Answer:
the frequency range of sound waves is about 70 Hz - 160 Hz
Hence, Option a) About 70 Hz -160 Hz is the correct answer
Explanation:
Given the data in the question;
L₁ = 10 micrometers = 0.00001 m
so
T₁ = 2π√(L/g)
g = 9.8 m/s
so we substitute
T₁ = 2π√(0.00001 /9.8) = 0.006347
⇒ f₁ = 1 / T₁ = 1 / 0.006347 = 157.55 ≈ 160 Hz
L₂ = 50 micrometers = 0.00005 m
T₂ = 2π√(L/g)
g = 9.8 m/s
so we substitute
T₂ = 2π√(0.00005 /9.8) = 0.0142
f₂ = 1 / T₂ = 1 / 0.0142 = 70.422 ≈ 70 Hz
Therefore, the frequency range of sound waves is about 70 Hz - 160 Hz
Hence, Option a) About 70 Hz -160 Hz is the correct answer
Explain how grality and electric charge are different
The plates of a capacitor are charged using a battery, and they produce an electric field across the separation distance d between them. The two plates are now to be pushed together to a separation of d/2. The pushing together can be done either with the battery connected or with it disconnected. In which case, with the battery connected or disconnected. Is the electric field magnitude greater with the battery connected with the battery disconnected?
Answer:
Explanation:
The equation of the capacitance of the capacitor can be represented as:
[tex]C = \dfrac{\varepsilon_oA}{d}[/tex]
Also, the electric field between the plates can be expressed as:
[tex]E = \dfrac{\sigma}{\varepsilon_o}[/tex]
[tex]= \dfrac{Q}{A \varepsilon _o} \ \ (surface \ charge\ density \ \sigma =\dfrac{Q}{\varepsilon_o})[/tex]
However;
when pushed to a distance d/2, the new capacitance of the capacitor is:
[tex]C = \dfrac{\varepsilon _oA}{(d/2))}[/tex]
[tex]=\dfrac{2 \varepsilon_oA}{d}[/tex]
[tex]=2C \\ \\ Q' = CV \\ \\ = 2CV \\ \\ =2Q[/tex]
SImilarly, the new electric field between the plates is:
[tex]E' = \dfrac{\sigma'}{\varepsilon_o} \\ \\ = \dfrac{Q'}{A \varepsilon_o} \\ \\ = \dfrac{2Q}{A \varepsilon_o} \\ \\ =2E[/tex]
For Battery disconnected:
The electric field between the plates doesn't rely upon the distance between the plates yet relies upon the magnitude of the charge. At the point when the battery is detached, the charge on the capacitor stays as before, so does the electric field.
Therefore, the magnitude of the electric field when the battery is associated is twice however much the magnitude of the electric field when the battery is separated and disconnected.
Hence, the ratio is :
[tex]\dfrac{E_{connected}}{E_{disconnected}} =\dfrac{2E}{E} \\ \\ \dfrac{E_{connected}}{E_{disconnected}} = 2[/tex]
Hence, the ratio is = 2
Convert the following angles in degrees to radians:
(a) 300°
(b) 18°
(c) 105°
Answer:
a) 5.23599
b) 0.314159
c) 1.8326
Explanation:
Answer:
a. 5.236 rad
b. 0.314 rad
c. 1.833 rad
Explanation:
300° × π/180 = 5.236rad
18° × π/180 = 0.3142rad
105° × π/180 = 1.833rad
An object A with mass 200 kg and an another object B with mass 1000 kg are moving with same speed. The ratio of kinetic energy of object A to B is
Answer:
Ratio of kinetic energy of object A to B = 1:5
Explanation:
Given:
Mass of object A = 200 kg
Mass of object B = 1,000 kg
Find:
Ratio of kinetic energy of object A to B
Computation:
Kinetic energy = (1/2)(m)(v²)
Kinetic energy of object A = (1/2)(200)(v²)
Kinetic energy of object A = (100)(v²)
Kinetic energy of object B = (1/2)(1,000)(v²)
Kinetic energy of object B = (500)(v²)
Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B
Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)
Ratio of kinetic energy of object A to B = 100 / 500
Ratio of kinetic energy of object A to B = 1/5
Ratio of kinetic energy of object A to B = 1:5
2. What does SCA stand for? (1 Point) Student Conservation Association h Securities Class Action Sudden Cardiac Arrest Solar Cell Array:',)
Answer:
sudden cardiac arrest
Explanation:
question is included in the picture!!! PUT REAL ANSWERS OR I WILL REPORT YOU
Answer:
Explanation:
this is like rubbing a balloon on your head to make your hair stand up. Do that to the can. The balloon is filled , ofc, and then just rub the balloon on the can. This will charge the can with static electricity. :P
Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the same. Compared to the pucks in the video, this pair of pucks will rotate View Available Hint(s) Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the same. Compared to the pucks in the video, this pair of pucks will rotate four times as fast. at the same rate. one-fourth as fast. twice as fast. one-half as fast.
