhhhhggghh
The molecules of a gas are in constant random motion. This means that they have energy in what type of energy store?
Answer:
Heat causes the molecules to move faster, (heat energy is converted to kinetic energy ) which means that the volume of a gas increases more than the volume of a solid or liquid.
Explanation:
Assume the angular momentum of a diatomic molecule is quantized according to the relation . What are the allowed rotational kinetic energies
Answer:
The answer to this question can be defined as follows:
Explanation:
In the given question, an equation is missing which can be defined as follows:
[tex]I \omega =\sqrt{J(J+1)}h[/tex]
solution:
Angular momentum:
[tex]L=I \omega =\sqrt{J(J+1)}h[/tex]
Convert Angular momentum in terms of kinetic energy:
[tex]K = \frac{L^2}{2I}[/tex]
[tex]= \frac{h^2(J(J+1))}{2I}[/tex]
A passenger jet flies from one airport to another 1,233 miles away in 2.4 h. Find its average speed. = ____ m/s
Speed = (distance) / (time)
Speed = (1,233 mile) / (2.4 hour)
Speed = 513.75 mile/hour
Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)
Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)
Speed = 229.7 meter/second
A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.
a. Where should a third charge, Q , be placed, so that all three charges will be in equilibrium? Express your answer in terms of d.
b. What should be its sign, so that all three charges will be in equilibrium?
c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.
Answer:
a) x = ⅔ d , b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.
(a) Find the magnitude of the angular acceleration of the wheel.
(b) Find the angle in radians through which it rotates in this time interval.
Explanation:
(a) Find the magnitude of the angular acceleration of the wheel.
angular acceleration = angular speed /timeangular acceleration = 12.9/2.98 = 4.329rad/s²(b) Find the angle in radians through which it rotates in this time interval.
angular speed = 2x3.14xf12.9rad = 2 x3.14rad = 6.28/12.9rad = 0.487Now we convert rad to angle
1 rad = 57.296°0.487 = unknown angleunknown angle =57.296 x 0.487 = 27.9°The angle in radians = 27.9°
The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20-m wide and averages 20-m deep.
(a) What is the average speed of the river in the gorge?
(b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m?
Answer:
(a) V = 0.75 m/s
(b) V = 0.125 m/s
Explanation:
The speed of the flow of the river can be given by following formula:
V = Q/A
V = Q/w d
where,
V = Speed of Flow of River
Q = Volume Flow Rate of River
w = width of river
d = depth of river
A = Area of Cross-Section of River = w d
(a)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 20 m
d = 20 m
Therefore,
V = (300 m³/s)/(20 m)(20 m)
V = 0.75 m/s
(b)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 60 m
d = 40 m
Therefore,
V = (300 m³/s)/(60 m)(40 m)
V = 0.125 m/s
What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?
Explanation:
The collision forces are equal and opposite. Therefore, the magnitudes are equal.
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
To maximize the percentage of the power from the emf of a battery that is delivered to a device external to the battery, what should the internal resistance of the battery be?
Answer:
the internal resistance of the battery be minimized
Explanation:
Because from
P= I²R+I²r
The amount of emf of battery delivered to the external load depends on internal resistance r such that the smaller the r the higher the emf
The internal resistance of the battery be :
-minimized.
Resistance of BatteryThe internal resistance of the battery be minimized.
Internal resistance refers to the restriction to the stream of current advertised by the cells and batteries themselves coming about within the era of warm.
Internal resistance is measured in Ohms.
The relationship between inside resistance (r) and emf (e) of cell s given by. e = I (r + R).
The sum of emf of battery conveyed to the outside stack depends on inside resistance r such that the littler the r the higher the emf.
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5. Sandor fills a bucket with water and whirls it in a vertical circle to demonstrate that the
water will not spill from the bucket at the top of the loop. If the length of the rope from his
hand to the centre of the bucket is 1.24 m, what is the minimum tension in the rope (at the
top of the swing)? How slow can he swing the bucket? Explain your answer.
Given that,
radius = 1.24 m
According to question,
The rope cannot push outwards. It must always have some slight tension or the bucket will fall.
We need to calculate the tension in the rope
At the top the force of gravity is
[tex]F=mg[/tex]
The force needed to move the bucket in a circle is centripetal force.
So, if mg is ever greater than centripetal force then the bucket and the contents will start to fall.
The rope have a tension of less than zero.
We need to calculate the velocity of swing bucket
Using centripetal force
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]mg=\dfrac{mv^2}{r}[/tex]
[tex]g=\dfrac{v^2}{r}[/tex]
[tex]v^2=gr[/tex]
[tex]v=\sqrt{gr}[/tex]
Put the value into the formula
[tex]v=\sqrt{9.8\times1.24}[/tex]
[tex]v=3.49\ m/s[/tex]
Hence, The minimum tension in the rope is less than zero .
The bucket swings with the velocity of 3.49 m/s.
A truck traveling at 60 km / h comes to a complete stop due to a traffic signal. Does this change in velocity violate the law of conservation of energy? Explain
Explanation:
No. Energy is still conserved. The kinetic energy of the truck is converted into heat.
