A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.

What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.

Explanation:

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Answer:

7.8% of the original volume.

Explanation:

From the given information:

Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C

Pressure [tex]P_1[/tex] = 240 kPa

Temperature [tex]T_2[/tex] = 45° C

At initial temperature and pressure:

Using the ideal gas equation:

[tex]P_1V_1 =nRT_1[/tex]

making V_1 (initial volume) the subject:

[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]

[tex]V_1 = \dfrac{nR*295}{240}[/tex]

Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

the final volume [tex]V_2[/tex] can be computed as:

[tex]V_2 = \dfrac{nR*318}{240}[/tex]

Now, the change in the volume ΔV =  V₂ - V₁

[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]

[tex]\Delta V = \dfrac{23nR}{240}[/tex]

∴

The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]

[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]

[tex]= 0.078[/tex]

= 7.8% of the original volume.

Answer:

2m

Explanation:

wavelength=speed/frequency

=340/170

=2m

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, [tex]\alpha=3.3 rad/s^2[/tex]

Diameter of the wheel, d=21 cm

Radius of wheel, [tex]r=\frac{d}{2}=\frac{21}{2}[/tex] cm

Radius of wheel, [tex]r=\frac{21\times 10^{-2}}{2} m[/tex]

1m=100 cm

Magnitude of total linear acceleration, a=[tex]1.7 m/s^2[/tex]

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,[tex]a_t=\alpha r[/tex]

[tex]a_t=3.3\times \frac{21\times 10^{-2}}{2}[/tex]

[tex]a_t=34.65\times 10^{-2}m/s^2[/tex]

Radial acceleration,[tex]a_r=\frac{v^2}{r}[/tex]

We know that

[tex]a=\sqrt{a^2_t+a^2_r}[/tex]

Using the formula

[tex]1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}[/tex]

Squaring on both sides

we get

[tex]2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}[/tex]

[tex]\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}[/tex]

[tex]v^4=r^2\times 2.7699[/tex]

[tex]v^4=(10.5\times 10^{-2})^2\times 2.7699[/tex]

[tex]v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}[/tex]

[tex]v=0.418 m/s[/tex]

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

As the spring is stretched, it exerts an upward restoring force f. At maximum extension, Newton's second law gives

∑ F = f - mg = 0   ==>   f = (2.0 kg) (9.8 m/s²) = 19.6 N

By Hooke's law, if k is the spring constant, then

f = kx   ==>   k = f/x = (19.6 N) / (0.15 m) ≈ 130 N/m

A 4.0 kg mass would cause the spring to exert a force of

f = (4.0 kg) (9.8 m/s²) = 39.2 N

which would result in the spring stretching a distance x such that

39.2 N = (130 N/m) x   ==>   x = (39.2 N) / (130 N/m) ≈ 0.30 m ≈ 30 cm

Answer:

A) 8GMm/d^2

Explanation:

We are given that

[tex]m_1=M[/tex]

[tex]m_2=3M[/tex]

[tex]m_3=m[/tex]

Distance between m1 and m2=d

Distance of object of mass m from m1 and m2=d/2

Gravitational force formula

[tex]F=\frac{Gm_1m_2}{r^2}[/tex]

Using the formula

Force acting between m and M is given by

[tex]F_1=\frac{GmM}{d^2/4}[/tex]

Force acting between m and 3M is given by

[tex]F_2=\frac{Gm(3M)}{d^2/4}[/tex]

Now, net force acting on  object of mass is given by

[tex]F=F_2-F_1[/tex]

[tex]F=\frac{Gm(3M)}{d^2/4}-\frac{GmM}{d^2/4}[/tex]

[tex]F=\frac{12GmM}{d^2}-\frac{4GmM}{d^2}[/tex]

[tex]F=\frac{12GmM-4GmM}{d^2}[/tex]

[tex]F=\frac{8GmM}{d^2}[/tex]

Hence, the magnitude of the force on the object with mass m=[tex]\frac{8GmM}{d^2}[/tex]

Option A is correct.

