Answer:
A. 1,000
Explanation:
Kilo is the prefix for thousands, thus, the multiplier that kilo represents is 1,000
A. 1,000
Answer:
B. 1,000
Explanation:
plato
A centripetal force of 190 N acts on a 1,550-kg satellite moving with a speed of 5,300 m/s in a circular orbit around a planet. What is the radius of its orbit?
Answer:
Radius, r is equal to 229.16×10^6m
Explanation:
Given the following parameters;
Centripetal force on the satellite, Fc = 190N.
Mass of the satellite, M = 1,550-kg.
Speed of the satellite, V = 5,300m/s.
The relationship between a satellite of mass (m) moving in a circular orbit of radius (r) with a speed (v) and a centripetal force (Fc) is given by the equation;
Fc = (MV²)/r
Since, we're solving radius, r; we make "r" the subject of formula;
Thus, r = (MV²)/Fc
Substituting into the above equation;
r = (1550 × [5300]²)/190
r = (1550 × 28090000)/190
r = 43539500000/190
r = 229155263.16
Radius, r is equal to 229.16 × 10^6m
A Chinook salmon can jump out of water with a speed of 6.50 m/s . How far horizontally can a Chinook salmon travel through the air if it leaves the water with an initial angle of =30.0° with respect to the horizontal? (Let the horizontal direction the fish travels be in the + direction, and let the upward vertical direction be the + direction. Neglect any effects due to air resistance.)
Answer:
3.73m
Explanation:
What we are asked to find is the range covered by the fish. This is given by the following equation (1). Range can simply be defined as the horizontal distance covered by a body whose motion freely under gravity is in two dimensions. The motion of the fish is in two dimensions, the vertical dimension and the horizontal dimension.
[tex]R=\frac{u^2sin2\theta}{g}.........(1)[/tex]
where u = 6.5m/s is the initial velocity, g is acceleration due to gravity which is taken as [tex]9.8m/s^2[/tex] and [tex]\theta=30.0^o[/tex].
Substituting these values into equation (1), we obtain the following;
[tex]R=\frac{6.5^2sin2(3)}{9.8}\\R=\frac{42.25sin60}{9.8}\\\\R=\frac{36.59}{9.8}\\R=3.73m[/tex]
The range should be 3.73m.
Important information:A Chinook salmon can jump out of water with a speed of 6.50 m/s . The initial angle of =30.0°
The range refers to the horizontal distance i.e. covered by a body whose motion freely under gravity should be in two dimensions. It can be the vertical dimension and the horizontal dimension.
calculation of the range:[tex]= 6.5sin^2\div 9.8\\\\= 36.59 \div 9.8[/tex]
= 3.73m
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the difference between how lenses and mirrors interact with light
Answer:
Light when strikes mirror, it bounces off where as it transmits through a lens. A mirror is glass with lustrous metal coated at its back. Light reflects off a mirror and image is formed behind it.
Explanation:
When an object is placed within the focal length of a convex lens, the image appears on the same side as the object itself. This situation is shown below. In this situation, which of the following statements is true?
Answer:
the image formed is virtual, this means that it is the prolongations of the rays that form the image. It is straight and magnified by the quantity
Explanation:
In a lens there is a relationship between the focal length, the distance to the object and to the image given by the constructor equation
1 / f = 1 / o + 1 / i
where is the focal length, or and i are the distance to the object and the image respectively.
In the case presented, the object is within the focal length, so the image formed is virtual, this means that it is the prolongations of the rays that form the image. It is straight and magnified by the quantity
m = - i / o
Answer:A
Explanation:Got it right :)
Suppose an automobile engine can produce 180 N*m of torque, and assume this car is suspended so that the wheels can turn freely. Each wheel acts like a 15.5 kg disk that has a 0.175 m radius. The tires act like 1.9-kg rings that have inside radii of 0.19 m and outside radii of 0.315 m. The tread of each tire acts like a 12-kg hoop of radius 0.335 m. The 14.5-kg axle acts like a solid cylinder that has a 1.95-cm radius. The 32.5-kg drive shaft acts like a solid cylinder that has a 2.9-cm radius.
(a) calculate the angular acceleration in radians per square second, produced by the motor if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car.
