What is a revolution?
a.one complete orbit of an object around another object
b.one complete spin of an object around an axis a change in the tilt of an object from one side to another
c.a full sequence of a planet through all of its seasons

Answer:

[tex]fraction = 0.11[/tex]

Explanation:

Linear kinetic energy of the bicycle is given as

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]K_1 = \frac{1}{2}(13) v^2[/tex]

[tex]K_1 = 6.5 v^2[/tex]

Now rotational kinetic energy of the wheels

[tex]K_2 = 2(\frac{1}{2}(I)(\omega^2))[/tex]

[tex]K_2 = (mR^2)(\frac{v^2}{R^2})[/tex]

[tex]K_2 = mv^2[/tex]

[tex]K_2 = 3.1 v^2[/tex]

now kinetic energy of the rider is given as

[tex]K_3 = \frac{1}{2}Mv^2[/tex]

[tex]K_3 = \frac{1}{2}(38) v^2 [/tex]

[tex]K_3 = 19 v^2[/tex]

So we have

[tex]fraction = \frac{K_2}{K_1 + K_2 + K_3}[/tex]

[tex]fraction = \frac{3.1 v^2}{6.5 v^2 + 3.1 v^2 + 19 v^2}[/tex]

[tex]fraction = 0.11[/tex]

Answer:

[tex]\theta = 20.98 degree[/tex]

Explanation:

As we know that the speed of the sound is given as

[tex]v = 332 + 0.6 t[/tex]

now at t = 273 k = 0 degree

[tex]v = 332 m/s[/tex]

so we have

[tex]a sin\theta = N\lambda[/tex]

[tex]a sin\theta = N(\frac{v_1}{f})[/tex]

now when temperature is changed to 313 K we have

[tex]t = 313 - 273 = 40 degree[/tex]

now we have

[tex]v = 332 + (0.6)(40)[/tex]

[tex]v_2 = 356 m/s[/tex]

[tex]a sin\theta' = N(\frac{v_2}{f})[/tex]

now from two equations we have

[tex]\frac{sin19.5}{sin\theta} = \frac{332}{356}[/tex]

so we have

[tex]sin\theta = 0.358[/tex]

[tex]\theta = 20.98 degree[/tex]

The diffraction angle of sound from a loudspeaker changes with temperature because the speed of sound varies with temperature. The higher the temperature, the faster the sound travels, resulting in a longer wavelength and, consequently, a larger diffraction angle.

The question involves the concept of wave diffraction, specifically as it pertains to sound waves exiting a diffraction horn loudspeaker. In this scenario, the diffraction angle θ of sound is observed to change with temperature, due to the dependence of the speed of sound on air temperature. To find the new diffraction angle when the temperature rises from 273 K to 313 K, we need to understand that the speed of sound in air increases with temperature. The formula v = 331.4 + 0.6T (with T in ℃) can be used to calculate the speed of sound at different temperatures, but here we'll use Kelvin and recall that v = 331.4 + 0.6(T - 273) to scale it correctly.

At 273 K, the speed of sound is approximately 331.4 m/s, and at 313 K, it is higher. Assuming the frequency of the sound remains constant, and using the relationship v = fλ (where v is the speed of sound, f is the frequency, and λ is the wavelength), the wavelength will increase with the speed of sound. According to the diffraction formula for a single slit, sin(θ) = mλ/d (where m is the order of the minimum, λ is the wavelength, and d is the width of the slit), if wavelength λ increases while d stays the same, the angle θ must also increase to satisfy the equation for the first minimum (m=1). Thus, the diffraction angle on a summer day when the temperature is higher would be larger than 19.5°.

Answer:

T= 210.15 N

F= 75.31 N

Explanation:

Let the tension in string be T newton.