In a Michelson interferometer, in order to shift the pattern by half a fringe, one of the mirrors at the end of an arm must be moved by:
Answer: Options not given.
The correct option is one quarter wavelengths.
Here are the options.
a. one wavelength
b. half a wavelength
c. one-quarter wavelength
d. it depends on the wavelength
e. it depends on which mirror is moved
It must be moved by one quarter wavelength
Explanation:
The standard way to use oMichelson apparatus is to first measure the wavelength of the incident light. One of the mirrors is moved nearer or farther away while observing the fringe pattern. one of the mirrors at the end of an arm must be moved by one quarter wavelength in order to shift the pattern by half a fringe,
A boat develops a leak and, after its passengers are rescued, eventually sinks to the bottom of a lake. When the boat is at the bottom, what is the force of the lake bottom on the boat?
Answer:
Explanation:
The force of the lake bottom on the boat = force the boat exerts on the bottom
force the boat exerts on the bottom = weight of the boat - buoyant force on the boat
so
The force of the lake bottom on the boat = weight of the boat - buoyant force on the boat
So
The force of the lake bottom on the boat will be less than its weight .
A 25-coil spring with a spring constant of 350 N/m is cut into five equal springs with five coils each. What is the spring constant of each of the 5-coil springs
1750N/m
Explanation:
According to Hooke's law, the spring constant (k) of an elastic material is the ratio of the force (F) applied to the material to the extension (x) of the material caused by this force. i.e
k = F / x --------------(i)
From the question, the elastic material (spring) of 25 coils has a spring constant of 350N/m. This means that for every 350N force applied to the spring, the spring extends by 1m.
Now if the spring is cut into five equal parts each with five coils, imagine they are attached together such that the same force of 350N is applied to still cause a total extension of 1m. Each spring contributes 1/5 of this extension.
Therefore, the extension caused by each spring is 1/5 of 1m = 0.2m
Since the same force of 350N is applied, substitute F = 350N and x = 0.2m into equation (i) as follows;
k = 350 / 0.2
k = 1750N/m
Therefore, the spring constant of each of the 5-coil spring is 1750N/m
The spring constant of each of the 5-coil springs is 1,750 N/m.
The given parameters;
spring constant, k = 350 N/mnumber of coils, N = 25The total spring constant of the 25 coils is calculated as follows;
[tex]K_t = 25 \times 350 \ N/m\\\\K_t = 8,750 \ N/m[/tex]
The spring constant of each of the given 5 coils is calculated as follows;
[tex]K = \frac{8,750 }{5} \\\\K = 1,750 \ N/m[/tex]
Thus, the spring constant of each of the 5-coil springs is 1,750 N/m.
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ou are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interferometer. The hydrogen discharge tube provides light of several different wavelengths (colors) in the visible range. The red light in the hydrogen spectrum has a wavelength of 656.3 nm and the blue light has a wavelength of 434.0 nm. When using the discharge tube and the red filter as the light source, you view a bright red spot in the viewing area of the interferometer. You now move the movable mirror away from the beam splitter and observe 146 bright spots. You replace the red filter with the blue filter and observe a bright blue spot in the interferometer. You now move the movable mirror towards the beam splitter and observe 122 bright spots. Determine the final displacement (include sign) of the moveable mirror. (Assume the positive direction is away from the beam splitter.)
Answer:
Explanation:
In interferometer , when the movable mirror is moved away by distance d , there is fringe shift on the screen . If n be number of fringes shifted
2 d = n λ
where λ is wavelength of light
Applying this theory for first case when no of fringes shifted is 146
2 d₁ = 146 x 656.3 nm
d₁ = 47909.9 x 10⁻⁹ m
= .048 x 10⁻³ m
= .048 mm
For second case n = 122
2d₂ = 122 x 434 x 10⁻⁹
d₂ = 26474 x 10⁻⁹ m
= .026 mm
So in the second case , mirror must have been moved towards the beam splitter by .048 - .026 = .022 mm
So movement = - 0 .022 mm ( negative displacement )
The resonance tube used in this experiment produced only one resonance tone. What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
Answer:
the length that would produce a sound tone under the same experimental contditions must be increased by Δl = [tex]\frac{v}{2f}[/tex]
Explanation:
Recall
V = f ×λ
where λ is ⁴/₃l₂ for second resonance
f = [tex]\frac{3v}{4l_{2} }[/tex]
l₂ = [tex]\frac{3v}{4f}[/tex]
where λ is 4l₁ for 1st resonance
f = [tex]\frac{v}{4l_{1} }[/tex]
l₁ = [tex]\frac{v}{4f}[/tex]
∴ Δl = l₂ - l₁ = [tex]\frac{3v}{4f}[/tex] ⁻ [tex]\frac{v}{4f}[/tex]
Δl= [tex]\frac{2v}{4f}[/tex]
Δl = [tex]\frac{v}{2f}[/tex]
Therefore, the length should increase by [tex]\frac{v}{2f}[/tex]
Un autocar que circula a 81 km/h frena uniformemente con una aceleración de -4,5 m/s2.
a) Determina cuántos metros recorre hasta detenerse.
b) Representa las gráficas v-t y s-t.