Answer:
Explanation:
Moment of inertia of each wheel = 1/2 m R²
m is mass and R is radius of wheel
= .5 x 15.5 x .175²
= .2373 kg m²
moment of inertia of tyre
1/2 m ( r₁² + r₂² )
= 1/2 x 1.9 x ( .315² + .19²)
= 1/2 x 1.9 x ( .099+ .036)
= .12825 kg m²
moment of inertia of tread
= 1/2 m r²
= .5 x 12 x .335²
= .67335 kg m²
moment of inertia of axle
= 1/2 m r ²
= .5 x 14.5 x .0195²
= .00275
moment of inertia of drive shaft
= 1/2 x 32.5 x .029²
= .0137 kg m ²
Total moment of inertia of one tyre
= 1.05535 kg m²
total moment of inertia of two rear wheels
= 2.1107 kg m²
95 % of torque
= .95 x 180
= 171 Nm
angular acceleration
= torque / moment of inertia
= 171 / 2.1107
= 81.01 radian /s²
A mass M subway train initially traveling at speed v slows to a stop in a station and then stays there long enough for its brakes to cool. The station's volume is V and the air in the station has a density rho_air and specific heat C_air.
Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, what is the expression for the change in the air temperature in the station? Make sure the quantities you enter match the ones given in the problem exactly.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The expression for the change in the air temperature is [tex]\Delta T = \frac{Mv^2}{2 \rho_{air} c_{air}* V}[/tex]
Explanation:
From the question we are told that
The mass of the train is M
The speed of the train is v
The volume of the station is V
The density of air in the station is [tex]\rho_{air}[/tex]
The specific heat of air is [tex]c_{air}[/tex]
The workdone by the break can be mathematically represented as
[tex]W =\Delta KE = \frac{1}{2} Mv^2[/tex]
Now this is equivalent to the heat transferred to air in the station
Now the heat capacity of the air in the station is mathematically represented as
[tex]Q = \rho_{air} * m_{air} * c_{air} (\Delta T)[/tex]
Now Since this is equivalent to the workdone by the breaks we have that
[tex]\frac{1}{2} Mv^2 = m_{air} * c_{air} (\Delta T)[/tex]
=> [tex]\Delta T = \frac{Mv^2}{2 \rho_{air} c_{air}* V}[/tex]
Whenever using a microscope slide, you should always:
A.
check for cracks or chips in the glass.
B.
make sure it is clean.
C.
handle it with care.
D.
all of these
Answer:
D.
all of these.......
Although electromagnetic waves can always be represented as either photons or waves, in the radio part of the spectrum we typically do not discuss photons (like we do in the visible) because they are at such a low energy. Nevertheless, they exist. Consider such a photon in a radio wave from an AM station has a 1545 kHz broadcast frequency.
Required:
What is the energy, in joules, of the photon?
Answer:
E = 1.02 x 10⁻²⁷ J
Explanation:
The energy of a photon is given by the Plank's formula. The formula is given as:
E = hυ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
υ = Frequency of the Wave = 1545 KHz = 1545000 Hz
Therefore, we can find the energy of photon in the given radio wave from an AM station, by substituting the known values in Plank's formula.
E = (6.626 x 10⁻³⁴ J.s)(1545000 Hz)
E = 1.02 x 10⁻²⁷ J
The energy of a photon, in a radio wave from an AM station has a 1545 kHz broadcast frequency, is found to be E = 1.02 x 10⁻²⁷ J
The energy , in joules, of the photon should be considered as the E = 1.02 x 10⁻²⁷ J.
Plank formula:The energy of a photon should be like
E = hυ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
υ = Frequency of the Wave = 1545 KHz = 1545000 Hz
Now
E = (6.626 x 10⁻³⁴ J.s)(1545000 Hz)
E = 1.02 x 10⁻²⁷ J
hence, The energy , in joules, of the photon should be considered as the E = 1.02 x 10⁻²⁷ J.