According to the question

⇒T×cos21°= mg

⇒T= mg/cos21°

⇒T=20×9.81/cos21

⇒T= 210.15 N

now, the magnitude of horizontal force

F= Tsin21°

⇒F= 210.15×sin21°

=75.31 N

The mass attached to the spring must be 0.72 kg

Explanation:

The frequency of vibration of a spring-mass system is given by:

[tex]f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}[/tex] (1)

where

k is the spring constant

m is the mass attached to the spring

We can find the spring constant by using Hookes' law:

[tex]F=kx[/tex]

where

F is the force applied on the spring

x is the stretching of the spring

When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have

[tex]mg=kx[/tex]

and using [tex]g=9.8 m/s^2[/tex] and [tex]x=0.0177 m[/tex], we find

[tex]k=\frac{mg}{x}=\frac{(2.82)(9.8)}{0.0177}=1561.3 N/m[/tex]

Now we want the frequency of vibration to be

f = 7.42 Hz

So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:

[tex]m=\frac{k}{(2\pi f)^2}=\frac{1561.3}{(2\pi (7.42))^2}=0.72 kg[/tex]

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Answer:

166 666 666.7 years

Explanation:

We start the question by making the units uniform. We are told that the continents move at 3 cm/year = 0.03 m/year.

We are also told that the continents are now 5000 km = 5 000 000 m apart

So to calculate the time it took for them to be this far apart

t = distance/speed

t = 5 000 000 m/(0.03 m/year) = 166 666 666.7 years

The equivalent resistance is calculated by taking the inverse of the sum of the reciprocals of each resistance.

The smallest resistance you can obtain by connecting a 36.0-ohm, a 50.0-ohm, and a 700-ohm resistor together can be found by connecting them all in parallel. When resistors are connected in parallel, the equivalent resistance (Requivalent) is smaller than the smallest individual resistance in the network. The formula for the equivalent resistance of parallel resistors is:

1/Requivalent = 1/R1 + 1/R2 + 1/R3

Substituting the given values gives us:

1/Requivalent = 1/36 + 1/50 + 1/700

Calculating the sum of inverses and then taking the inverse of that sum gives us the equivalent resistance for our parallel network. The smallest resistance value, which is Requivalent, can be calculated using this method.

Answer:

[tex]Re=1992.24[/tex]

Explanation:

Given:

vertical height of oil coming out of pipe, [tex]h=25\ m[/tex]

diameter of pipe, [tex]d=0.1\ m[/tex]

length of pipe, [tex]l=50\ m[/tex]

density of oil, [tex]\rho = 900\ kg.m^{-3}[/tex]

viscosity of oil, [tex]\mu=1\ Pa.s[/tex]

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

Using the equation of motion:

[tex]v^2=u^2-2gh[/tex]

where:

v = final velocity

u = initial velocity

Putting the respective values:

[tex]0^2=u^2-2\times 9.8\times 25[/tex]

[tex]u=22.136\ m.s^{-1}[/tex]

For Reynold's no. we have the relation as:

[tex]Re=\frac{\rho.u.d}{\mu}[/tex]

[tex]Re=\frac{900\times 22.136\times 0.1}{1}[/tex]

[tex]Re=1992.24[/tex]

Answer:

Angular acceleration will be [tex]1135.5rad/sec^2[/tex]

Explanation:

We have given initial angular speed of a particular disk [tex]\omega _i=646.1rad/sec[/tex]

And it finally comes to rest so final angular speed [tex]\omega _f=0rad/sec[/tex]

Time is given as t = 0.569 sec

From third equation of motion we know that

[tex]\omega _f=\omega _i+\alpha t[/tex]

So angular acceleration [tex]\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-646.1}{0.569}=1135.5rad/sec^2[/tex]

The required height of the tin can is given by the height at the minimum value of the surface area function of the tin can

Reason:

The given information on the tin can are;

Volume of the tin can = 16·π in.³

The amount of tin to be used = Minimum amount

The height of the can in inches required

Solution;

Let h, represent the height of the can, we have;

The surface area of the can, S.A. = 2·π·r² + 2·π·r·h

The volume of the can, V = π·r²·h

Where;

r = The radius of the tin can

h = The height of the tin can

Which gives;

16·π = π·r²·h

16 = r²·h

[tex]h = \dfrac{16}{r^2}[/tex]

Which gives;

[tex]S.A. = 2 \cdot \pi \cdot r^2 + 2 \cdot \pi \cdot r \cdot \dfrac{16}{r^2} = 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r}[/tex]

When the minimum amount of tin is used, we have;

[tex]\dfrac{d(S.A.)}{dx} =0 = \dfrac{d}{dx} \left( 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r} \right) = \dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2}[/tex]

Therefore;

[tex]\dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2} = 0[/tex]

4·π·(r³ - 8) = r² × 0

r³ = 8

r = 2

The radius of the tin can, r = 2 inches

The

[tex]h = \dfrac{16}{2^2} = 4[/tex]

The height of the tin can, h = 4 inches

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The required height of can is 4 inches.

