Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. If the incoming ball has a speed of 200 m/s . Part APart complete What is the final speed of the incoming ball if it is much more massive than the stationary ball?

The displacement of the object between t = 1.0 s and t = 3.0 s, using the equation x = –4t + t², is found to be 0 m.

The displacement of the object between t = 1.0 s and t = 3.0 s can be found by substituting these time values into the given equation x = –4t + t2 and then subtracting the earlier position from the later one.

For t = 1.0 s; x1 = -4(1) + (1)² = -3 m
For t = 3.0 s; x2 = -4(3) + (3)² = -3 m

The displacement can be calculated as Δx = x2 - x1 = -3 - (-3) = 0 m. So, the displacement of the object between t = 1.0 s and t = 3.0 s is 0 m.

https://brainly.com/question/32279145

#SPJ3

The percent decrease in the number of errors committed by the baseball player is 33.33%, calculated by finding the difference in errors between the two years and dividing by the number of errors in the first year.

To calculate the percent decrease in the number of errors committed by the baseball player, we need to first find the difference in the number of errors between the two years and then divide that by the number of errors made in the first year. Finally, we'll multiply the result by 100 to get the percentage.

Here's the step-by-step calculation:

Therefore, the baseball player had a 33.33% decrease in the number of errors committed compared to the previous year.

Answer:

Option b

b. Vega

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax (see the image below).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (AU).

[tex]\tan{p} = \frac{1AU}{d}[/tex]

Where d is the distance to the star.

[tex]p('') = \frac{1}{d}[/tex]  (1)  

Equation (1) can be rewritten in terms of d:

[tex]d(pc) = \frac{1}{p('')}[/tex]  (2)

Equation (2) represents the distance in a unit known as parsec (pc).

For the case of Sirius  ([tex]p('') = 0.38[/tex]):

[tex]d(pc) = \frac{1}{0.38}[/tex]

[tex]d(pc) = 2.63[/tex]

Hence, the object is 2.63 parsecs away from Earth.  

For the case of Procyon  ([tex]p('') = 0.29[/tex]):

[tex]d(pc) = \frac{1}{0.29}[/tex]

[tex]d(pc) = 3.44[/tex]

Hence, the object is 3.44 parsecs away from Earth.  

For the case of Vega  ([tex]p('') = 0.13[/tex]):

[tex]d(pc) = \frac{1}{0.13}[/tex]

[tex]d(pc) = 7.69[/tex]

Hence, the object is 7.69 parsecs away from Earth.  

Therefore, Vega is the star farther away.

Summary:

Notice how a small parallax angle means that the object is farther away.

Answer:

1. Electromagnetic Radiation

2. Visible light

3. Electromagnetic Spectrum

4. Wavelength

5. Frequency

6. Telescope

7. Bigger

8. Refractor telescope

9. Reflector telescope

10. Astronomical Interferometer

Explanation:

The light that we see and can not see is actually Electromagnetic radiation. It consists of various rays like X-rays, gamma, radio, IR etc. Out of the complete EM spectrum we can only see 7% i.e. the visible part of the spectrum. These waves are categorized basis their wavelength and frequency. Telescopes are used to study the EM spectrum. Bigger the opening of the telescope more light can be gathered and the image will be brighter and better. In refractor telescopes, convex lenses are used to collect the light while Reflector telescopes use concave mirrors are used to gather the light. When multiple telescopes, mirror segments are used in sync to work as one, the arrangement is called as Astronomical Interferometer.

The question is about the properties and tools used to study electromagnetic radiation. It introduces terms like electromagnetic spectrum, wavelength, frequency, telescopes, refracting telescope, reflecting telescope, and interferometry.

(1) Electromagnetic radiation consists of electric and magnetic disturbances, or waves, that travel through space. Human eyes see one form of this energy, called (2) visible light. All forms of electromagnetic radiation, including X-rays and radio waves, make up the (3) electromagnetic spectrum. Each type of radiation can be classified in two ways. (4) Wavelength measures the distance between the peaks on a wave and (5) frequency is the number of waves that occur each second. Scientists study radiation with (6) telescopes, which collect and focus light. The (7) larger the opening that gathers light in a telescope, the more light that can be collected. A(n) (8) refracting telescope uses lenses to bring light to a focus, and a(n) (9) reflecting telescope uses mirrors to do the same thing. The process of linking several telescopes together so that they can act as one is called (10) interferometry.

https://brainly.com/question/2334706

#SPJ3

The weight of an oil tank with the mass of 50 slugs on the Earth's surface would be 1,610 pounds or 7164.2732 Newtons. On the moon, the weight would be 268.5 pounds, but its mass would remain the same (50 slugs) because mass is independent of location.