Answer:
a) [tex]\Delta x=56.25 m[/tex]
b) imagen adjunta
Explanation:
a) Primero debemos hacer la conversión de 81 km/h a m/s, esto es 22.5 m/s.
Ahora, usando la ecuacion cinemática, en un movimiento acelerado tenemos:
[tex]v_{f}^{2}=v_{0}^{2}+2a \Delta x [/tex]
Queremos encontrar la posición hasta detenerse, osea vf = 0.
[tex]\Delta x=\frac{-v_{0}^{2}}{2a}[/tex]
[tex]\Delta x=\frac{-22.5^{2}}{-2*4.5}[/tex]
[tex]\Delta x=56.25 m[/tex]
b) Para este caso el gráfico se encuentra adjunto.
Espero que te sirva de ayuda!
a 1000 kg car accelerates from 0 to 15 m/s in 5.0 s with negligible friction and air resistance. what is the average power delivered by the engine
Answer:
30.16 hp
Explanation:
Given :
m= 1000 kg
v=15 m/s
t=5.0 s
The average power delivered by the engine can be determined by using the given formula
[tex]Average\ power\ =\ \frac{0.5*m*v^2\ }{t}[/tex]
where m=,mass, v=velocity and t=time
Now putting the value of m,v and t in the previous equation we get
[tex]Average \ power\ =\ \frac{0.5*1000*15*15}{5} \\Average \ power\ =\ 22,500\ w\\Average \ power\ =\frac{22,500}{746} \\Average \ power\ =30.16\ hp[/tex]
To convert w to hp we divide by 746
Therefore 30.16 hp is the answer.
You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum
Answer:
The width is [tex]Z = 0.0424 \ m[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]
The wavelength of the light is [tex]\lambda = 721 \ nm[/tex]
The position of the screen is [tex]D = 2.83 \ m[/tex]
Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as
[tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]
Where m which is the order of the interference is 1
substituting values
[tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]
[tex]\theta = 0.5317 ^o[/tex]
Now the width of first minimum of the interference pattern is mathematically evaluated as
[tex]Y = D sin \theta[/tex]
substituting values
[tex]Y = 2.283 * sin (0.5317)[/tex]
[tex]Y = 0.02 12 \ m[/tex]
Now the width of the pattern's central maximum is mathematically evaluated as
[tex]Z = 2 * Y[/tex]
substituting values
[tex]Z = 2 * 0.0212[/tex]
[tex]Z = 0.0424 \ m[/tex]
The springs of a 1700-kg car compress 5.0 mm when its 66-kg driver gets into the driver seat. If the car goes over a bump, what will be the frequency of oscillations?
Answer:
The frequency of oscillations is 7 Hz
Explanation:
Given;
mass of car, = 1700 kg
mass of driver, = 66 kg
compression of the spring, x = 5mm = 0.005 m
The frequency of the oscillation is given as;
[tex]F = \frac{1}{2 \pi } \sqrt{\frac{k}{m} }[/tex]
where;
k is force constant
m is the total mass of the car and the driver
m = 1700 kg + 66 kg = 1766 kg
Weight of the car and the driver;
W = mg
W = 1766 x 9.8
W = 17306.8 N
Apply hook's law, to determine the force constant;
F = kx
W = F
Thus, k = W/x
k = 17306.8 / 0.005
k = 3461360 N/m
Now, calculate the frequency
[tex]F = \frac{1}{2\pi}\sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi}\sqrt{\frac{3461360}{1766} }\\\\F = 7 \ Hz[/tex]
Therefore, the frequency of oscillations is 7 Hz
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 43 m/s (about 96 mph, the speed of a good major league fastball)."
Required:
a. Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
b. Now calculate the maximum height using universal gravitation.
c. How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.
Answer:
a)[tex]h_{max}=14536.16 m[/tex]
b)[tex]h = 15687.9 m[/tex]
c)[tex]PD=7.62\%[/tex] The estimate is low.
Explanation:
a) Using the energy conservation we have:
[tex]E_{initial}=E_{final}[/tex]
we have kinetic energy intially and gravitational potential energy at the maximum height.
[tex]\frac{1}{2}mv^{2}=mgh_{max}[/tex]
[tex]h_{max}=\frac{v^{2}}{2g}[/tex]
[tex]h_{max}=\frac{43^{2}}{2*0.0636}[/tex]
[tex]h_{max}=14536.16 m[/tex]
b) We can use the equation of the gravitational force
[tex]F=G\frac{mM}{R^{2}}[/tex] (1)
We have that:
[tex] F = ma [/tex] (2)
at the surface G will be:
[tex]G=\frac{gR^{2}}{M}[/tex]
Now the equation of an object at a distance x from the surface.
is:
[tex]F=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
[tex]m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
Using that dv/dt is vdx/dt and integrating in both sides we have:
[tex]v_{0}=\sqrt{\frac{2gRh}{R+h}}[/tex]
[tex]h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}[/tex]
[tex]h=15687.9[/tex]
c) The difference is:
So the percent difference will be:
[tex]PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%[/tex]
[tex]PD=7.62\%[/tex]
The estimate is low.
I hope it helps you!