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A parallel plate capacitor is attached to a battery which stores 3 C of charge. A dielectricmaterial is inserted to fill the gap. There is now 9 C of charge stored.1. What is the dielectric constant of the material?2. As a fraction of the original how much energy is stored in the capacitor after thedielectric is inserted?3. If we pull the dielectric half way out how much charge is stored on the capacitor?Hint:we could imagine our capacitor now as 2 in parallel, each with half the area and onewith the dielectric.
Answer:
A) 3
B) fraction is 2/1 = 2
C) 3 C
Explanation:
Initial capacitance with air U is 3 C
Final charge with dielectric Ud material is 9 C
Dielectric constant = capacitance with dielectric/capacitance with air
= 9/3 = 3
Since it is connected to a battery, the potential difference at the plate will be constant.
P.d = V
Also energy stored in a capacitor is given as 0.5CV^2
For capacitance with air, energy is 0.5 x 3 x V^2 = 9V^2
For capacitance with dielectric, energy is 0.5 x 9 x V^2 = 18V^2
Fraction of energy stored in capacitance with dielectric to that with air is 18V^2/9V^2 = 2
From C = eA/d
Where C is the capacitance,
e is the dielectric constant
A is the area of the dielectric
d is the distance between plates of the capacitor.
For initial, assuming the distance to be of unit distance, area will be given as
9 = (3 x A)/1
9 = 3A
A = 2 m^2. If we pull dielectric half way out, area becomes
C = (3 x 1)/1
C = 3 C
What is the answer to the question?
Answer:
Explanation:
The y component is measured by the horizontal component and the vertical component. Together they determine the magnitude of the vector. In this case, the y or vertical component is found by using the sine function.
Formula
Sin(angle) = vector resultant / y component component.
Givens
angle = 42 degrees.
vector = 419 degrees
Solution
sin(42) = y / 419 Multiply both sides by 419
419 * sin(42) = 419 * y / 419
y = 419 * 0.6691
y = 280.37
Note
The vector is pointing downward so technically the vertical component should be negative. I'm not sure what to tell you to answer. I would try - 280.37, but if the computer marks you wrong, try 280.37 (no minus sign).
When an electric current flows through a wire a
magnetic field is created. To increase the power
of the magnetic field you would
A. decrease the number of coils in the wire
Bkeep the same number of coils in the wire
Cincrease the number of coils in the wire
Dremove the coils from the wire
Answer:
Option C. is correct
Explanation:
The magnetic field is the area around a magnet in which there is magnetic force. When an electric current flows through a wire a magnetic field is created. A single wire does not produce a strong magnetic field. So, to increase the power of the magnetic field, increase the number of coils in the wire.
The surface tension of water was determuned in a laboratory by using the drop weight method. 100 drops were released from a burette the inner diameter of whose opening is 1.8mm. The mass of the droplets was 3.78g dertermine the surface tension of the water and comparing it with the tabulatef value
Answer:
The surface tension of the water is 6.278×10⁻² N/m
error = 13.65%
Explanation:
The surface tension of water is given by
[tex]$ \gamma = \frac{F}{L} $[/tex]
Where F is the force acting on water and L is the length over which is force is acted.