To begin, the weight can be determined using the formula Weight = mass x gravitational force. On the Earth's surface, the gravitational force is approximately 32.2 ft/s2.

A) The weight of the oil tank on Earth can be calculated by multiplying its mass (50 slugs) by the gravitational force. Therefore, it would be 50 slugs * 32.2 ft/s2 = 1,610 pounds. To convert to Newtons, since 1 pound = 4.44822 Newtons, the weight is 1,610 pounds * 4.44822 = 7164.2732 Newtons.

B) On the moon, the gravitational force is a sixth of that on Earth, i.e., 32.2 ft/s2 / 6 = 5.37 ft/s2. The mass of the oil tank would remain the same, 50 slugs, because mass doesn't change with location. However, its weight on the moon would be 50 slugs * 5.37 ft/s2 = 268.5 pounds.

Learn more about Weight and Mass here:

https://brainly.com/question/12271021

#SPJ3

To find the weight of 50 slugs on Earth, multiply by 32.17 ft/s^2 to get 1608.5 pounds and convert to 7157.48 Newtons. On the Moon, the weight is 1/6th, resulting in 268.083 pounds, with the mass still being 50 slugs.

To determine the weight of a tank of oil with a mass of 50 slugs on Earth's surface, we use the U.S. Customary system. The acceleration due to gravity (g) in this system is 32.17 feet per second per second. The formula to convert mass in slugs to weight in pounds is: Weight (lb) = mass (slugs) \\u00d7 acceleration due to gravity (ft/s^2).

A) On Earth, the weight in pounds is:
Weight (lb) = 50 slugs \\u00d7 32.17 ft/s^2 = 1608.5 pounds.
To convert this to Newtons, we use the conversion factor 1 pound = 4.44822 Newtons, resulting in:
Weight (N) = 1608.5 lb \\u00d7 4.44822 N/lb = 7157.48 Newtons.

B) On the Moon, where gravity is 1/6th that of Earth, the mass remains the same. Therefore, the weight in pounds is:
Weight (lb) = 50 slugs \\u00d7 (32.17 ft/s^2 / 6) = 268.083 pounds.
The mass in slugs remains the same at 50 slugs as mass is independent of gravity.

Answer:

1.714 s

Explanation:

The equation of a moving wave,

v = λf.................................. Equation 1

Where v = velocity of the wave, λ = wave length of the wave, f = frequancy of the wave,

Given: λ = 1.0 m, f = 3.5 Hz.

Substitute into equation 1

v = 1.0(3.5)

v = 3.5 m/s.

From motion,

velocity = distance/time.

v = d/t

t = d/v................................. Equation 2

Where distance of the elastic chord from the wall, t = time taken to travel to the other end.

Given: d = 3.0 m, v = 3.5 m/s

Substitute into equation 2

t = 3.0/3.5

t = 0.857 s.

Note: for the wave to form a standing wave, it has to travel twice.

Thus,

Time taken for standing wave to fill the entire length of the string = 2t

= 2×0.857

1.714 s

Answer: 0.05470atm

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = initial pressure at  = 1 atm (standard atmospheric pressure

[tex]P_2[/tex] = final pressure at [tex]15^oC[/tex] = ?

[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 38.56 kJ/mol = 38560 J/mol

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex]= initial temperature = [tex]78.4^oC=273+78.4=351.4K[/tex]

[tex]T_2[/tex] = final temperature =[tex]15^oC=273+15=288K[/tex]

Now put all the given values in this formula, we get

[tex]\log (\frac{P_2}{1atm})=\frac{38560}{2.303\times 8.314J/mole.K}[\frac{1}{351.4K}-\frac{1}{288K}][/tex]

[tex]\log (\frac{P_2}{1atm})=-1.262[/tex]

[tex]P_2=0.05470atm[/tex]

Thus the vapor pressure of ethanol at [tex]15^0C[/tex] is 0.05470atm

Answer:

714s

Explanation:

t=H/v=500000m/700m/s=714s

The space shuttle edeavor is launched to altitude of 500,000 m above the surface of earth. The shuttle travels at an average rate of 700 m/s. Time taken by space shuttle is 714.28 m/s to reach its orbit.

Given the data to find the time taken by space shuttle,

Distance covered by the space shuttle = 500,000 meter

Speed of space shuttle = 700 m/s

Mathematically, it can be calculated as follows :

distance = speed × time

As we have to find the time taken, the formula will be altered.

