3.) [15 points] A physics teacher is on the west side of a small lake and wants to swim across and up at a point directly across from his starting point. He notices that there is a current in the lake and
that a leaf floating by him travels 4.2m [S] In 5.0s. He is able to swim 1.9 m/s in calm water,
(a) What direction will he have to swim in order to arrive at a point directly across from his position?
Answer:
The teacher should swim in a direction 29.24° North of East
Explanation:
Given that the there is a water current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;
The speed of the water current is given by the speed of the floating leaf traveling with the water current
Distance traveled by the leaf = 4.2 m South
Time of travel of the leaf = 5.0 s
Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current
Swimming peed of the teacher, v = 1.9 m/s
To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.
Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim
Therefore;
1.9 × sin(θ) = 0.84
sin(θ) = 0.84/1.9 = 0.44
θ = 26.24°
That is the teacher should swim in a direction 29.24° North of East.
To cross the lake the teacher has to swim in a direction 29.24° North of the East
Finding the direction of speed required:
The speed of the water current can be derived from the speed of the floating leaf :
The distance traveled by the leaf L = 4.2 m South
Time taken T = 5s
So, the speed of the leaf is:
u = 4.2/5
u = 0.84 m/s South
So, the speed of the current is 0.84 m/s South
Now, it is given that the speed of the teacher is, v = 1.9 m/s East
To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.
Let, the speed of the teacher makes an angle of θ from the EAST.
So, the North component is given by:
v(north) = vsinθ
it must be equal to the speed of the current:
vsinθ = u
1.9 × sinθ = 0.84
sinθ = 0.84/1.9
sinθ = 0.44
θ = 26.24°
The teacher should swim in a direction 29.24° North of East.
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a) A conductor carrying a current I = 12.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A magnetic force per unit length of 0.110 N/m acts on the conductor in the negative y direction. Determine the magnitude of the magnetic field in the region through which the current passes.
Answer:
8.8 mT
Explanation:
Current through the conductor = 12.5 A
Magnetic force per unit length on the wire = 0.110 N/m
Recall that the magnetic force per unit length on a current carrying conductor is in a uniform magnetic field is
[tex]\frac{F}{l}= IBsin\alpha[/tex]
where B is the magnetic field magnitude
[tex]I[/tex] is the current in the conductor
α is the angle the conductor makes with the magnetic field ( since it is perpendicular in this case, α is 90°)
imputing values into the equation, we'll have
0.11 = 12.5 x B x sin 90°
but sin 90° = 1, therefore, we have
0.11 = 12.5B
B = 0.11/12 = 8.8 x 10^-3 T
or rather = 8.8 mT
Stopping distance of vehicles When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (v0) and the braking capacity, or deceleration that is caused by braking. A car travelling at speed 72km/hr suddenly applies the brake with the deceleration of 5m/s2. Find the stopping distance of the car.
Answer:
Stopping distance = 40m
Explanation:
Given the following :
Initial speed of vehicle before applying brakes = 72km/hr
Converting km/hr to m/s:
72km/hr = [(72 * 1000)m] / (60 * 60)
72km/hr = 72,000m / 3600s
72km/hr = 20m/s
Deceleration after applying brakes (-a) (negative acceleration) = - 5m/s^2
From the 3rd equation of motion:
v^2 = u^2 + 2as
Where v = final Velocity ; u= Initial Velocity ; a = acceleration and s = distance
Final velocity when the car stops will be 0
Therefore ;
v^2 = u^2 + 2as
0 = 20^2 + 2(-5)(s)
0 = 400 - 10s
10s = 400
s = 400/10
s = 40m
Therefore, the stopping distance of the car = 40 meters
Formulating a Hypothesis: Part II
Now you will focus on a second hypothesis. This
hypothesis can be very similar to the first, but this time
you want to focus only on the second variable in
question, speed. What could be a hypothesis that would
illustrate the relationship between speed and kinetic
energy? Use the format of "if.. then... because..." when
writing your hypothesis.
If we have a ball rolling its speed could determine the amount of
kinetic energy in it while in movement
Explanation:
The kinetic energy of an object is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
m is mass of the object
v is the speed of the object
It is clear from the above relation that the kinetic energy of the object is directly proportional to the product of mass and velocity.