Given data:

The volume of cylindrical tin with top and bottom is, [tex]V = 16 \pi \;\rm in^{3}[/tex].

The volume of tin is,

[tex]V = \pi r^{2}h\\16 \pi = \pi r^{2}h\\[/tex]

here, r is the radius of can and h is the height of can.

[tex]16 =r^{2}h\\\\h=\dfrac{16}{r^{2}}[/tex]

The surface area of tin is,

[tex]SA = 2\pi r(r+h)\\SA = 2\pi r(r+\dfrac{16}{r^{2}})\\SA = 2\pi r^{2}+\dfrac{32 \pi}{r})[/tex]

For minimum amount of tin used, we have,

[tex]\dfrac{d(SA)}{dr} =0\\\dfrac{ d(2\pi r^{2}+\dfrac{32 \pi}{r})}{dr} = 0\\4\pi r-\dfrac{32 \pi}{r^{2}}=0\\4\pi r=\dfrac{32 \pi}{r^{2}}\\r^{3}=8\\r =2[/tex]

So, height of can is,

[tex]h=\dfrac{16}{2^{2}}\\h=4[/tex]

Thus, the required height of can is 4 inches.

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Answer:12.12 m/s

Explanation:

Given

mass of child [tex]m=49 kg[/tex]

height of hill [tex]h=7.5 m[/tex]

inclination [tex]\theta =26^{\circ}[/tex]

Conserving Energy at top and bottom Point of hill

Potential Energy at Top =Kinetic Energy at bottom

[tex]mgh=\frac{mv^2}{2}[/tex]

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8\times 7.5}[/tex]

[tex]v=\sqrt{147}[/tex]

[tex]v=12.12 m/s[/tex]

Answer:

[tex]v\approx 12.129\,\frac{m}{s}[/tex]

Explanation:

The final speed of the child-sled system is determined by means of the Principle of Energy Conservation:

[tex]U_{1} + K_{1} = U_{2} + K_{2}[/tex]

[tex]K_{2} = (U_{1}-U_{2})+K_{1}[/tex]

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot \Delta h[/tex]

[tex]v = \sqrt{2\cdot g \cdot \Delta h}[/tex]

[tex]v = \sqrt{2\cdot\left(9.807\,\frac{m}{s^{2}} \right)\cdot (7.50\,m)}[/tex]

[tex]v\approx 12.129\,\frac{m}{s}[/tex]

Answer:

Total load = 2999.126 kg

Explanation:

Let the spring constant of the shock absorber be k.

We know that the force applied on a spring is directly proportional to elongated length and the constant of proportionality is called spring constant.

Thus

Force, F = kx

where,

x = elongation = 9.1 cm 0.091 m

mass of the people, m = 127 kg

F = weight of the people = mg = 127 x 9.8 = 1244.6 N

substituting these values in the first equation,

1244.6 = k x 0.091

thus, k = 13,676.923 N/m

Now we know that the time period, T of an oscillating spring with a load of mass m is

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

[tex]\frac{m}{k} = \frac{T^{2} }{4\pi ^{2} }[/tex]

thus,

[tex]m = k\frac{T^{2} }{4\pi ^{2}}[/tex]

T = 1.66s

substituting these values in the equation,

[tex]m =13,676.923\frac{1.66^{2} }{4\pi ^{2}  }[/tex]

m = 2999.126 kg

Answer:

Equilibrium temperature will be [tex]T=52.2684^{\circ}C[/tex]

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So [tex]2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)[/tex]

[tex]0.334T-3.67=1688-32.031T[/tex]

[tex]T=52.2684^{\circ}C[/tex]

By equating the heat gained by the lead to the heat lost by the water, the final temperature is calculated to be approximately 41.5°C.