Computation:

Time taken = Distance / Speed

Time taken by space shuttle = Distance cover by space shuttle / Speed of space shuttle

Time taken by space shuttle = 500,000 / 700

Time taken by space shuttle = 714.28 m/s

To know more about the distance,

https://brainly.com/question/23395651

#SPJ5

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

[tex]KE_r = \dfrac{1}{2}I\omega^2[/tex]

I of the moment of inertia of the sphere

[tex]I = \dfrac{2}{5}MR^2[/tex]

 v = R ω

using conservation of energy

[tex]KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2[/tex]

[tex]KE_r = \dfrac{1}{5}MV^2[/tex]

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

[tex]\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h[/tex]

[tex] 0.7 V^2 = 0.7 V'^2 + gh[/tex]

[tex] 0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53[/tex]

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

The linear speed of the ball when it reaches the top of the ramp is 0.84 m/s.

The given parameters;

Apply the principle of conservation of energy to determine the linear speed of the ball when it reaches the top of the ramp.

[tex]K.E_i + P.E_i = K.E_f + P.E_f \\\\(K.E + K.E_r)_i + P.E_i = (K.E + K.E_r)_f \ + P.E_f\\\\(\frac{1}{2} mv^2 + \frac{1}{2} I \frac{V^2}{R^2} )_i + 0 = (\frac{1}{2} mv^2 + \frac{1}{2} I \frac{V^2}{R^2} )_f + mgh[/tex]

where;

[tex](\frac{1}{2} mv^2 + \frac{1}{2} \times \frac{2}{5} mR^2\times \frac{v^2}{R^2} )_i = (\frac{1}{2} mv^2 + \frac{1}{2} \times \frac{2}{5} MR^2\times \frac{v^2}{R^2} )_f + mgh\\\\(\frac{1}{2} mv^2 + \frac{1}{5} mv^2)_i = (\frac{1}{2} mv^2 + \frac{1}{5} mv^2)_f + mgh\\\\0.7 mv_i^2 = 0.7mv_f^2+ mgh\\\\0.7 v_i^2 = 0.7v_f^2+ gh\\\\0.7v_f^2 = 0.7v_i^2 -gh\\\\v_f= \sqrt{\frac{0.7v_i^2 -gh}{0.7} } \\\\v_f = \sqrt{\frac{0.7(2.85)^2 -9.8(0.53)}{0.7} }\\\\v_f = 0.84 \ m/s[/tex]

Thus, the linear speed of the ball when it reaches the top of the ramp is 0.84 m/s.

Learn more here:https://brainly.com/question/20626677

The man's centre of mass will shift upward when he raises his arms from down to up. However, the exact amount of shift for the man plus his arms as a system would need more detailed information and would involve calculations including both the body's and the arms' centres of mass.

The subject matter of this question is high school-level physics, specifically, mechanics and the concepts of center of mass and system dynamics. The question asks, 'When a 70 kg man raises both his arms, from hanging down to straight up, by how much does he raise his centre of mass?' To answer this, we need to consider the centre of mass of the person's body separately and the arms as separate objects as well.

Let's take the arm's centre of mass to be at its mid-point due to it being modelled as a uniform rod. When the arms are hanging down, the centre of mass of each arm is 75 cm from the ground. When the arms are raised straight up, the centre of mass of each arm is moved to a height equivalent to the man's height plus 75/2 cm.

However, to get the overall shift in the centre of mass of the man plus his arms as a whole system, we would need more specific details about the man's body proportions. Generally, the human body's centre of mass with arms hanging down by the sides is around 56% of the person's height from the ground. When the arms are raised, the body's individual centre of mass doesn't shift much, but the overall system's centre of mass (of the man and his arms together) does shift upward, though how much exactly requires more detailed anthropomorphic data.

https://brainly.com/question/33607198

#SPJ3

When a 70 kg man raises both his arms from hanging down to straight up, his center of mass raises by 0.0375 meters (3.75 cm).

To determine how much a 70 kg man raises his center of mass when he raises both arms from hanging down to straight up, we model his arms as uniform rods. Given:

Length of each arm ( L) = 75 cm = 0.75 m

Mass of each arm ( M) = 3.5 kg

Total mass of the man ( M_{total}) = 70 kg

The center of mass of each arm when it is hanging down is at half its length, i.e., 0.75 m / 2 = 0.375 m from the shoulder. When the arms are raised straight up, the center of mass of each arm moves to a point 0.75 m above the shoulder.

The vertical displacement ( Δh) of the center of mass of each arm is thus:

Δh = 0.75 m - 0.375 m = 0.375 m

To find the overall shift in the man's center of mass, we calculate the contribution of the arms' displacement to the man's center of mass:

The change in center of mass ( ΔCM) is given by:

ΔCM = (2 x M x Δh) / M_{total}

Substituting the values, we get:

ΔCM = (2 x 3.5 kg x 0.375 m) / 70 kg = 0.0375 m

Thus, the man's center of mass raises by 0.0375 meters (3.75 cm).