We are given the mass of 100 droplets of water
M = 3.78 g
n = 100
The mass of 1 droplet is given by
[tex]m = \frac{M}{n} \\\\m = \frac{3.78}{100}\\\\m = 0.0378 \: g \\\\m = 3.780\times10^{-5} \: kg[/tex]
The force acting on a single droplet of water is given by
[tex]F = m \cdot g[/tex]
Where m is the mass of water droplet and g is the acceleration due to gravity
[tex]F = 3.780\times10^{-5} \cdot 9.81[/tex]
[tex]F = 3.708\times10^{-4} \: N[/tex]
The circumferential length of the droplet is given by
[tex]L = \pi \cdot d[/tex]
Where d is the diameter
[tex]L = \pi \cdot 1.88\times10^{-3}\\\\L = 5.906 \times10^{-3} \: m[/tex]
Now we can find out the required surface tension of the water
[tex]\gamma = \frac{3.708\times10^{-4} }{5.906 \times10^{-3}} \\\\\gamma = 0.06278\: N/m\\\\\gamma = 6.278 \times10^{-2} \: N/m\\\\[/tex]
Therefore, the surface tension of the water is 6.278×10⁻² N/m
The tabulated value of the surface tension of water at 20 °C is given by
[tex]$ \gamma_t = 0.0727 \: N/m $[/tex]
The percentage error between tabulated and calculated surface tension is given by
[tex]$ error = \frac{\gamma_t - \gamma }{\gamma_t} $[/tex]
[tex]$ error = \frac{ 0.0727 - 0.06278}{0.0727} \times 100\% $[/tex]
[tex]$ error = 13.65 \%[/tex]
a student stands several meters in front of a smooth reflecting wall, holding a board on which a wire is fixed at each end. the wire, vibrating in its third harmonic, is 75.0cm long, has a mass of 2.25g, and is under a tension of 400 N. a second student, moving towards the wall, hears 8.30 beats per second. what is the speed of the student approaching the wall? (solve without calculus)
Answer:
Explanation:
From the question we are told that
The length of the wire is [tex]L = 75.0cm = \frac{75}{100} = 0.75 \ m[/tex]
The mass of the wire is [tex]m = 2.25 \ g = \frac{2.25}{1000} = 0.00225 \ kg[/tex]
The tension is [tex]T = 400 \ N[/tex]
The frequency of the beat heard by the second student is
[tex]f_b = 8.30\ beat/second[/tex]
The speed of the wave generated by the vibration of the wire is mathematically represented as
[tex]v = \sqrt{\frac{TL}{m}}[/tex]
substituting values
[tex]v = \sqrt{\frac{400 *0.75}{0.00225}}[/tex]
[tex]v = 365.15 m/s[/tex]
The wire is vibrating in its third harmonics so the wavelength is
[tex]\lambda = \frac{2L}{3}[/tex]
substituting values
[tex]\lambda = \frac{2*0.75}{3}[/tex]
[tex]\lambda = 0.5 \ m[/tex]
The frequency of this vibration is mathematically represented as
[tex]f = \frac{v}{\lambda }[/tex]
substituting values
[tex]f = \frac{365.15}{0.5 }[/tex]
[tex]f = 730.3 Hz[/tex]
The speed of the second student (Observer) is mathematically represented as
[tex]v_o = [\frac{f_b}{2f} ] * v[/tex]
substituting values
[tex]v_o = [\frac{8.30}{2* 730.3} ] * 365.15[/tex]
[tex]v_o = 2.08 \ m/s[/tex]
Give reasons for the following:
a. Knives and swords are supposed to have extremely thin blades.
b. Lorries and trucks carrying heavy loads have 8 tires instead of four, and the
tires are broader.
c. Camels can walk easily in desserts but humans cannot.
d. In a flight, the human ear pain during take-off and landing.
e. A suction cup does not stick on a rough surface.
Answer:
a. lower surface area, less resistence
b. more surface area, the load is split so no single tire overstrained
c. more surface area, more resistance against the sand. human steps sink down in the sand.
d. rapid change in air pressure on eardrums lead to somewhat-painful tension
e. air would always find its way in so no pressure difference can be achieved
(would indeed appreciate the brainliest if you appreciate the work)
A child in an inner tube is bobbing up and down in the ocean and notices that after a wave crest passes, four more crests pass in a time of 38.4 s and the distance between the crests is 32 m. If possible, determine the following properties for the wave. (If not possible, enter IMPOSSIBLE.)(a) Period (s)(b) Frequency (Hz)(c) Wavelength (m)(d) Speed ( m/s)(e) Amplitude (m)
Answer:
Explanation:
Given:
Four more crests pass in a time of 38.4 s and the distance between the crests is 32 m.
We have to determine five terms.
Lets start with one-one basis.
a.
Period = Time taken by a wave to pass though.
⇒ [tex]P = \frac{Total\ time}{No.\ of\ waves}[/tex]
⇒ [tex]P = \frac{38.4}{4}[/tex]
⇒ [tex]P=9.6 s[/tex]
b.
Frequency = Reciprocal of time period in Hertz.
⇒ [tex]f=\frac{1}{T}[/tex]
⇒[tex]f=\frac{1}{9.6}[/tex]
⇒ [tex]f=0.104 Hertz[/tex]
c)
Wavelength = Distance between two consecutive trough and crest.