Answer:

Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils

melting point (intensive): the temperature at which a substance melts

Explanation:

Answer:

Explanation:

For starters begin with a warning not to touch the brake drums. All of the KE is transferred to the brake drums. The result is a large rise in temperature. Heat. If you press hard on the brakes, rubber is left on the road and there is heat involved in that too.

Answer:

KInetic energy reduces.

Explanation:

Application of breaks reduces velocity. Reduction of velocity constitutes velocity reduction.

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In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 seconds, assuming that he ran this race with constant acceleration, then the required constant acceleration would have been

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

By using the second equation of motion given by Newton,

S = ut + 1/2at²

100= 0 + 0.5*a*9.58²

a = 2.17 meters / second²

Thus,the required constant acceleration of Usain Bolt would have been 2.17 meters / second².

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Answer:

W = 4.75 KW

Explanation:

First, we will calculate the heat to be removed:

Q = (No. of students)(Metabolic Power of Each Student)

Q = (190)(125 W)

Q = 23750 W = 23.75 KW

Now the formula of COP is:

[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]

W = 4.75 KW

Explanation:

No matter where you are in the universe, your mass is always the same: mass is a measure of the amount of matter which makes up an object. Weight, however, changes because it is a measure of the force between an object and body on which an object resides (whether that body is the Earth, the Moon, Mars, et cetera).

Explanation:

Hence, weight of a body will change from one place to another place because the value of g is different in different places. For example, the value of g on moon is 1/6 times of the value of g on earth. As mass is independent of g , so it will not change from place to place.

Answer:

The velocities of the skaters are [tex]v_{1} = 3.280\,\frac{m}{s}[/tex] and [tex]v_{2} = 0.024\,\frac{m}{s}[/tex], respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex] (1)

Second skater

[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex] (2)

Where:

[tex]m_{1}[/tex] - Mass of the first skater, in kilograms.

[tex]m_{2}[/tex] - Mass of the second skater, in kilograms.

[tex]v_{1,o}[/tex] - Initial velocity of the first skater, in meters per second.

[tex]v_{1}[/tex] - Final velocity of the first skater, in meters per second.

[tex]v_{b}[/tex] - Launch velocity of the meter, in meters per second.

[tex]v_{2}[/tex] - Final velocity of the second skater, in meters per second.

If we know that [tex]m_{1} = 70\,kg[/tex], [tex]m_{b} = 0.043\,kg[/tex], [tex]v_{b} = 32\,\frac{m}{s}[/tex], [tex]m_{2} = 58.5\,kg[/tex] and [tex]v_{1,o} = 3.30\,\frac{m}{s}[/tex], then the velocities of the two people after the snowball is exchanged is:

By (1):

[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex]

[tex]m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}[/tex]

[tex]v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}[/tex]

[tex]v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)[/tex]

[tex]v_{1} = 3.280\,\frac{m}{s}[/tex]

By (2):

[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex]

[tex]v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}[/tex]

[tex]v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}[/tex]

[tex]v_{2} = 0.024\,\frac{m}{s}[/tex]

Answer:

the speed of the nerve impulse in miles per hour is 201.59 mi/hr

Explanation:

Given;

the speed of the nerve impulse, v = 90.1 m/s

To convert this speed in meters per second to miles per hour, we use the following method;

1,609 meter = 1 mile

3,600 s = 1 hour

[tex]v(mi/h) = 90.1 \ \frac{m}{s} \times \frac{1 \ mile}{1,609 \ m} \times \frac{3,600 \ s}{1 \ hour} = (\frac{90.1 \times 3,600}{1,609} )\frac{mi}{hr} = 201.59 \ mi/hr[/tex]

Therefore, the speed of the nerve impulse in miles per hour is 201.59 mi/hr

Explanation:

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If an object of a constant mass experiences a constant net force, it will have a constant acceleration.

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

According to Newton's second law of motion:

Applied force = mass × acceleration.

Hence, if an object of a constant mass experiences a constant net force, it will have a constant acceleration.

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Answer:

5.78amps

Explanation:

Given data

Time t= 57 seconds

Charge Q= 330C

Current I= ??