Answer:

θ=5.65°

Explanation:

Given Data

Mass m=1.5 kg

Length L=0.80 m

First spring constant k₁=35 N/m

Second spring constant k₂=56 N/m

To find

Angle θ

Solution

As the both springs take half load so apply Hooks Law:

Force= Spring Constant ×Spring stretch

F=kx

x=F/k

as

[tex]d=x_{1}-x_{2}\\ as \\x=F/k\\so\\d=\frac{F_{1} }{k_{1}} -\frac{F_{2}}{k_{2}}\\ Where \\F=1/2mg\\d=\frac{(1/2)mg}{k_{1}} -\frac{(1/2)mg}{k_{2}}\\ d=\frac{mg}{2}(\frac{1}{k_{1}} -\frac{1}{k_{2}} )\\ And\\Sin\alpha=d/L\\\\alpha =sin^{-1}[\frac{mg}{2L}(1/k_{1}-1/k_{2})]\\\alpha =sin^{-1}[\frac{(1.5kg)(9.8m/s^{2} )}{2(0.80m)}(1/35Nm-1/56Nm) ]\\\alpha =5.65^{o}[/tex]

θ=5.65°

Answer:

5.00 N

Explanation:

Force: This is defined as the product of mass and acceleration. The S.I unit of Force is Newton ( N).

Mathematically,

F = ΔM/t...................... Equation 1

Where F = Force needed to stop the snowball, ΔM = change in momentum, t = time.

But

ΔM = mΔv

ΔM = m(v-u)......................... Equation 2

Where m = mass of the snowball, Δv = change n velocity of the snowball, u = initial velocity, v = final velocity.

Substituting equation 2 into equation 1

F = m(v-u)/t...................... Equation 3.

Given: m = 120 g = 120/1000 = 0.12 kg, v  = 0 m/s, u = 7.5 m/s, t = 0.18 s

Substituting into equation 3,

F = 0.12(0-7.5)/0.18

F = -0.9/0.18

F = - 5.00 N.

From newtons third law of motion,

The Force need to stop the ball is equal and opposite to the force on the wall( Average force by the ball)

Therefore,

The average force on the wall = 5.00 N in opposite direction to the motion of the snowball

Answer:

<-0.45, 0, 0> kgm/s

Explanation:

The change in momentum would equal to the impulse generated by force F = <-0.3, 0, 0> over time duration of Δt = 1.5 second. The impulse is defined as the product of force F and the duration Δt

[tex]\Delta p = F \Delta t = (-0.3) * 1.5 = -0.45 kgm/s[/tex]

So the change in momentum is <-0.45, 0, 0> kgm/s

Final answer:

The change in momentum of the fancart is < -0.45, 0, 0 > kgm/s, which is calculated using the impulse-momentum theorem by multiplying the force applied by the time interval over which it was applied.

Explanation:

The question pertains to change in momentum of a fancart when a constant force is applied to it. The change in momentum (delta p) can be calculated using the impulse-momentum theorem, which states that the change in momentum is equal to the impulse applied to the object. The impulse is the product of the force and the time over which the force is applied. Given a force of < -0.3, 0, 0 > N applied for 1.5 seconds to a cart with an initial mass of 0.8 kg and initial velocity of < 0.9, 0, 0 > m/s, we can calculate the change in momentum:

Impulse = Force times Time

Impulse = < -0.3, 0, 0 > N
times 1.5 s = < -0.45, 0, 0 > Ns

Since delta p = Impulse, the change in momentum is also < -0.45, 0, 0 > kgm/s.

Answer:

It corresponds to a distance of 100 parsecs away from Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

It is defined in a analytic way as it follows:

[tex]\tan{p} = \frac{1AU}{d}[/tex]

Where d is the distance to the star.

[tex]p('') = \frac{1}{d}[/tex] (1)  

Equation (1) can be rewritten in terms of d:

[tex]d(pc) = \frac{1}{p('')}[/tex] (2)

Equation (2) represents the distance in a unit known as parsec (pc).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).

For the case of   ([tex]p('') = 0.01[/tex]):

[tex]d(pc) = \frac{1}{0.01}[/tex]

[tex]d(pc) = 100[/tex]

Hence, it corresponds to a distance of 100 parsecs away from Earth.

Summary:

Notice how a small parallax angle means that the object is farther away.

Key terms:

Parsec: Parallax of arc second

A stellar parallax of 0.01 arcsecond measured using telescopes on Earth corresponds to a distance of 100 parsecs, or about 326 light-years.