The hypothesis can be :
If the speed of the object increased then its kinetic energy increase because kinetic energy is directly proportional to ist speed.
The kinetic energy of an object is given by :
m is mass of the object
v is the speed of the object
It is clear from the above relation that the kinetic energy of the object is directly proportional to the product of mass and velocity.
The hypothesis can be :
If the speed of the object increased then its kinetic energy increase because kinetic energy is directly proportional to its speed.
A very thin film of soap, of thickness 170 nm, in between air seems dark. On the other hand, when placed on top of glass some visible light is seen to shine from the film. How can this happen and what is the smallest visible light that creates constructive interference when we place the film on top of glass
Answer:
λ₀ = 2 d n
Explanation:
A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.
Let's analyze the general conditions of this interference,
* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º
* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.
λ = λ₀ / n
where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains
2d = (m + ½) λ
the expression for destructive interference remains
2d = m λ
2d = m λ₀ / n
When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.
Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.
Find us the wavelength that this constructive interference creates
2d n = m λ₀
λ₀ = 2 d n / m
To find the minimum wavelength, suppose we observe the first interference pattern m = 1
λ₀ = 2 d n
where d is the thickness of the film and n the index of refraction of the same
On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the average speed for the entire journey is 40km/h?
Answer:
v = 46.67 km/h
Explanation:
We will use the following formula throughout this numerical:
s = vt
where,
s = distance covered
v = speed
t = time taken
FOR FIRST 30 km:
s = 30 km
v = 30 km/h
t = t₃₀ = ?
Therefore,
30 km = (30 km/h)(t₃₀)
t₃₀ = (30 km)/(30 km/h)
t₃₀ = 1 h
FOR TOTAL 100 km:
s = 100 km
v = 40 km/h (Average Speed)
t = total time = ?
Therefore,
100 km = (40 km/h)(t)
t = (100 km)/(40 km/h)
t = 2.5 h
FOR LAST 70 km:
s = 70 km
t₇₀ = t - t₃₀ = 2.5 h - 1 h = 1.5 h
v = v₇₀ = ?
Therefore,
70 km = v(1.5 h)
v = 70 km/1.5 h
v = 46.67 km/h
Radar uses radio waves of a wavelength of 2.5 m . The time interval for one radiation pulse is 100 times larger than the time of one oscillation; the time between pulses is 10 times larger than the time of one pulse. What is the shortest distance to an object that this radar can detect
Answer:
The minimum distance to the object that the radar can detect is 124.995 m
Explanation:
Here, we are to calculate the shortest distance to an object the radar in the question can detect.
Mathematically;
v = c/λ
Where v is the frequency, c is the speed of light and λ is the wave length
Thus;
v = (3 * 10^8)/2.5 = 1.2 * 10^8 Hz
Mathematically, the time period
t = 1/v = 1/(1.2 * 10^8) = 0.000000008333 = 8.333 * 10^-9 sec
From the question, we are told that the transmitting time is 100 times a single oscillation
Transmitting time = 100 * one oscillation
Hence Transmitting time = 100 * 8.33 * 10^-9 = 8.33 * 10-7
Mathematically;
Minimum distance =( Transmitting time * speed of light)/2 =
(8.33 * 10^-7 * 3 * 10^8)/2 = 124.995 m
Two trains run in the opposite direction with speeds of v1 = 15 m / s and v2 = 20 m / s. A passenger on the first train (the one on v1) notes that train 2 takes 6 s to pass on its side. What is the length of the second train? (The passenger is supposed to be immobile looking through the window)
Answer:
210 m
Explanation:
The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.