⇒ [tex]\lambda = 32 m[/tex]
d.
Speed (v) = Product of frequency and wavelength.
⇒ [tex]v=f\times \lambda[/tex]
⇒ [tex]v=0.104\times 32[/tex]
⇒[tex]v = 3.33 ms^-1[/tex]
e)
Amplitude = The maximum displacement or half the distance from crest to trough.
⇒ Here it can't be determined.
⇒ Impossible.
Josh has a helium-filled balloon. He wants to calculate the speed of his balloon as it rises to the ceiling. What two measurements should he take to calculate the average speed of his balloon?
Answer:Distance and time
Explanation:
We know that average speed is the ratio of distance traveled by the particle over the time taken
So average velocity of balloon is
[tex]v_{avg}=\frac{\text{Distance}}{\text{Time taken}}[/tex]
So, Josh should note the distance traveled by balloon and time taken in doing so.
The two measurements required by Josh to calculate the average speed of the helium-filled balloon are distance and time.
The average speed of an object signifies and explains the total distance an object travels within a specified time. The average speed is a scalar quantity since it doesn't have a definite direction.
The average speed can be estimated if the measurement of the distance and time is being known because of its formula:
[tex]\mathsf{v_{avg} = \dfrac{distance}{time}}[/tex]
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Correct Number Formats Numbers may be entered in several formats - including scientific notation and numerical expressions.
WebAssign uses standard scientific or "e" notation for "times 10 raised to the power." For example, 1e3 is the scientific notation for 1000.
O You cannot have a space in a number.
O You cannot substitute the letter O for zero or the letter l for 1.
O You cannot include the units in the number unless specifically asked for.
O You can include the sign + or - of the number.
1. Which of the entries below will be interpreted as a valid numeric answer in WebAssign?
(Select all that apply.)
a. 1.56e-9
b. 3.25E4
c. 2.54 m
d. 1.56 e-9
e. 1.9435
f. -4.99
g. 1.23 inches
Answer:
a. Valid
b. Invalid
c. Invalid
d. Invalid
e. Valid
f. Valid
g. Invalid
Explanation:
a.
Since, it follows all the four rules,
Therefore, this number is Valid
b.
It uses upper case E, instead of lower case e.
Therefore, it is Invalid
c.
It violates third rule for including unit.
Therefore, it is Invalid
d.
It violates the first rule. As, there is a space between 6 and e. While in the example given in question, there is no gap.
Therefore, it is Invalid
e.
Since, it follows all the four rules,
Therefore, this number is Valid
f.
Since, it follows all the four rules,
Therefore, this number is Valid
g.
It violates third rule for including unit.
Therefore, it is Invalid
A manufacturing firm has hired your company, Acoustical Consulting, to help with a problem. Their employees are complaining about the annoying hum from a piece of machinery. Using a frequency meter, you quickly determine that the machine emits a rather loud sound at 1100 Hz. After investigating, you tell the owner that you cannot solve the problem entirely, but you can at least improve the situation by eliminating reflections of this sound from the walls. You propose to do this by installing mesh screens in front of the walls. A portion of the sound will reflect from the mesh; the rest will pass through the mesh and reflect from the wall.How far should the mesh be placed in front of the wall for this scheme to work?
Answer: 7.7cm
Explanation:
If i want to set up destructive interference, meaning that the reflection will be half (½) of the wavelength out of phase, or the distance is ¼ of a wavelength.
v.s = Speed of sound = 340 m/s ( some books may use 343 m/s, this change varies due to change in temperature and humidity.)
f = 1100 hz
λ = v.s / f
λ = 340 m/s / 1400 Hz
λ = .3091 m
Recall that the distance is ¼ of the wavelength,
.3091/4 = 0.0773 m
Distance used would be = 7.7 cm
A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 10.8 m/s . A 1.0-kg stone is thrown from the basket with an initial velocity of 14.3 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 18.8 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 10.8 m/s.
(a) How high was the balloon when the rock was thrown out?
(b) How high is the balloon when the rock hits the ground?
(c) At the instant the rock hits the ground, how far is it from the basket?
Answer:
a) -1529m
b)-1326m
c)268.84 meters.