To determine the final temperature of the lead weight and water, we can use the principle of calorimetry, which states that the heat gained by one object is equal to the heat lost by another object in a closed system.

Given:

Mass of lead[tex](m_l_e_a_d) = 2.61 g[/tex]

Specific heat capacity of lead [tex](c_l_e_a_d)[/tex] = 0.128 J/g°C

Initial temperature of lead[tex](T_l_e_a_d_i_n_i_t_i_a_l)[/tex]) = 11.1°C

Mass of water [tex](m_w_a_t_e_r)[/tex]= 7.67 g

Specific heat capacity of water[tex](c_w_a_t_e_r)[/tex] = 4.184 J/g°C

Initial temperature of water [tex](T_w_a_t_e_r_i_n_i_t_i_a_l)[/tex] = 52.6°C

Let T_final be the final equilibrium temperature.

Heat gained by lead [tex](Q_lead) = m_lead * c_lead * (T_final - T_lead_initia[/tex]l)

Heat lost by water [tex](Q_water) = m_water * c_water * (T_water_initial - T_final)[/tex]

Since the system is thermally insulated, Q_lead = -Q_water

[tex]m_lead * c_lead * (T_final - T_lead_initial) = - m_water * c_water * (T_water_initial - T_final)[/tex]

Solving for T_final:

[tex]T_final = (m_lead * c_lead * T_lead_initial + m_water * c_water * T_water_initial) / (m_lead * c_lead + m_water * c_water)[/tex]

Plugging in the values:

T_final = (2.61 g * 0.128 J/g°C * 11.1°C + 7.67 g * 4.184 J/g°C * 52.6°C) / (2.61 g * 0.128 J/g°C + 7.67 g * 4.184 J/g°C)

T_final ≈ 41.5°C

Therefore, the final temperature of both the lead weight and the water at thermal equilibrium is approximately 41.5°C.

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The speed of the mass when t = 3 s in simple harmonic motion is approximately 6.283 m/s.

The speed of the mass in simple harmonic motion can be determined by using the formula v = ωA sin(ωt), where v is the speed, ω is the angular frequency, A is the amplitude, and t is the time. In this case, the frequency of the motion is given as 4.0 Hz, which corresponds to an angular frequency of ω = 2πf = 2π(4.0 Hz) = 8π rad/s. The amplitude is given as 4.0 cm, which corresponds to A = 0.04 m. Plugging these values into the formula, we get:

v = (8π rad/s)(0.04 m) sin((8π rad/s)(3 s))

v ≈ 6.283 m/s

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The speed of the mass when[tex]\( t = 3 \) s is \( v = 8\pi \)[/tex] cm/s.

To find the speed of the mass at any given time \( t \) during its simple harmonic motion, we can use the following equation for velocity in SHM:

[tex]\[ v(t) = -A\omega \sin(\omega t + \phi) \][/tex]using the relation [tex]\( \omega = 2\pi f \).[/tex]Substituting the given frequency:

[tex]\[ \omega = 2\pi (4.0 \, \text{Hz}) = 8\pi \, \text{rad/s} \][/tex]

 The phase constant \( \phi \) is 0 because the timer is started when the mass is at its maximum displacement, which corresponds to \[tex]( x = A \)[/tex]when [tex]\( t = 0 \).[/tex]

 Now, we can calculate the velocity at [tex]\( t = 3 \)[/tex] s:

[tex]\[ v(3) = -A\omega \sin(\omega \cdot 3 + \phi) \][/tex]

[tex]\[ v(3) = -(4.0 \, \text{cm})(8\pi \, \text{rad/s}) \sin(8\pi \, \text{rad/s} \cdot 3 \, \text{s} + 0) \][/tex]

[tex]\[ v(3) = -32\pi \, \text{cm/s} \sin(24\pi \, \text{rad}) \][/tex]

Since \[tex]( \sin(24\pi \, \text{rad}) = 0 \)[/tex] (because [tex]\( 24\pi \)[/tex] is a multiple of[tex]\( 2\pi \)[/tex]), the velocity at [tex]\( t = 3 \)[/tex] s is:

[tex]\[ v(3) = -32\pi \, \text{cm/s} \cdot 0 = 0 \, \text{cm/s} \][/tex]

 However, this result corresponds to the instantaneous speed at a point where the mass changes direction, and thus its speed is momentarily zero. To find the speed at a point where the mass is actually moving, we need to consider the time just before or after this point of maximum displacement.