It takes 6 seconds for the train to pass, so the length of the train is:
(35 m/s) (6 s) = 210 m
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What far vertically will it travel before hiting the ground A. 40 m B. 30 m C. 60 m D. 50 m
Answer:
First, let's think in the vertical problem:
The acceleration will be the gravitational acceleration:
g = 9.8 m/s^2
a = -9.8 m/s^2
For the velocity, we integrate over time:
v(t) = (-9.8 m/s^2)*t + v0
Where v0 is the initial velocity, in this case v0 = 30m/s.
v(t) = (-9.8 m/s^2)*t + 30m/s
Now, for the position we integrate again over time, and get:
P(t) = (1/2)*(-9.8 m/s^2)*t^2 + 30m/s*t + p0
Where p0 is the initial position, as the ball is launched from the ground, we can use p0 = 0m
p(t) = (-4.9m/s^2)*t^2 + 30m/s*t
Now, the maximum vertical height is reached when:
v(t) = 0m/s = -9.8m/s^2*t + 30m/s
t = 30m/s/9.8m/s^2 = 3.06s
Now we can evaluate the vertical position in t = 3.06s
p(3.06s) = (-4.9m/s^2)*(3.06)^2 + 30m/s*3.06 = 62m
So, rounding down, the correct option is: C. 60 m
a 6 letter word a way of explaining an object or event using a set of facts
Explanation:
A theory is a way of explaining an object or event using a set of facts.
A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air has an index of refraction of 1.00. At what minimum angle will light in the glass tube be totally reflected at the glass/air interface?
Answer:
θ = 41.8º
Explanation:
This is an internal total reflection exercise, the equation that describes this process is
sin θ = n₂ / n₁
where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂
θ = sin⁻¹ n₂ / n₁
let's calculate
θ = sin⁻¹ (1.00 / 1.50)
θ = 41.8º
A square loop 24.0 cm on a side has a resistance of 6.10Ω. It is initially in a 0.665-T magnetic field, with its plane perpendicular to magnetic field B but is removed from the field in 40.0ms. Calculate the electric energy dissipated in this process.
Answer:
[tex]E=6.01\times 10^{-3}\ J[/tex]
Explanation:
It is given that,
Side of a square loop, l = 24 cm = 0.24 m
Resistance of loop, R = 6.1 ohms
Initial magnetic field is 0.665 T and final magnetic field is 0 as the field is removed in 40 ms
We need to find the electrical energy dissipated in this process.
Due to change in magnetic field, the loop will induce a voltage. The induced voltage is given by :
[tex]V=-\dfrac{dB}{dt}\\\\V=\dfrac{BA}{t}[/tex]
If I is induced current then,
[tex]V=IR[/tex]
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{BA}{tR}[/tex]
Power is given by voltage times current. So,
[tex]P=\dfrac{(BA)^2}{(t^2R)}[/tex]
Now, energy is given by the product of power and time. So,
[tex]E=\dfrac{(BA)^2}{(t^2R)}\times t\\\\E=\dfrac{(BA)^2}{(tR)}[/tex]
Now putting all the values in above formula. So,
[tex]E=\dfrac{(0.665\times (0.24)^2)^2}{(40\times 10^{-3}\times 6.1)}\\\\E=6.01\times 10^{-3}\ J[/tex]
So, the electrical energy of [tex]6.01\times 10^{-3}\ J[/tex] is dissipated in this process.
The electrical energy dissipated throughout this process will be "6.01 × 10⁻³ J".
Magnetic fieldAccording to the question,
Square loop's side, l = 24 cm or,
= 0.24 m
Loop's resistance, R = 6.1 ohms
Initial magnetic field = 0.665 T
Final magnetic field = 0
We know the relation,
→ V = - [tex]\frac{dB}{dt}[/tex]
= [tex]\frac{BA}{t}[/tex]
Also we know,
Current, V = IR
I = [tex]\frac{V}{R}[/tex]
= [tex]\frac{BA}{tR}[/tex]
Now, Energy, E = [tex]\frac{(BA)^2}{t^2R}[/tex] × t or,
= [tex]\frac{(BA)^2}{tR}[/tex]
By substituting the values,
= [tex]\frac{(0.665\times (0.24)^2)^2}{40\times 10^{-3}\times 6.1}[/tex]
= 6.01 × 10⁻³ J
Thus the response above is correct.
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A railroad car having a mass of 15 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass of 12 Mg is coasting at 0.75 m/s in the opposite direction. If the cars meet and couple together, determine the speed of both cars just after the coupling. Find the difference between the total kinetic energy before and after coupling has occurred. Explain qualitatively what happened to this energy.