Answer:
w = w₀ / 2 the angular velocity is half the initial value.
Explanation:
We can analyze this exercise as if we added another disk to obtain a disk with twice the mass, for which if the system is two disks, the angular tidal wave is conserved
initial instant.
L₀ = I₀ w₀
final moment
L_f = I w
the moment is preserved
L₀ = L_f
I₀ w₀ = I w
the moment of inertia of a disk is
I = ½ m R²
we substitute
½ m R² w₀ = ½ (2m) R² w
w = w₀ / 2
for the case of a disk with twice the mass, the angular velocity is half the initial value.
The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.3 m/s . Initially, the car is also traveling at a speed 19.3m/s and its front bumper is a distance 25.0m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.560m/s^2 , then pulls back into the truck's lane when the rear of the car is a distance 26.5m ahead of the front of the truck. The car is of length 4.50m and the truck is of length 20.7m .
Part A) How much time is required for the car to pass the truck?
Part B ) What distance does the car travel during this time?
Part C) What is the final speed of the car?
Answer:
A) t = 10.56 s, B) x = 235 m, C) v = 25.2 m / s
Explanation:
A) We can solve this problem using kinematics expressions.
The distance traveled by the truck is
x_c = v_c t
Distance traveled by the car.
The car must travel the distance that separates them from the truck x₀=25.0. Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2 4.5 = 9 m, therefore the total length traveled by the car is
x_t = x₁ + x₂ + x₃
x_t = 26.5 + 20.7 +9 = 56.2 m
the distance traveled by the car when it returns to the lane is
x_c + x_t = x₀ + v₀ t + ½ a t²
when the car passes the car the distance traveled by the two vehicles is the same, we substitute
v_c t + x_t = x₀ + v₀ t + ½ a t²
½ a t² + t (v₀ -v_c) + (x₀ - x_t) = 0
we substitute the values
½ 0.560 t² + t (19.3 -19.3) + (25.0 - 56.2) =
0.28 t² -31.2 = 0
t = [tex]\sqrt{ \frac{31.2}{0.28} }[/tex]
t = 10.56 s
This is the time it takes for the car to pass the truck and back into the lane.
B) the distance traveled is
x = v₀ t + ½ a t²
x = 19.3 10.56 + ½ 0.560 10.56²
x = 235 m
C) the final velocity is
v = v₀ + a t
v = 19.3 + 0.560 10.56
v = 25.2 m / s
I need help please someone !!!!! Would appreciate it
Answer:
Yes, it would make it back up.
Explanation:
If it has 100,000 Joules of gravitational potential energy at the top of the hill, by the time the cart gets to the bottom, it will become PE = 0, KE = 90,000 since 10% of 100,000 is 10,000. The cart only requires 80,000J to climb back up so it should easily do so.
I didn't quite understand if the 10% energy loss is total, or every time it goes up or down, but it isn't a problem because 10% of 90,000 is 9,000, which means it would have 81,000J of energy on the way back up IF it loses energy due to friction on the way back up also.
The only physical law you need to prove this is the Law of Conservation of Energy: no energy is lost, only transformed; 10% of the energy becomes heat, the rest remains mechanical energy, which is the reason why the reasoning above works.