When measuring stellar parallax from Earth with a precision of about 0.01 arcsecond, we can calculate the corresponding distance in space using the concept that a parallax of 1 arcsecond is equivalent to a distance of 1 parsec. To find the distance for a parallax of 0.01 arcsecond, we use the formula: distance (parsecs) = 1 / parallax angle (arcseconds).

Thus, a parallax of 0.01 arcsecond corresponds to:
distance = 1 / 0.01 arcseconds = 100 parsecs. This means the star is 100 parsecs away, which is approximately 326 light-years, since 1 parsec equals 3.26 light-years.

Additional Context

Parallax measurements are vital for determining distances to the nearest stars in astronomy, and with ground-based telescopes, we can measure out to about 100 parsecs. However, due to atmospheric turbulence, accuracy diminishes beyond 40 parsecs, and ground-based parallax measurement becomes unfeasible beyond 100 parsecs. Space missions like Hipparcos have improved our ability to measure stellar parallax, pushing accurate distance measurements out beyond 500 parsecs.

Answer:

Time of ascent is greater than time of descent.

Explanation:

The gravitational force always acts in the downward direction. The air drag always opposes the motion.

During ascent, the gravitational force and air drag act in opposite direction to the motion where as during descent, only air drag acts in opposite direction to the motion of the ball while gravitational force acts in the same direction. Thus, the time of ascent and descent become unequal with time of ascent being greater than time of descent.

Answer:

15 MPH or less than that

Explanation:

You shouldn't drive faster than 15 MPH so if a truck unexpectedly pulls out you'll have time to stop.

When parked cars, branches of the tree, fences, houses, or other things block your vision of an open intersection, you can pause before approaching the junction and move slowly ahead until you can see if there is cross-traffic before moving through the junction.

Answer:

A=0.1024 m

Explanation:

Given Data

M=3.60 kg

m=42.0 g=0.042 kg

v₀=140 m/s

k=560 N/m

To find

Amplitude of Simple Harmonic Motion

Solution

Use conservation of momentum to solve for velocity

[tex]mv=(m+M)V\\V=\frac{mv}{m+M}\\ V=\frac{0.042kg*140m/s}{0.042kg+3.60kg}\\ V=1.6145 m/s[/tex]

Use conservation of energy to solve for amplitude

ΔKE+ΔUg+ΔUs=0

ΔKE+0+ΔUs=0

1/2MV²-1/2kA²=0

[tex]A=\sqrt{\frac{MV^{2} }{k} }\\ A=\sqrt{\frac{(3.642kg)*(1.6145m/s)}{560N/m} }\\ A=0.1024m[/tex]

Answer:

Work energy theorem: [tex]W=\Delta KE=\frac{1}{2} m.v_f^2-\frac{1}{2} m.v_i^2[/tex]

Conservation of energy:

This theorem states that energy can neither be created nor be destroyed it can only get converted from one form to another.

[tex]PE=m.g.h[/tex]

Explanation:

Work energy theorem:

It states that the total work done by the sum of all the forces acting on a particle is equal to the change in kinetic energy of the particle.

Mathematically:

[tex]W=\Delta KE=\frac{1}{2} m.v_f^2-\frac{1}{2} m.v_i^2[/tex]

Conservation of energy:

This theorem states that energy can neither be created nor be destroyed it can only get converted from one form to another.

Relation between the Forces and potential energy:

Force here implies the force of gravitation. Potential energy in general means the energy which is contained in a body due to virtue of its position in a field of influence.

Here the field of influence is gravity. So,

[tex]PE=m.g.h[/tex]

where:]

m = mass of the body

g = acceleration due to gravity

h = height of the object from the earth surface

Answer:

The Scenario:

On a normal Sunday afternoon Mr. Golanski is sitting in his living room reading his book. He decides after a while to turn on the Television to see what’s on the news, (Mr. Golanski is using Radio waves when he turns his television on by signaling the TV from his remote control). After a few hours Mr. Golanski decides it’s time to have dinner. He heats up a quick meal in his microwave because he doesn’t have the patience for cooking. (He is using microwave radiation to heat his food because water molecules in food absorb the radiation). He sits down for his meal, and halfway through he starts to choke! In a panicked frenzy he runs to his bathroom to try and dislodge the obstacle from his throat. By doing so he switched on the fluorescent lights in his bathroom exposing himself to small amounts of ultraviolet radiation. (Fluorescent lights absorb UV radiation and transmit visible light along with small amounts of UV light). Unable to dislodge the obstacle from his throat Mr. Golanski seeks help from his neighbor who drives him straight to the ER. To treat him properly the physicians opt for a fluoroscopy to examine Mr. Golanski’s esophageal tract. (Thus he is making use of X-ray imaging to obtain a visual of his internal esophageal structure to check for the obstruction). Once treated and discharged from the hospital Mr. Golanski returns home grateful to have survived this ordeal with minimum damage.