 Now, we calculate the velocity at [tex]( t = 3 - \Delta t \):[/tex]

[tex]\[ v(3 - \Delta t) = -A\omega \sin(\omega (3 - \Delta t) + \phi) \][/tex]

[tex]\[ v(3 - \Delta t) = -(4.0 \, \text{cm})(8\pi \, \text{rad/s}) \sin(8\pi (3 - \frac{1}{8\pi}) \, \text{rad}) \][/tex]

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \sin(24\pi - 1) \, \text{rad}) \][/tex]

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \sin(-1) \, \text{rad}) \][/tex]

 Since [tex]\( \sin(-1) \approx -1 \)[/tex] for a very small angle in radians:

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \cdot (-1) \][/tex]

[tex]\[ v(3 - \Delta t) = 32\pi \, \text{cm/s} \][/tex]

 Therefore, just before the mass reaches its maximum displacement at [tex]\( t = 3 \) s[/tex], its speed is [tex]\( 32\pi \)[/tex] cm/s.

Answer:

-No

Explanation:

Given that

Mass of box = 100 kg

Coefficient of friction ,μ= 0.23

We know that friction force depends on the normal force acts on the box

Fr= μ N

When we pull the box then :

Normal force N= mg

Friction force Fr=  μ mg                  

When we push the box :

Lets take pushing force = F

θ =Angle make by pushing force from the vertical line

Normal force

N = mg + F cosθ

Fr= μ ( mg + F cosθ )

The friction force is more when we push the box.That is why this is not easier to push the box.

Therefore answer is ---No

Answer:

The answer is to insulate Earth's surface temperature.

Explanation:

Clouds inhibit the outflow of terrestrial radiation. This acts to insulate Earth's surface temperature, keeping it warmer at night and cooler in the day.

Answer:

See explanation

Explanation:

To do this, you need to use energy conservation.  The sum of kinetic and potential energies is the same at all points along the path so, you can write the expression like this:

1/2mv1² + mgh1 = (1/2)mv2² + mgh2

Where:

v1 = 0 because it's released from rest

h1 = 18.6 m

v2 = speed we want to solve.

h2 = height at point A. In this case, you are not providing the picture or data, so, I'm going to suppose a theorical data to solve this. Let's say h2 it's 12 m.

Now, let's replace the data in the above expression (assuming h2 = 12 m). Also, remember that we don't have the mass of the bead, but we don't need it to solve it, because it's simplified by the equation, therefore the final expression is:

1/2v1² + gh1 = 1/2v2² + gh2  

Replacing the data we have:

1/2*(0) + 9.8*18.6 = 1/2v² + 9.8*12

182.28 = 117.6 + 1/2v²

182.28 - 117.6 = v²/2

64.68 * 2 = v²

v = √129.36

v = 11.37 m/s

Now, remember that you are not providing the picture to see exactly the value of height at point A. With that picture, just replace the value in this procedure, and you'll get an accurate result.

Answer:

seismograph or seismometer.

Explanation:

Seismograph is an instrument used to detect and record earthquakes.

It has a mass attached to a fixed base. During an earthquake, the base moves and the mass does not. The motion of the base with respect to the mass is commonly transformed into an electrical voltage. The electrical voltage is recorded. This record is proportional to the motion of the seismometer mass relative to the earth, but it can be mathematically converted to a record of the absolute motion of the ground.

Answer and Explanation:

Tropical storm:

The point at which the tropical depression intensifies and can sustain a maximum wind speed in the range of 39-73 mph, it is termed as a tropical storm.

Hurricane:

A hurricane is that type of tropical cyclone which includes with it high speed winds and thunderstorms and the intensity of the sustained wind speed is 74 mph or more than this.

Hurricanes are the most intense tropical cyclones.

The only difference between the tropical storm and hurricane is in the intensity.