Explanation:
a) Since the stone is thrown horizontally, its initial vertical velocity is the same as the basket. Let’s use the following equation to determine the vertical distance it moves in 18.8 seconds.
According to kinematic equation the displacement is given by
[tex]h_1 = v_{0y} y - \frac{1}{2}gt^2[/tex]
[tex]h_1[/tex]=10.8 * 18.8 - ½ * 9.8 * 18.8² => -1529m
The negative sign is due to the direction.
b)while the stone was travelling [tex]h_1[/tex] for 18.8s the balloon was also travelling the displacement [tex]h_b[/tex] with [tex]v_o_y[/tex]. so [tex]h_b[/tex] is given by
[tex]h_b=v_o_y t= - 10.8\times18.8[/tex]=> -203.04m
The height above the eart is given by,
[tex]h_2=h_1-h_b = -1529+203.04 => -1326[/tex]m
c)At the instant the rock hits the ground, how far is it from the basket?
This is the product of its initial horizontal velocity and the time.
d = 14.3 * 18.8 = 268.84 meters.
What is the speed of a wave that has a frequency of of 3.7 * 10 ^ 3 Hz and a wavelength of 1.2*10^ -2 m?
Answer:
44.4m/s
Explanation:
v = (3.7*10 ^3) ( 1.2 * 10 ^ -2)
= 44.4 m/s
Answer:
44.4 m/s
Explanation:
By using the wave speed formula, v = fλ ( speed= frequency × wavelength)
v = 3.7×10³ × 1.2×10^-2
= 44.4 m/s
Hope that's the answer you're looking for:)
Ball A is negatively charged and ball B is electrically neutral. What happens to the charges of both balls when they come into contact with each other if:
a) they are made of an electrical conductor
b) they are made of insulator
Explanation:
They will repel, thus made of electrical conductor.
An object is moving at a velocity of 30 m/s. It accelerates to a velocity of 55 m/s over a time of 12.5 s. What acceleration did it experience? SHOW WORK. ONLY SERIOUS RESPONSES. DUMB COMMENTS WILL BE DELETED. THINGS THAT ARE N/A WILL BE DELETED.
Answer:
Acceleration, [tex]a=2\ m/s^2[/tex]
Explanation:
We have,
Initial speed of an object is 30 m/s
It accelerates to a velocity of 55 m/s over a time of 12.5 s
It is required to find the acceleration experienced by it. The rate of change of velocity is called acceleration of an object. It is given by :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{55-30}{12.5}\\\\a=2\ m/s^2[/tex]
So, the acceleration of the object is [tex]2\ m/s^2[/tex].
If R= 200 , C= 15 MF , L=230 MH ,f= 60 HZ find XL :
XL = 2π•f•L = 86.7 ohms
If Newton's 1st Law says, it should never stop, then why has every ball you have ever kicked, thrown or rolled stopped moving?
Answer:
The ball stops due to friction when it rolls on something such as grass.
Explanation:
Newton's first law is that an object will move at a constant velocity or stay at rest, unless it is acted upon by an external force.
When a ball on Earth is kicked or rolled, it is subject to external forces that oppose it's motion (like the force due to gravity and air resistance, or friction when rolling on the ground) - so the ball will not keep moving and will eventually stop.
A sled plus passenger with total mass m = 53.1 kg is pulled a distance d = 25.3 m across a horizontal, snow-packed surface for which the coefficient of kinetic friction with the sled is μk = 0.155. The pulling force is constant and makes an angle of φ = 28.3 degrees above horizontal. The sled moves at constant velocity.
Required:
a. Write an expression for the work done by the pulling force in terms of m, g (acceleration due to gravity), φ, μk, and d.
b. What is the work done by the pulling force, in joules?
c. Write an expression for the work done on the sled by friction in terms of m, g (acceleration due to gravity), φ, μk, and d.
d. What is the work done on the sled by friction, in joules?
Answer:
Explanation:
Force of friction
F = μ mg
μ is coefficient of friction , m is mass and g is acceleration due to gravity .