Explanation:

The Electromagnetic Spectrum is the range of frequencies and wavelengths for different light waves. They range with increasing frequency from Radio waves, to Microwaves, to Infrared waves, to Visible waves, to Ultraviolet waves, to Infrared waves, to X-rays, and to Gamma rays. Several of which we use in our daily lives such as Radio waves when operating our television or using our cellular phones. We also use microwaves to heat our food or for communication with satellites. We are also exposed to natural Ultraviolet radiation from the sun; however, we can also get exposed to other forms such as from certain types of light bulbs. We see visible light in the form of all the colors we can detect around us. We make use of x-rays for imaging techniques widely used in medicine for diagnostics, as well as Infrared waves in our home security systems. The electromagnetic spectrum is always used as a part of our everyday life.

Answer:

spring constant will be 547.619 N /m

Explanation:

We have given that force exerted if uniformly from 0 to 230 N

So exerted force F = 230 N

String is stretched by 0.420 m due to applying force

So x = 0.420 m

We have to find the spring constant in N/m

We know that stretched force is given by

[tex]F=Kx[/tex] , here K is spring constant and x is stretched length

So [tex]230=K\times 0.420[/tex]

[tex]K=\frac{230}{0.420}=547.619N/m[/tex]

So spring constant will be 547.619 N /m

The equivalent spring constant of the bow is 547.62 N/m.

To determine the equivalent spring constant of the bow, we can use Hooke's law which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

Given that the bow string is pulled back by 0.420 m and the force increases uniformly from 0 to 230 N, we can use the formula for spring force: F = kx, where F is the force, k is the spring constant, and x is the displacement.

Plugging in the given values, we have:

230 N = k * 0.420 m

Solving for k, we divide both sides of the equation by 0.420 m:

k = 230 N / 0.420 m = 547.62 N/m

Therefore, the equivalent spring constant of the bow is 547.62 N/m.

Answer:

Royal proclamation was issued in 1763 by the king George III which forbade the american colonist from settling west of Appalachia. This proclamation was issued to avoid the conflicts between the native Americans and the new settlers.

Explanation:

Answer:

Swallowed fish speed=4 m/s

Explanation:

Given

m₁(fish mass)=20 kg

v₁(fish speed)=1 m/s

m₂(Swallowed fish mass)=5 kg

To find

v₂(Swallowed fish speed)

Solution

From law of conservation of momentum

[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (20kg)(1 m/s)+(5 kg)v_{2}=(20kg+5kg)0m/s\\20kg.m/s+5kgv_{2}=0\\v_{2}=(20kg.m/s)/5kg\\v_{2}=4m/s[/tex]

As I neglected the negative sign because it only shows that swallowed fish is moving in opposite direction

The smaller fish was moving at a velocity of 4 m/s towards the larger fish before being eaten

The question involves the principle of conservation of momentum to solve how fast a 5-kg fish was moving prior to being eaten by a 20-kg fish to bring both to a halt. In the scenario, the 20-kg fish is moving at 1 m/s towards the smaller fish. To find the speed of the smaller fish before being swallowed, we assume no external forces are acting on the system.

Momentum is calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity. Initially, the momentum of the 20-kg fish is (20 kg)(1 m/s) = 20 kg·m/s, and let's say the smaller fish has a velocity v. Since they come to a halt, the final momentum is 0. By setting the total initial momentum equal to the final momentum, we have:

(20 kg)(1 m/s) + (5 kg)(-v) = 0
⇒ 20 - 5v = 0
⇒ v = 4 m/s

Answer: 3.02kg

Explanation:

According to the law of conservation of momentum, the sum of momentum of bodies before collision is equal to the sum of momentum of the bodies after collision.

Note that this bodies will move with a common velocity after collision.

Since momentum = mass of a body × its velocity

Let m1 and m2 be the masses of the spheres

u1 and u2 be their velocities

v be their common velocity after collision

Mathematically

m1u1 + m2u2 = (m1+m2)v

From the question, the second sphere is initially at rest i.e u2 = 0m/s and the composite system moves with a speed equal to one-third the original speed of 1st sphere i.e V = u1/3

Substituting this conditions into the formula, we have;

m1u1 + m2(0) = (m1+m2)u1/3

m1u1 = u1/3(m1+m2)

Given m1 = 1.51kg

m1 = 1/3(m1+m2)

m1 = m1/3 + m2/3

m1-m1/3 = m2/3

2m1/3 = m2/3

2m1 = m2

2(1.51) =m2

m2 = 3.02kg

The mass of the second sphere is 3.02kg

Answer: Each half life is 8 minutes long.