Answer:
Explanation:
We shall apply law of conservation of momentum to know velocity after collision . Let it be v .
total momentum before collision = total momentum after collision
15 x 1.5 - 12 x .75 = ( 15 + 12 ) v
v = .5 m /s
kinetic energy before collision
1/2 x 15 x 1.5² + 1/2 x 12 x .75²
= 16.875 + 3.375
= 20.25 J
kinetic energy after collision
= 1/2 x ( 15 + 12 ) x .5²
= 3.375 J
Loss of energy = 16.875 J
This energy appear as heat and sound energy that is produced during collision .
calculate the energy dissipated by an electric iron which draws a current of 5A from a240v power supply for 25minutes
Explanation:
Given,
I = 5 A
V = 240 V
T = 25 mins = 1500 sec
Now,
Energy dissipated = IVT= 5×240×1500 = 1800000 J
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 7 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 17 m2 and 0.20 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Answer:
88 %
Explanation:
The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by
P = KAΔT/d
The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m
The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m
The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂
Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT
So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂
Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is
P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)
= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)
= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)
= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]
= 1/(0.003927/0.0288 + 1)
= 1/(0.1364 + 1)
= 1/1.1364
= 0.88.
The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %
The percentage of heat lost by window of the total heat is 88 %
A load of 500N is carried by 200N effort in a simple machine having load distance 3m Calculate effort distance.
Answer:
2.5 mExplanation:
Load ( L ) = 500 N
Effort ( E ) = 200 N
Load distance ( LD ) = 3 m
Effort distance ( ED ) = ?
Now, Let's find the Effort distance ( ED )
We know that,
Output work = Input work
i.e L × LD = E × ED
plug the values
[tex]500 \times 3 = 200 \times ED[/tex]
multiply the numbers
[tex]1500 = 200 \times ED[/tex]
Swipe the sides of the equation
[tex]200 \: ED \: = 500[/tex]
Divide both sides of the equation by 200
[tex] \frac{200 \: ED}{200} = \frac{500}{200} [/tex]
Calculate
[tex]ED\: = 2.5 \: m[/tex]
Hope this helps..
best regards!!
The following passage has not been edited. There is an error in each line. Write the
incorrect word and the correction in your answer sheet against the correct question
number. The first one has been done as an example. ( 1 x 4 = 4 )
Community service sensitize people to Error: sensitize ; Correction: sensitizes
other‟s needs and supports inclusive (a) Error: _______ ; Correction: ______
development to the underprivileged (b) Error: _______ ; Correction: ______
sections with society. Courses about social (c) Error: ______ ; Correction: _______
work prepares frontline workers to (d) Error: _______ ; Correction: ______
Answer:
(a) Error: Other's ; Correction: Others'
(b) Error: to ; Correction: for
(c) Error: with ; Correction: of
(d) Error: prepares ; Correction: prepare
Explanation:
a) The error is in the word "other's" as the position of apostrophe is wrong, so the correct word will be "others'", it shows plural nouns.
b) The error is in the word "to", so the correct word will be "for" as for is use to talk about a purpose.
c) The error is in the word "with" and the correct word will be "of" as of indicates relationships between other words including things that made of other things.
d) The error is in the word "preapres" and the correct word will be "prepare".
PLEASE HELP ME ASAP. IT'S VERY IMPORTANT
Answer:
1) a. 52.41 m/s
b. The skier will be going 15.35 m/s slower
2) 103.68 m
3) 35,127 J
4) a. 88.825 kJ
(b) 16.36 %
5) 3,071.12 J
Explanation:
1) a. The given height of the hill, h = 140.0 m
The mass of the skier at the top of the hill, m = 85.0 kg
The acceleration due to gravity, g = 9.81 m/s²
The initial potential energy, P.E of the skier = m×g×h = 85.0×140.0×9.81 = 116739 J
From the principle of conservation of energy, we have;
The potential energy, P.E. lost = The gain in kinetic energy, K.E.
m×g×h = 1/2×m×v²
116739 J = 1/2×85.0×v²
v² = 116739/(1/2*85.0)= 2746.8 m²/s²
v = √(2746.8 m²/s²) = 52.41 m/s
b. From 70 m up, we have;
The initial potential energy, P.E., of the skier is now = 85.0×70×9.81 = 58,369.5 J
The potential energy, P.E. lost = The gain in kinetic energy, K.E.