A tropical storm becomes a hurricane when its sustained wind speeds exceed 74 miles per hour, with both systems featuring low pressure centers and heavy rains, but hurricanes pose a greater threat due to their higher wind speeds.

The primary difference between a tropical storm and a hurricane lies in their wind speeds. When a storm's sustained winds reach between 39 to 73 miles per hour, it is classified as a tropical storm. Once those winds exceed 74 miles per hour, the storm is then classified as a hurricane. Both types of storm systems form over warm ocean waters, typically warmer than 80 °F and involve low pressure centers, strong winds, and heavy rains. However, a hurricane's increased wind speed is what gives it the potential to cause significantly more damage upon making landfall.

The Coriolis force is responsible for the cyclonic rotation of these storms—with hurricanes in the Northern Hemisphere rotating counterclockwise and those in the Southern Hemisphere rotating clockwise. The terms hurricane, typhoon, and tropical storm are regional names for these cyclones. Regardless of the name, these storms are dangerous due to the combination of high winds, heavy rains, and consequential risks such as flooding and structural damage.

Answer:

Tension in the string, F = 6.316 N

Explanation:

It is given that,

Mass of the Yo - Yo, m = 0.2 kg

Length of the string, l = 0.8 m

It makes a complete revolution each second, angular velocity, [tex]\omega=2\pi\ rad[/tex]

Let T is the tension exist in the string. The tension acting on it is equal to the centripetal force acting on it. Its expression is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]F=m\omega^2 r[/tex]

[tex]F=0.2\times (2\pi )^2 \times 0.8[/tex]

F = 6.316 N

So, the tension must exist in the string is 6.316 N. Hence, this is the required solution.

Answer:

7.30523 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 2.72 m

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 2.72+0^2}\\\Rightarrow v=7.30523\ m/s[/tex]

The speed which a quarter would have reached before contact with the ground is 7.30523 m/s

Final answer:

The speed at which the quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head is approximately 7.3 m/s.

Explanation:

To determine the speed at which a quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head, we can use the principle of conservation of mechanical energy.

Let's denote the height of Robert Wadlow's head as h and the mass of the quarter as m. We are given that h = 2.72 meters.

The potential energy (PE) of the quarter at the top of Robert Wadlow's head is:

PE = m * g * h

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

The kinetic energy (KE) of the quarter just before it hits the ground is:

KE = (1/2) * m * v^2

where v is the speed of the quarter just before it hits the ground.

According to the principle of conservation of mechanical energy, the total mechanical energy at the top of Robert Wadlow's head is equal to the total mechanical energy just before the quarter hits the ground:

PE = KE

Substituting the expressions for PE and KE, we get:

m * g * h = (1/2) * m * v^2

Solving for v^2, we get:

v^2 = 2 * g * h

Substituting the given values for g and h, we get:

v^2 = 2 * 9.8 m/s^2 * 2.72 m

v^2 ≈ 53.2 m^2/s^2

Taking the square root of both sides, we get:

v ≈ 7.3 m/s

So, the speed at which the quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head is approximately 7.3 m/s.

Answer:

If we assume that the temperature remains constant. And if the intermolecular forces are increased, then there would be an increase in the boiling point, the viscosity, and surface tension of the substance.

Explanation:

The intermolecular forces are interactions between particles. And decrease as we go from solid >liquid >gas. And on gases we have weak intermolecular forces.

The most common 3 types of intermolecular forces occurs between neutral molecules are known as Van der Walls forces (Dipole-Dipole,Hydrogen bonding, London Dispersion forces).

If the intermodelucar attraction forces increases we have:

1. Decreasing vapor pressure (pressure of the vapor that is in equilibrium with the liquid)

2. Increasing boiling point (temperature at which the vapor pressure is equal to the pressur exerted on the surface of the liquid)

3. Increasing surface tension (resistance of a liquid to spread out and increase the surface area)

4. Increasing viscosity (resistance of a liquid to flow)

If we assume that the temperature remains constant. And if the intermolecular forces are increased, then there would be an increase in the boiling point, the viscosity, and surface tension of the substance.