If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional force
The vertical component of applied force will reduce the normal force or reaction force from the ground
Reaction force R = mg - f sin28.3
frictional force = μ R where μ is coefficient of friction
frictional force = μ x (mg - f sin28.3 )
This force should be equal to horizontal component of f
μ x (mg - f sin28.3 ) = f cos 28.3
μ x mg = f μsin28.3 + f cos 28.3
f = μ x mg / (μsin28.3 + cos 28.3 )
a )
work done by pulling force = force x displacement
f cos28.3 x d
μ x mg d cos28.3 / (μsin28.3 + cos 28.3 )
b ) Putting the given values
= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )
= 1796.76 / (.073 + .88 )
= 1885.37 J
c )
Work done by frictional force
= frictional force x displacement
= - μ x (mg - f sin28.3 ) x d
= - μ x mgd + f μsin28.3 x d
= - μ x mgd + μsin28.3 x d x μ x mg / (μsin28.3 + cos 28.3 )
d )
Putting the values in the equation above
- .155 x 53.1 x 9.8 x 25.3 +
.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)
= -2040.67 + 149.92 / .95347
= -2040.67 + 157.23
= -1883.44 J .
Two blocks with mass M and 3M on a horizontal frictionless surface are pushed together and compress a spring of negligible mass that is squeezed between them. The spring remains in place because the compression leads to a sufficient amount of friction with the sides of the blocks. It is important to understand that the spring is NOT attached to the blocks, in other words, that it will fall down as soon as the distance between the blocks exceeds the natural length of the spring. After the blocks are pushed together, a horizontal rope then secures the blocks in place.Later the rope is cut with scissors and the heavier block is launched with a speed of 2 m/s in the positive x-direction.
Define very precisely the system, the interaction(s) and the interaction time(s) of interest, and justify very precisely the principle(s) you apply to calculate the launching speed of the lighter block.
Answer:
launching speed of the lighter block = -6 m/s
Explanation:
We are given;
Mass of light block; M
Mass of heavy block; 3M
Speed of launched block: v = 2m/s
We are told that the two blocks are sitting on the horizontal frictionless surface. Thus, we can say that no external force is being applied on the system and so, the momentum of the whole system is conserved accordingly to that condition.
We are also told that when the rope is cut with scissors, that the heavier block attains the speed of 2 m/s in the positive x-direction which is horizontal direction.
We know that formula for momentum is; M = mass x velocity.
Thus, the momentum of the heavier block is calculated as;
M_1 = 3M × 2
M_1 = 6M kg.m/s
Since no external force is applied on the object, the initial momentum will be zero.
Hence, to conserve the system, the momentum of the lighter block will be equal and opposite to the momentum of heavier block.
So, momentum of lighter block is;
M_2 = -6M kg.m/s
Since mass of lighter block is M and formula for momentum = mass x velocity.
Thus;
-6M = Mv
Where v is speed of lighter block.
So, v = -6M/M
v = -6 m/s
how does the force of gravity affects an objects acceleration
Answer:
When objects fall to the ground, gravity causes them to accelerate.
Explanation:
Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.
A parachutist is falling with terminal velocity. Which of the following statement is not correct?
a) Gravitational potential energy is converted into kinetic energy of the air
b) G.P.E is converted into K.E of the parachutist
c) G.P.E is converted into thermal energy of the air
d) G.P.E is converted into thermal energy of the parachutist
Answer: D i think
Explanation:
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see figure below). One such pendulum is constructed with a string of length
L =10.7 cm and bob of mass 0.344 kg. The string makes an angle = 5.58° with the vertical.
(a) What is the radial acceleration of the bob?
magnitude
(b) What are the horizontal and vertical components of the tension force exerted by the string on the bob? (Assume radially inward to be the positive x axis and vertically upward to be the
positive y axis. Express your answer in vector form.)