58,369.5 J = 1/2×85.0×v²
v² = 58,369.5/(1/2*85.0) = 1373.4 m²/s²
v = 37.06 m/s
The skier will be going 52.41 - 37.06 = 15.35 m/s slower
The skier will be going 15.35 m/s slower
2) From the principle of conservation of energy, the amount of work done (energy used) = The (potential) energy gained by the load
The amount of work done by the electric hoist = 356,000 J
The mass of the load = 350.0 kg
The height to which the load is raised = h
The potential energy gained by the load = m×g×h = 350.0×9.81×h
356,000 J = 350.0×9.81×h
h = 356,000/(350.0*9.81) = 103.68 m
The height to which the load is lifted= 103.68 m
3) The initial potential energy of the roller coaster cart = 600*35.0*9.81 = 206010 J
The final potential energy = 600*28.0*9.81= 164808 J
The velocity at point 3 = 4.5 m/s
The kinetic energy at point 3 = 1/2*600*4.5^2 = 6075 J
The total energy at point 3 = 164808 + 6075 = 170,883 J
The energy loss = The initial potential energy at point 1 - Total energy at point 3
The energy loss = 206010 - 170,883 = 35,127 J
The heat energy due to friction that must have been produced between points 1 and 3 = 35,127 J
4) a. The heat energy absorbed = mass × specific heat capacity for water, [tex]C_{water}[/tex] × Temperature change
The mass of the water = 2.5×10² g = 0.25 kg
[tex]C_{water}[/tex] = 4,180 J/(kg·°C)
Initial temperature = 10.0°C
Final temperature = 95°C
The temperature change = 95.0°C - 10.0°C = 85.0°C
The heat energy absorbed = 0.25*4,180* 85 = 88,825 J = 88.825 kJ
(b) The percentage efficiency = (Heat absorbed/(Heat supplied)) × 100
The heat supplied = 543 kJ
The efficiency = (88.825/543) × 100 = 16.36 %
5) The mass of the box = 115 kg
Force acting on the rope = 255 N
The angle of inclination of the force to the horizontal = 24.5°
The distance the box is displaced = 15.0 m to the right
The work done = Force applied × distance moved in the direction of the force
The work done = Force applied × distance moved in the direction of the force
Given that the load moves a distance 15.0 m to the right,we have;
The component of the force acting in the direction of the movement of the load (to the right) is 225 × cos(24.5°) = 204.74 N
The work done = 204.7*15 = 3071.12 J
The amount of work done = 3,071.12 J
PLEASE HELP ASAP. IT'S URGENT
Answer:
Q1 acceleration = 16m/s²
Q2 Net force = 9N North
Explanation:
Q1 Using the formula F=ma
Q2 R = F1 + F2
a bicycle has a momenum of 36kg•m/s and a velocity of 4m/s. what is the mass of the bicycle?
In the school cafeteria, a trouble-making child blows a 12.0 g spitball through a 25.0 cm straw. The force (F) in Newtons, of his breath as a function of the distance along the straw (x) in meters, can be modeled as a linearly decreasing function for the first half of the straw, followed by a constant force through the rest of the straw. The force decreases by half along the first half of the straw. Assume there is negligible friction and the straw is held horizontally.
(a) Sketch a plot of the force of his breath as a function of position along the straw, labeling the force at x = 0 as F0.
(b) If the spitball begins from rest and leaves the straw with a speed of 16 m/s, how much work is done on the spitball?
(c) What is the maximum force F0, that acts on the spitball?
Answer:
A.) Find the attached file
B.) 1.536 J
C.) 123N
Explanation:
Given that the force (F) in Newtons, of his breath as a function of the distance along the straw (x) in meters, can be modeled as a linearly decreasing function for the first half of the straw,
Half of the straw = 12.5 cm
Let Force F = dependent variable
And distance x = independent variable
A.) Find the attached file for the graph.