The expression for the electric current is given as

Q= It

Substituting we have

330= I*57

I= 330/57

I=5.78 amps

Hence the current is 5.78amps

Answer:

If the mass of B is m and the temperature change is the same, the mass of B will be 2m.

Explanation:

Q = mcT

T = mc/Q

M = 4Q/2cT........... (1)

T = Q/mc

Plug this in equation 1.

M = 4Q/(2c × Q/mc)  = 4Q ÷ 2Q/m  = 4Q × m/2Q = 2m

This above answer helps a lot.

Answer:

Block A velocity is 23.33 m/s and the collission is not elastic.

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

Plug in givens

90+60=3v1'+80

solve for v1'= 23.33m/s

b) Find the initial and final kinetic energy of Block B

Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J

Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J

Since Ki does not equal Kf the collision is not elastic

Answer:

Answer is LC Cirenit (seres)

Answer:

The mass and atomic numbers don't change

Explanation:

An excited atom relaxes to the ground state emitting a photon...called a gamma ray.

The answer is that the mass and atomic numbers don't change.

In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.

To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.

During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.

The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).

For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.

It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).

Thus, the product nucleus remains unchanged in terms of atomic number and mass number.

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Answer:

the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm

Explanation:

Given the data in the question;

Diameter of bright central maxima;

⇒ 2 × ( 1.22 × (λD/d) ) ⇒ 2.44( λD/d )

where D is the distance from the the droplet to the screen ( 30 cm )

d is the diameter of the droplet

λ is the wavelength of light ( 690 nm  = 690 × 10⁻⁷ cm )

since the central maximum of the pattern is 0.24 cm in diameter,

we substitute

0.24 cm = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / d )

solve for d

d = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / 0.24 cm

d = 0.0050508 cm² / 0.24 cm

d = 0.021045 cm or 2.1 × 10⁻² cm

Therefore, the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm

Answer:

I think it would be (-7 C )..

Answer:

c)  N_s / N_p = 115.15

Explanation:

Let's look for the voltage in the secondary, they do not indicate the power dissipated

          P = V_s i

          V_s = P / i

          V_s = 76 / 5.5 10⁻³

          V_s = 13.818 10³ V

the relationship between the primary and secondary of a transformer is

           [tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]

           [tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]

           Ns / Np = 13,818 10³ /120

           N_s / N_p = 115.15

Answer:

Explanation:

Given:

volume of urine discharged, [tex]V=400~mL=0.4~L=4\times 10^{-4}~m^3[/tex]

time taken for the discharge, [tex]t=30~s[/tex]

diameter of cylindrical urethra, [tex]d=4\times10^{-3}~m[/tex]

length of cylindrical urethra, [tex]l=0.2~m[/tex]

density of urine, [tex]\rho=1000~kg/m^3[/tex]

a)

we have volume flow rate Q:

[tex]Q=A.v[/tex] & [tex]Q=\frac{V}{t}[/tex]

where:

[tex]A=[/tex] cross-sectional area of urethra

[tex]v=[/tex] velocity of flow

[tex]A.v=\frac{V}{t}[/tex]

[tex]\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}[/tex]

[tex]v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}[/tex]

[tex]v=1.06~m/s[/tex]

b)

The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:

[tex]P=\rho.g.l[/tex]

[tex]P=1000\times 9.8\times 0.2[/tex]

[tex]P=1960~Pa[/tex]

Answer:

[tex]F=208.83N[/tex]

Explanation:

From the question we are told that:

Acceleration [tex]a=2.4m/s^2[/tex]

Force of Branch [tex]F=260N[/tex]

Generally the Newton's equation second law for Force is mathematically given by

 [tex]ma=F-mg[/tex]

 [tex]m=\frac{260}{2.4+9.8}[/tex]

 [tex]m=21.31kg[/tex]

Therefore

 [tex]F=mg[/tex]

 [tex]F=(21.31)(9.8)[/tex]

 [tex]F=208.83N[/tex]

Answer:

A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).