Explanation:

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  

t = age of sample  = 24 minutes

a = initial amount of the reactant  = 10 grams

a - x = amount left after decay process  = 1.25 grams

[tex]24min=\frac{2.303}{k}\log\frac{10}{1.25}[/tex]

[tex]k=\frac{2.303}{24}\log\frac{10}{1.25}[/tex]

[tex]k=0.0866min^{-1}[/tex]

for completion of half life:  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]t_{\frac{1}{2}}=\frac{0.693}{0.0866min^{-1}}=8.00min[/tex]

Thus half life is 8 minutes long

Answer:

In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force

Explanation:

Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Using the impulse-momentum theorem as a guide, the statement in bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force is correct.

The impulse-theorem defined as the change in momentum of an object will be equal to the impulse exerted on it.

Mathematically, the theorem can be described as:

The impulse is defined as the product between the force exerted (F) and the duration of the collision Δt,

I = FΔt

The change in momentum is equal to the product between the mass of the object (m) and the change in velocity (Δυ):

Δp = mΔv

So, the theorem can be written as,

FΔt = mΔv

This theorem can also be proved using Newton's second law, we know that,

F = ma

where, F - Force,

m - mass

a - acceleration

Here, we have to alter it with the acceleration change.

a = Δv /Δt,

when we substitute this with the force,

F = mΔv /Δt,

FΔt =mΔv.

The above formula is the exact formula of impulse- momentum theorem.

By using this concept, according to the force and velocity, If you land on sand your stopping time is much shorter than if you land on water, in bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.

Hence, Option B is the correct answer.

To know more about impulse and momentum:

brainly.com/question/9484203

#SPJ5

The distance covered by the proton is 48.4 m

Explanation:

The electric field produced by an electrically charged infinite plane is given by

[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]

where in this case,

[tex]\sigma = -1.10\cdot 10^{-6} C/m^2[/tex] is the surface charge density

[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity

Substituting,

[tex]E=\frac{-1.10\cdot 10^{-6}}{2(8.85\cdot 10^{-12})}=-5.65\cdot 10^4N/C[/tex]

And the direction is towards the plane (because the charge is negative).

The electric force on the proton due to this field is

[tex]F=qE[/tex]

where

[tex]q=1.6\cdot 10^{-19}C[/tex] is the proton charge

Substituting,

[tex]F=(1.6\cdot 10^{-19})(-5.65\cdot 10^4)=-9.0\cdot 10^{-15} N[/tex]

where the direction is toward the plane.

Now we can calculate the proton's acceleration using Newton's second law:

[tex]a=\frac{F}{m}[/tex]

where

[tex]m=1.67\cdot 10^{-27}kg[/tex] is the proton mass

Substituting,

[tex]a=\frac{-9.0\cdot 10^{-15}}{1.67\cdot 10^{-27}}=-5.4\cdot 10^{-12} m/s^2[/tex]

Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:

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

where

v = 0 is the final velocity

[tex]u=2.40\cdot 10^7 m/s[/tex] is the initial velocity

a is the acceleration

s is the distance covered

And solving for s,

[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(2.4\cdot 10^7)^2}{2(-5.4\cdot 10^{12})}=53.3 m[/tex]

Therefore, the closest answer is 48.4 m.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s  

Acceleration, a = 11.9 m/s²  

Final velocity, v = ?

Displacement,s = 400 m

Substituting  

v² = u² + 2as

v² = 0² + 2 x 11.9 x 400

v = 97.57 m /s

Speed of the race car at the end of the run is 97.57 m/s

Answer

given,

initial speed of merry-go-round = 0 rad/s

final speed of merry-go-round = 1.5 rad/s

time = 7 s

Radius of the disk = 6 m

Mass of the merry-go-round = 25000 Kg

Moment of inertia of the disk

[tex]I = \dfrac{1}{2}MR^2[/tex]

[tex]I = \dfrac{1}{2}\times 25000\times 6^2[/tex]

   I = 450000 kg.m²

angular acceleration

[tex]\alpha = \dfrac{\omega_f-\omega_0}{t}[/tex]

[tex]\alpha = \dfrac{1.5-0}{7}[/tex]

[tex]\alpha =0.214\ rad/s^2[/tex]

we know,

[tex]\tau= I \alpha[/tex]

[tex]\tau= 450000\times 0.214[/tex]

[tex]\tau=96300\ N.m[/tex]

The required value of frictional torque produced by the bearings is 96300 N-m.