Sinnce the force decreases by half along the first half of the straw. Assume there is negligible friction and the straw is held horizontally
B.) Given that the mass M = 12g = 12/1000 = 0.012kg
And Velocity V = 16m/s
The workdone = the kinetic energy of the split ball
WD = 1/2mv^2
Substitutes m and V into the formula
WD = 1/2 × 0.012 × 16^2
WD = 1.536 Joule
C.) The maximum force acts on the spit ball will be at maximum kinetic energy.
Since work done = force × distance
Where distance = 12.5 cm
F × S = 1/2mv^2
Substitutes all the parameters
F × 0.0125 = 1.536
Make F the subject of formula
F = 1.536/0.0125
F = 122.88 N
The maximum force F0, that acts on the spitball is therefore equal to 123 N approximately.
We have that for the Question "(a) Sketch a plot of the force of his breath as a function of position along the straw, labeling the force at x = 0 as F0.
(b) If the spitball begins from rest and leaves the straw with a speed of 16 m/s, how much work is done on the spitball?
(c) What is the maximum force F0, that acts on the spitball?"
It can be said that
a) The sketch is attachedb) The workdone = [tex]1.536J[/tex]c) Maximum force = [tex]6.4N[/tex]From the question we are told
mass of spitball = 12g through a 25.0 cm straw
Generally the equation for workdone is mathematically given as
[tex]= \frac{1}{2}mv^2 - 0\\\\= \frac{1}{2} * 12*10^{-3} * 16^2\\\\= 1.536J[/tex]
Maximum force
[tex]w = \frac{1}{2}*\frac{x_0}{2}*\frac{F_0}{2} + x_0*\frac{F_0}{2}\\\\w = \frac{5}{8}x_0F_0[/tex]
Therefore,
[tex]F_0 = \frac{8w}{5x_0}\\\\F_0 = \frac{8*1.536}{5*25*10^[-2]}\\\\F_0 = 6.4N[/tex]
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A piano has a mass of 99 kg. What is the weight of the piano?
A. 1030 N
B. 842 N
C. 1129 N
D. 970 N
Help!!
Answer:
[tex]D.\ 970\ N[/tex]
Explanation:
Given:
Mass of Piano: 99kg
Required:
Calculate its weight
The weight of an object is calculate as thus;
[tex]Weight = Mass\ (m) * Acceleration\ due\ to\ gravity\ (g)[/tex]
[tex]m = 99kg[/tex] and [tex]g = 9.8m/s^2[/tex]
The formula becomes
[tex]Weight = 99kg * 9.8m/s^2[/tex]
[tex]Weight = 970.2\ kgm/s^2[/tex]
[tex]Weight = 970.2\ N[/tex]
[tex]Weight = 970\ N[/tex] Approximated
Hence, the weight of the piano is 970N
The graph shows a wave that oscillates with a frequency of 60 Hz. Based on the information given in the diagram, what is the speed of the wave?
Answer:
900 cm/s or 9 m/s.
Explanation:
Data obtained from the question include the following:
Length (L) = 30 cm
frequency (f) = 60 Hz
Velocity (v) =.?
Next, we shall determine the wavelength (λ).
This is illustrated below:
Since the wave have 4 node, the wavelength of the wave will be:
λ = 2L/4
Length (L) = 30 cm
wavelength (λ) =.?
λ = 2L/4
λ = 2×30/4
λ = 60/4
λ = 15 cm
Therefore, the wavelength (λ) is 15 cm
Now, we can obtain the speed of the wave as follow:
wavelength (λ) = 15 cm
frequency (f) = 60 Hz
Velocity (v) =.?
v = λf
v = 15 × 60
v = 900 cm/s
Thus, converting 900 cm/s to m/s
We have:
100 cm/s = 1 m/s
900 cm/s = 900/100 = 9 m/s
Therefore, the speed of the wave is 900 cm/s or 9 m/s.
Select the correct answer.
According to the Universal Law of Gravitation, every object attracts every other object in the universe. Why can’t you feel the force of attraction between you and Mars?
A.
There is no force of attraction between you and Mars.
B.
Your mass is too low.
C.
Mars is a larger planet than Earth.
D.
Mars is a long distance away.
Answer:
D. Mars is a long distance away
Iron man wears an awesome ironsuit.He is flying over high current carrying wire. Will he be affected?