A tennis ball with mass 57 g is travelling at 25 m/s [S] when it is intercepted by a tennis racquet for 4.0 ms after which the ball travels at 32 m/s [N]. What is the average force applied to the ball by the racquet?
Answer:
F = 812.25 N
Explanation:
Given (convert to SI units):
m = 57 g = 0.057 kg
[tex]v_{1}[/tex] = -25 [tex]\frac{m}{s}[/tex]
[tex]v_{2}[/tex] = 32 [tex]\frac{m}{s}[/tex]
t = 4 ms = 0.004 s
Find acceleration:
a = [tex]\frac{v_{2}-v_{1} }{t}[/tex] = [tex]\frac{32-(-25)}{0.004}[/tex] = 14250
Find force:
F = ma = 0.057(14250) = 812.25 N
I will mark you brainlist! Use your own word
What is the weathering, erosion, and deposition?
Weathering is the breaking down or dissolving of rocks and minerals on Earths surface.
Erosion is the process by which the surface of the Earth gets worn down. Erosion can be caused by natural elements such as wind and glacial ice
Deposition is the dropping of sediment by wind, water, ice, or gravity.
Answer:
Weathering: refers to the process of breaking down and disintegrating rocks, minerals, soils, as well as several other materials.
Erosion: refers to the process of wearing down the surface of the earth due to glacial ice, wind and other natural elements.
Deposition: refers to the geological process, of sediments and soil, added to landforms due to wind, ice and other natural elements to build up layers od sediments.
Explanation:
Which of the following is NOT part of the grain group?
Answer:
Any food made from wheat, rice, oats, cornmeal, barley, or another cereal grain is a grain product. Anything else is not
Explanation:
Which of the following would be most likely to contribute to molecules
experiencing intermolecular forces?
O A. Containing oxygen
B. Containing charged regions
O C. Electric neutrality
O D. Being made of atoms
Answer:
B. Containing charged regions
Explanation:
The term i.e. intermolecular forces would be used to explain the attraction forces. Here the interaction would be done between molecules etc that acts between the acts & the other types of particles i.e. neighboring like atoms or ions
So in the given case, the option b would be contributed to the molecules that have intermolecular forces
hence, the option b is correct
How much kinetic energy does an object have that is moving at a rate of 30 m/s and has a mass of 4000 kg ?
Answer:
K = 1800 kJ
Explanation:
Given that,
The speed of the object, v = 30 m/s
Mass of the object, m = 4000 kg
We need to find the kinetic energy of the object. The formula for the kinetic energy is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 4000\times 30^2\\\\K=1800000\\\\or\\\\K=1800\ kJ[/tex]
So, the required kinetic energy is equal to 1800 kJ.
Which one of the statements below is true about mechanical waves?
They must travel in empty space.
They can travel in a vacuum.
Both sound and light are examples of mechanical waves.
They require a medium to travel through.
Coherent monochromatic light of wavelength l passes through a narrow slit of width a, and a diffraction pattern is observed on a screen that is a distance x from the slit. On the screen, the width w of the central diffraction maximum is twice the distance x. What is the ratio a>l of the width of the slit to the wavelength of the light
Answer:
λ = a
Explanation:
This is a diffraction exercise that is described by the expression
a sin θ = m λ
sin θ = m λ/ a
the first zero of the diffraction occurs for m = 1
sin θ = λ / a
angles are generally very small and are measured in radians
sin θ = θ = y / x
we substitute
[tex]\frac{y}{x} = \frac{\lambda}{a}[/tex]
the width of the central maximum is twice the distance to zero
w = 2y
in the exercise indicate that this width is equal to twice the distance to the screen (2x)
W = 2x
2y = 2x
we substitute
1 = λ/ a
λ = a
we see that the width of the slit is equal to the wavelength used.
a student practicing for a track meet ran 250 m in 30 seconds. What was her average speed?
Answer:
8.33 meters/sec.
time = 30 sec. 30 sec. = 8.33 meters/sec.
Guys can you please help me with this
How much mechanical work is required to catch a 14.715N ball traveling at a velocity of 37.5m/s?
To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.
What is work?Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.
In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that
Work is the product of the component of the force acting in the displacement's direction and its magnitude.