T= N
Answer:
a
The radial acceleration is [tex]a_c = 0.9574 m/s^2[/tex]
b
The horizontal Tension is [tex]T_x = 0.3294 i \ N[/tex]
The vertical Tension is [tex]T_y =3.3712 j \ N[/tex]
Explanation:
The diagram illustrating this is shown on the first uploaded
From the question we are told that
The length of the string is [tex]L = 10.7 \ cm = 0.107 \ m[/tex]
The mass of the bob is [tex]m = 0.344 \ kg[/tex]
The angle made by the string is [tex]\theta = 5.58^o[/tex]
The centripetal force acting on the bob is mathematically represented as
[tex]F = \frac{mv^2}{r}[/tex]
Now From the diagram we see that this force is equivalent to
[tex]F = Tsin \theta[/tex] where T is the tension on the rope and v is the linear velocity
So
[tex]Tsin \theta = \frac{mv^2}{r}[/tex]
Now the downward normal force acting on the bob is mathematically represented as
[tex]Tcos \theta = mg[/tex]
So
[tex]\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}[/tex]
=> [tex]tan \theta = \frac{v^2}{rg}[/tex]
=> [tex]g tan \theta = \frac{v^2}{r}[/tex]
The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as
[tex]a_c = \frac{v^2}{r}[/tex]
=> [tex]a_c = gtan \theta[/tex]
substituting values
[tex]a_c = 9.8 * tan (5.58)[/tex]
[tex]a_c = 0.9574 m/s^2[/tex]
The horizontal component is mathematically represented as
[tex]T_x = Tsin \theta = ma_c[/tex]
substituting value
[tex]T_x = 0.344 * 0.9574[/tex]
[tex]T_x = 0.3294 \ N[/tex]
The vertical component of tension is
[tex]T_y = T \ cos \theta = mg[/tex]
substituting value
[tex]T_ y = 0.344 * 9.8[/tex]
[tex]T_ y = 3.2712 \ N[/tex]
The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is
[tex]T = T_x i + T_y j[/tex]
substituting value
[tex]T = [(0.3294) i + (3.3712)j ] \ N[/tex]
The radical acceleration of the bob is 0.9575 m/s². The horizontal and vertical components of the tension force exerted by the string on the bob are 0.329 N and 3.37 N respectively.
Taking the vertical component of the tension where the weight mass is balanced, then:
T sin θ = mg
[tex]\mathbf{T = \dfrac{mg}{sin \theta}}[/tex]
However, the centripetal force of the system is given by the horizontal component of the tension which can be expressed as:
T cos θ = m[tex]\mathbf{a_r}[/tex]
Making [tex]\mathbf{a_r}[/tex] the subject, we have:
[tex]\mathbf{a_r = \dfrac{Tcos \theta }{m}}[/tex]
replacing the value of tension (T), we have:
[tex]\mathbf{a_r = \dfrac{ \dfrac{mg}{sin \theta}cos \theta }{m}}[/tex]
[tex]\mathbf{a_r=g tan \theta}[/tex]
where;
angle θ = 5.58°[tex]\mathbf{a_r=9.8 m/s^2 \times tan 5.58}[/tex]
[tex]\mathbf{a_r=9.8 m/s^2 \times 0.0977}[/tex]
[tex]\mathbf{a_r=0.9575 \ m/s^2}[/tex]
Thus, the radical acceleration of the bob is 0.9577 m/s²
On the positive x-axis, the horizontal component of the tension force is:
[tex]\mathbf{T_x =Tcos \theta}[/tex]
[tex]\mathbf{T_x =ma_r}[/tex]
[tex]\mathbf{T_x =0.344 \ kg \times 0.9575 \ m/s^2}[/tex]
[tex]\mathbf{T_x =0.329 \ N}[/tex]
On the positive y-axis, the vertical component of the tension force is:
[tex]\mathbf{T_y =Tsin \theta}[/tex]
[tex]\mathbf{T_y =mg}[/tex]
[tex]\mathbf{T_y=0.344 \ kg \times 9.8 \ m/s^2}[/tex]
[tex]\mathbf{T_x =3.37 \ N}[/tex]
Learn more about simple pendulum here:
https://brainly.com/question/12886217?referrer=searchResults
Two radio antennas A and B radiate in phase. Antenna B is a distance of 140 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
a) What is the longest wavelength for which there will be destructive interference at point ?
b) What is the longest wavelength for which there will be constructive interference at point ?
Explanation:
a) 120m
b) since Q is the first order constructive interference, the distance between mid point of antennas and Q is 0.5 wavelengths (appear in A-Level question). so the wavelength should be (2)(30) =60m
60m