Answer:
According to super hero logic , nothing will happen to him.
But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.
He may / may not be affected but his suit will be damaged for sure as it is made of metal.
HOPE THIS HLEP AND PLSSSSS MARK AS BRAINLIEST AND THNXX :)
A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of aluminum is 2698.4 kg / m3
Answer:
W =1562.53 N
Explanation:
It is given that,
Radius of the aluminium ball, r = 24 cm = 0.24 m
The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]
We need to find the thrust and the force. The mass of the liquid displaced is given by :
[tex]m=dV[/tex]
V is volume
Weight of the displaced liquid
W = mg
[tex]W=dVg[/tex]
So,
[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]
So, the thrust and the force is 1562.53 N.
Allocate birr 5000 among the three workers in the ratio 1/3 :1/6 and 5/12.
Answer:
1666.7 ETB (birr)
833.3 ETB (birr)
2083.3 ETB (birr)
Explanation:
The first worker
5000*1/3=1666.7
The second worker
5000*1/6=833.3
The third worker
5000*5/12=2083.3
Hope this helps :) ❤❤❤
please help me with this
Answer:
12 Ω
Explanation:
Data obtained from the question include:
Resistor 1 (R1) = 50 Ω.
Resistor 2 (R1) = 30 Ω.
Resistor 3 (R3) = 20 Ω.
Resistor 4 (R4) = R
Galvanometer reading = Zero deflection.
The Resistor 4 (R4) in the Wheatstone bridge can be obtained as follow:
Since the galvanometer gives zero deflection, it means the bridge is balanced. Therefore,
R1/R2 = R3/R4
50/30 = 20/R
Cross multiply
50 x R = 30 x 20
Divide both side by 50
R = (30 x 20)/50
R = 12 Ω
Therefore, the value of R in the wheatstone bridge is 12 Ω.
A train travels 120 km at a speed of 60 km/h, makes a stop for 0.5 h, and then travels the next 180 km at a speed of 90 km/h. What is the average speed of the train for this trip? 9th grade level pls FFFFFFFFFFFAAAAAAAAAAASSSSSSSSSSSSSTTTTTTTTTTTTT
Answer:
average speed = 66.67 km/h
Explanation:
In order to find the average speed of the train, you need to calculate the total distance traveled, divided by the time it took to cover that distance. So for the total distance:
Distance= 120 km + 180 km = 300 km
For the total time we need to add three different quantities, two of which we need to derived based on the information provided:
time for first part of the trip:
[tex]time_1=\frac{D_1}{v_1} =\frac{120}{60} \,h= 2\,h[/tex]
for the time of the stop:
[tex]time_2=0.5\,\,h[/tex]
for the last part of the trip:
[tex]time_3=\frac{180}{90} \,h= 2 \,\,h[/tex]
Which gives a total of 4.5 hours
Then, the average speed is: 300/4.5 km/h = 66.67 km/h
Let us treat a helicopter rotor blade as a long thin
rod, as shown in Fig. 8–49. (a) If each of the three rotor
helicopter blades is 3.75 m long and has a mass of 135 kg,
calculate the moment of inertia of the three rotor blades
about the axis of rotation. (b) How much torque must the
motor apply to bring the blades from rest up to a speed
of 6.0 rev/s in 8.0 s?
Rotor
Answer:
(a) 1900 kg m²
(b) 8950 Nm
Explanation:
(a) The moment of inertia of a rod about its end is I = ⅓mL².
For 3 rods of mass m = 135 kg and length L = 3.75 m, the total moment of inertia is:
I = 3 (⅓ (135 kg) (3.75 m)²)
I = 1900 kg m²
(b) Net torque = moment of inertia × angular acceleration
∑τ = Iα
First, find the angular acceleration.
ω₀ = 0 rad/s
ω = 6.0 rev/s (2π rad/rev) = 37.7 rad/s
t = 8.0 s
α = (37.7 rad/s − 0 rad/s) / 8.0 rad/s = 4.71 rad/s²
∑τ = Iα
∑τ = (1900 kg m²) (4.71 rad/s²)
∑τ = 8950 kg m² / s²
∑τ = 8950 Nm