Weight of the ball = 14.715 N.
Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.
Velocity of the ball = 37.5 m/s
Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.
Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.
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What do you mean by physics?
Answer:
Explanation:
Physics is the branch of science that studies the natural world and its rules and orders. It is one of the oldest sciences as ancient people study the stars and astronomy is considered part of Physics.
assuming weightless pulleys and 100% efficiency, what is the minimum input force required to lift a 120 N weight using a single fixed pulley?
A. 21 N
B. 61 N
C. 121 N
D. 241 N
create a poem that incorporates those ten words. Feel free to make it as silly as you like! MINIMUM of 6 lines with a MINIMUM of 5 words and 10 should come from your book. These do not have to rhyme, but can if you wish.
I could make a poem for you if you actually gave the words...... what 10 words do i need to incorporate???☹︎
A 2.50 MHz sound wave travels through a pregnant woman's abdomen and is reflected from the fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 78 beats per second are detected. The speed of sound in body tissue is 1500 m/s. Part A Calculate the speed of the fetal heart wall at the instant this measurement is made.
Answer:
the speed of the fetal heart wall at the instant is 0.0325 m/s
Explanation:
Given the data in the question;
f₀ = 2.50 MHz = 1.80 × 10⁶ Hz
f[tex]_B[/tex] = 78 beat per sec ( Hz )
V = 1500 m/s
the speed of the fetal heart wall at the instant this measurement is made = ?
now, if v is the magnitude of hart wall speed, V is speed of sound and f[tex]_h[/tex] the frequency the heart receives( and reflects), f₀ is original frequency and f, is reflected back;
so
heart is moving observer and device is stationary source
f[tex]_h[/tex] = ((V + v)/v )f₀
Heart is moving source and device is stationary observer
f' = (V/(V-v ))f[tex]_h[/tex]
Beats
f[tex]_B[/tex] = f₀ - f' = f₀ - f₀( V+v / V-v ) = f₀( 2v / V-v )
so we solve for v
v = V( f[tex]_B[/tex] / ( 2f₀ + f[tex]_B[/tex] )
so we substitute
v = 1500 ( 78 / ( (2×1.80 × 10⁶) + 78 )
v = 1500 ( 78 / ( 3,600,000 + 78 )
v = 1500 ( 78 / 3,600,078 )
v = 1500 ( 2.1666 × 10⁻⁵ )
v = 0.0325 m/s
Therefore, the speed of the fetal heart wall at the instant is 0.0325 m/s
A light bulb has a resistance of 360 . What is the current in the bulb when it has a potential difference of 120 V across it? 0.33 A 3 A 480 A 43,200 A
Answer:
I=R×V
360×120 V
=43,200 A
Answer:
A.)0.33 A
Explanation:
i just took the quiz 2021
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. Bifocal lenses are used to correct his vision. A diverging lens is used in the top part of the lens to allow the person to clearly see distant objects.
1. What power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?
2. What power lens (in diopters) should be used in the bottom half of the lens to allow him to clearly see objects 25 cm away?
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
[tex]\frac{1}{f_1} = \frac{1}{q}[/tex]
q = f₁
2) for an object located at p = 25 cm
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]
[tex]\frac{1}{f_2}[/tex] = 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = [tex]\frac{1}{f}[/tex]
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
a man in a lift is moving upwards in a constant speed.the weight of the man is W.Acc
ording to Newtons third law,the reaction force of the weight W is the force of ?
Answer:
Normal force=mg
Explanation:
The reaction force of weight is the normal force.
in order to find the normal for we need to write all the forces and set it equal to the net force:
N-mg=ma (since it is a constant speed the a=0)
N=mg
We say that the mass comes to rest if, after a certain time, the position of the mass remains within an arbitrary small distance from the equilibrium position. Will the mass ever come to rest
Answer:
No, the mass will never come to rest
Explanation:
It is so because even at arbitrarily small distance it will experience some amount of force (irrespective of how small the value of force is).
This does not allow the mass to become stationary or in a equilibrium state as it is still subject to some amount of force.
Hence, the the mass will never come to rest
In which direction does the magnetic field in the center of the coil point?