Answer:

the volume is 0.253 cm³

Explanation:

The pressure underwater is related with the pressure in the surface through Pascal's law:

P(h)= Po + ρgh

where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)

replacing values

P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa

Also assuming that the bubble behaves as an ideal gas

PV=nRT

where

P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature

therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have

at the surface) PoVo=nRTo

at the depth h) PV=nRT

dividing both equations

(P/Po)(V/Vo)=(T/To)

or

V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³

V = 0.253 cm³

Final answer:

Using the given formula and solving for the constant using the temperature 10 minutes after pouring, the temperature of the coffee 30 minutes after it was poured is found to be 60 degrees Fahrenheit.

Explanation:

The question asks us to determine the temperature of coffee 30 minutes after it was poured, given the formula F = 120(2–at) + 60, where F is in degrees Fahrenheit and t is the time in minutes since the coffee was poured. Since the temperature of the coffee 10 minutes after it was poured was 120 degrees Fahrenheit, we first need to find the value of the constant a using the given formula. Substituting F with 120 and t with 10 gives us 120 = 120(2-10a) + 60. Solving for a, we find that a is equal to 0.1. Now, to find the temperature of the coffee 30 minutes after it was poured, we substitute t with 30 in the formula, getting F = 120(2-0.1*30) + 60, which simplifies to F = 60 degrees Fahrenheit.

Answer:[tex]24.70 ^{\circ}C[/tex]

Explanation:

Given

mass of lead piece [tex]m_l=234 gm\approx 0.234 kg[/tex]

mass of water in calorimeter [tex]m_w=611 gm\approx 0.611 kg[/tex]

Initial temperature of water [tex]T_w=24^{\circ}C[/tex]

Initial temperature of lead piece [tex]T_l=24^{\circ}C[/tex]

we know heat capacity of lead and water are [tex]125.604 J/kg-k[/tex] and [tex]4.184 kJ/kg-k[/tex] respectively

Let us take [tex]T ^{\circ}C[/tex] be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

[tex]m_lc_l(T_l-T)=m_wc_w(T-T_w)[/tex]

[tex]0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)[/tex]

[tex]86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)[/tex]

[tex]86-T=86.97T-2087.49[/tex]

[tex]T=\frac{2173.491}{87.97}=24.70^{\circ}C[/tex]

Answer:

[tex]v = 0.085 m/s[/tex]

Explanation:

First when man throws the ball then the speed of the man is given as

[tex]m_1v_1 = m_2v_2[/tex]

[tex]97 v = 0.365 \times 11.3[/tex]

[tex]v = 0.042 m/s[/tex]

now ball will rebound after collision with the wall

so speed of the ball will be same as initial speed

now again by momentum conservation we will have

[tex]m_1v_1 + m_2v_2 = (m_1 + m_2) v[/tex]

[tex]97 \times 0.042 + 0.365 \times 11.3 = (97 + 0.365) v[/tex]

[tex]v = 0.085 m/s[/tex]

Answer:

Explanation:

The tendency of the automobile running on a circular path at high speed to turn towards left or right around one of its wheels , is due to torque by centripetal force acting at its centre of mass about that wheel .

Suppose the automobile tends to turn in clockwise direction about its wheel on the outer edge . A rotational angular momentum is created . To counter this effect , we can take the help of a rotating wheel or flywheel . We shall have to keep its rotation in anticlockwise direction so that it can create rotational angular momentum in direction opposite to that created by centrifugal force on fast moving automobile. Each of them will nullify the effect of the other . In this way , rotating flywheel will save the automobile from turning upside down .

Final answer:

To counteract the tendency for a car to roll over when rounding curves at high speed, a flywheel should be mounted horizontally and perpendicular to the travel direction, with a sense of rotation that produces gyroscopic precession force downward on the inside wheels.

Explanation:

When an automobile rounds a curve at high speeds, the loading on the wheels changes, with the tendency to lighten the load on the inside wheels, which can lead to the car rolling over. To combat this, a large spinning flywheel can help equalize the loading on the wheels. The flywheel should be mounted such that its axis of rotation is horizontal and perpendicular to the direction of the car's travel. The sense of rotation should be such that when the car turns, the gyroscopic precession of the flywheel produces a force that pushes down on the inside wheels. This force counteracts the effect of the weight transfer to the outside wheels, reducing the risk of the car tipping over.