Given data:

The final angular speed of merry-go round is, [tex]\omega_{2}=1.5 \;\rm rad/s[/tex].

The time interval is, t = 7.0 s.

The radius of merry-go round is, r = 6.0 m.

The mass of merry-go round is, m = 25,000 kg.

The frictional torque of an object undergoing rotational motion is the product of moment of inertia and angular acceleration. So,

[tex]T_{f} = I \times \alpha[/tex] ....................................................(1)

Here, I is the moment of inertia of merry-go round and its value is,

[tex]I =\dfrac{mr^{2}}{2}\\\\I =\dfrac{25000 \times (6.0)^{2}}{2}\\\\I =450000 \;\rm kg.m^{2}[/tex]

And angular acceleration is,

[tex]\alpha = \dfrac{\omega_{2}-\omega_{1}}{t}\\\\\alpha = \dfrac{1.5-0}{7}\\\\\alpha = 0.214 \;\rm rad/s^{2}[/tex]

Then the frictional torque is calculated from equation (1) as,

[tex]T_{f} = 450000 \times 0.214\\\\T_{f}=96300 \;\rm N.m[/tex]

Thus, we can conclude that the required value of frictional torque produced by the bearings is 96300 N-m.

Learn more about the frictional torque here:

https://brainly.com/question/10250469

Answer:

Would life be different if the electron were positively charged and the proton were negatively charged?

A) Yes.

B) No.

Answer is NO (Option B)

Does the choice of signs have any bearing on physical and chemical interactions?

A) Yes.

B) No.

Answer is NO (Option B)

Explanation:

Electric charges are of two general types: positive and negative, electrons are designated to be the carriers of negative charge. The charge on the proton and electron are exactly the same size but opposite. Opposite charges would still attract, and like charges would still repel in other words electrons would still repel other electrons, but protons and electrons would still attract each other. The designation of charges as positive and negative is merely a definition.

The choice of signs have no bearing on physical and chemical interactions because, proton and electron is just a name assigned to the sign as like charges repel and unlike charges attract. An atom is built with a combination of three distinct particles: electrons, protons, and neutrons. Electric charge, which can be positive or negative,  is neither created nor destroyed. The atom that has lost its equal share in the bonding electron pair acquires a partial positive charge because its nuclear charge is no longer fully canceled by its electrons.

Answer:

[tex]\frac{k_eQ}{2h}[/tex]

Explanation:

The question is missing an image. I have added this as an attachment to my answer.

Given;

Q = total charge

R = radius of cylindrical shell

h = height of cylindrical shell

d = distance of point from the right side of the cylinder

Let the thickness of the cylindrical shell be [tex]dx[/tex] , and the charge  [tex]\frac{Qdx}{h}[/tex],

Now, using the formula for finding the electric field due to a ring at a chosen point:

[tex]dE = \frac{k_ex}{(x^2 + R^2)^{\frac{3}{2}}} {\frac{Qdx}{h}i}[/tex]

where [tex]x[/tex] = center of the ring to the point

[tex]k_e[/tex] = electrostatic constant

We integrate on both sides from the limits [tex]d[/tex] to [tex]d + h[/tex]  in order to determine the electric field at the point [tex]E[/tex]

[tex]\int\limits dE[/tex] = [tex]\int\limits^{d + h}_d {\frac{k_eQxdx}{h(x^2 + R^2)^{\frac{3}{2}}}i}[/tex]

[tex]E = \int\limits^{d + h}_d {\frac{k_eQdx}{h(x^2 + R^2)^{\frac{3}{2}}}i} = \frac{k_eQ}{2h}[/tex]  

The electric field at a point a distance d from the right side of the cylinder is :   E = [tex]\frac{K_{e}Q }{2h}[/tex]

Let's' assume thickness of cylindrical shell = dx and charge = [tex]\frac{Qdx}{h}[/tex]  

Next step : calculate the electric field due to a ring at a point

dE = [tex]\frac{kex}{(x^2+ R^2)^{\frac{3}{2} } } \frac{Qdx}{h} i[/tex]  --- ( 1 )

where : x = centre of the ring to the specific point

           ke = electrostatic constant

To determine the electric field constant by integrating equation within limits d to d + h

E = [tex]\int\limits^d_d {\frac{keQdx}{h(x^{2}+R^2)^{\frac{3}{2} } } } \, i = \frac{KeQ}{2h}[/tex]

Hence we can conclude that The electric field at a point a distance d from the right side of the cylinder is :   E = [tex]\frac{K_{e}Q }{2h}[/tex].

Learn more about electric field : https://brainly.com/question/14372859