Answer:
Right
Explanation:
Coil move right yes
It takes 20 seconds to fill a two-liter bottle with water from your kitchen faucet. What is the mass flow rate from the faucet if water has a density of 1000 fraction numerator k g over denominator m cubed end fraction?
a. 0.1kg/sec.
b. 0.01kg/sec.
c. 1g/sec.
d. 1kg/sec.
Answer:
0.1 kg/s.
Explanation:
The density of water, d = 1000 kg/m³
Volume, V = 2 L
Time, t = 20 s
We need to find the mass flow rate from the faucet. We know that the density of an object is given by :
[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]
So, the mass flow rate is equal to 0.1 kg/s.
The function s(t)=−t3+3t+3 gives the distance from a starting point at time t of a particle moving along a line. Find the velocity and acceleration functions. Then find the velocity and acceleration at t=0 and t=1. Assume that time is measured in seconds and distance is measured in centimeters. Velocity will be in centimeters per second (cm/sec) and acceleration in centimeters per second per second (cm/sec2).
Answer:
v(t)=3t²+3, and a(t)=6t
v(0)=3cm/sec, and a(0)=0cm/sec²
v(1)=6cm/sec, and a(1)=6cm/sec²
Explanation:
Find velocity and acceleration functions ( v(t) and a(t) )
(Relevant background: Let's say f(t) is a function of y with respect to x. Think of the derivative of this function, f'(t) , as representing the rate at which y changes with respect to x)
s(t) is a function of distance with respect to time. Therefore, s'(t) - the derivative of this - represents the rate at which distance changes with time, which is just the definition of velocity. So we can say velocity v(t) = s'(t).
s(t)=t³+3t+3
s'(t)=v(t)=3t²+3
Similarly, if v(t) is a function of speed with respect to time, then v'(t) represents the rate at which speed changes with time, which is acceleration. So we can say that acceleration a(t)=v'(t)=s''(t)
v(t)=3t²+3
v'(t)=a(t)=6t
Find velocity and acceleration at t=0 and t=1
t=0
v(t)=3t²+3
v(0)=3(0²)+3
v(0)=3 cm/sec
a(t)=6t
a(0)=6(0)
a(0)=0 cm/sec²
t=1
v(t)=3t²+3
v(1)=3(1²)+3
v(1)=3+3
v(t)=6 cm/sec
a(t)=6t
a(1)=6(1)
a(1)=6 cm/sec²
The function s(t)=−t3+3t+3 gives the distance from a starting point at time t of a particle moving along a line.
v(t)=3t²+3, and a(t)=6t
v(0)=3cm/sec, and a(0)=0cm/sec²
v(1)=6cm/sec, and a(1)=6cm/sec²
What is velocity and acceleration functions ( v(t) and a(t)?Relevant background: Let's say f(t) is a function of y with respect to x. Think of the derivative of this function, f'(t) , as representing the rate at which y changes with respect to x.
s(t) is a function of distance with respect to time. Therefore, s'(t) - the derivative of this - represents the rate at which distance changes with time, which is just the definition of velocity. So we can say velocity v(t) = s'(t).
s(t)=t³+3t+3
s'(t)=v(t)=3t²+3
Similarly, if v(t) is a function of speed with respect to time, then v'(t) represents the rate at which speed changes with time, which is acceleration. So we can say that acceleration a(t)=v'(t)=s''(t)
v(t)=3t²+3
v'(t)=a(t)=6t
Find velocity and acceleration at t=0 and t=1
t=0
v(t)=3t²+3
v(0)=3(0²)+3
v(0)=3 cm/sec
a(t)=6t
a(0)=6(0)
a(0)=0 cm/sec²
t=1
v(t)=3t²+3
v(1)=3(1²)+3
v(1)=3+3
v(t)=6 cm/sec
a(t)=6t
a(1)=6(1)
a(1)=6 cm/sec²
Therefore, The function s(t)=−t3+3t+3 gives the distance from a starting point at time t of a particle moving along a line.
v(t)=3t²+3, and a(t)=6t
v(0)=3cm/sec, and a(0)=0cm/sec²
v(1)=6cm/sec, and a(1)=6cm/sec²
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