This concept exploits the conservation of angular momentum and the phenomenon known as gyroscopic precession. In a closed system, any attempt to change the direction of the axis of rotation of the spinning flywheel produces a force perpendicular to the direction of the applied force (precession). If the top of the flywheel is tilted outward, the precession force acts downward on the side of the flywheel closest to the inside of the turn, creating a stabilizing effect on the inside wheels.

Answer:

Radius, r = 1.5 meters

Explanation:

It is given that,

When an object is moving in an amusement park it will exert centripetal force. The centripetal force acting on the person, F = 560 N

Mass of the person, m = 83 kg

Speed of the wall, v = 3.2 m/s

The centripetal force acting on the person is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]r=\dfrac{mv^2}{F}[/tex]

[tex]r=\dfrac{83\times (3.2)^2}{560}[/tex]

r = 1.51 meters

or

r = 1.5 meters

So, the radius of the chamber is 1.5 meters.

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma              

[tex]a = \frac{F}{m}[/tex]      ...... (1)

acceleration (a) [tex]= \frac{dv}{dt}[/tex]   ......(2)

substituting (2) into (1)

Hence, F [tex]= \frac{mdv}{dt}[/tex]

[tex]\frac{dv}{dt} = \frac{F}{m}[/tex]

[tex]dv = \frac{F}{m} dt[/tex]

[tex]dv = \frac{1}{m}Fdt[/tex]

Integrating both sides

[tex]\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt[/tex]

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

[tex]v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt[/tex]     ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

[tex]v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt[/tex]

[tex]v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}[/tex]

[tex]v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|[/tex]

[tex]v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |[/tex]

[tex]v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |[/tex]

[tex]v = \frac{1}{5} | 250 - 50 + 15 |[/tex]

[tex]v = \frac{215}{5}[/tex]

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

Answer:

[tex]x=0.0049\ m= 4.9\ mm[/tex]

[tex]d=0.01153\ m=11.53\ mm[/tex]

Explanation:

Given:

(a)

Let x be the maximum distance of stretch without moving the mass.

The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.

[tex]f=k.x[/tex]

[tex]\mu_s.N=k.x[/tex]

where:

N = m.g = the normal reaction force acting on the body under steady state.

[tex]0.1\times (9.8\times 0.75)=150\times x[/tex]

[tex]x=0.0049\ m= 4.9\ mm[/tex]

(b)

Now, according to the question:

Let d be the total distance the object travels before stopping.

Now, the energy stored in the spring due to vibration of amplitude:

[tex]U=\frac{1}{2} k.A^2[/tex]

This energy will be equal to the work done by the kinetic friction to stop it.

[tex]U=F_k.d[/tex]

[tex]\frac{1}{2} k.A^2=\mu_k.N.d[/tex]

[tex]0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d[/tex]

[tex]d=0.01153\ m=11.53\ mm[/tex]

is the total distance does it travel before stopping.

To find the maximum distance the spring can be stretched without moving the mass, calculate the force of static friction and equate it to the spring force.

(a) How far can the spring be stretched without moving the mass?

To find the maximum distance the spring can be stretched without moving the mass, we need to calculate the force of static friction. The force of static friction can be calculated using the equation:

Fs = μs * m * g

where μs is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity.

Once we have the force of static friction, we can equate it to the spring force:

Fs = k * x

where k is the force constant of the spring and x is the maximum distance the spring is stretched without moving the mass.

Answer:

FALSE

Explanation:

The earth's lithosphere is divided into 2 types, namely the continental and the oceanic crust. The continental crust is comprised of lighter rock minerals such as silicate minerals, alkali and potash feldspar, and some oxide minerals, whereas, the oceanic crust is comprised of olivine, pyroxene and some feldpars that are comparatively denser than the minerals present in the continental crust. Due to this composition, the oceanic lithosphere is denser than the continental lithosphere.

During the convergent plate motion, the oceanic lithosphere (plate) subducts below the continental lithosphere due to its greater density.

Thus, the above given statement is False.