A uniform stick has a mass of 4.42 kg and a length of 1.23 m. It is initially lying flat at rest on a frictionless horizontal surface and is struck perpendicularly by a puck imparting a horizontal impulsive force of impulse 12.8 N s at a distance of 46.4 cm from the center. Determine the subsequent motion of the stick.

A [tex]1000[/tex] kg car travelling with velocity [tex]70[/tex]m/s takes [tex]3[/tex] to stop under full braking. And the same car under similar road conditions travelling with velocity [tex]140m/s[/tex] takes [tex]12m[/tex] to stop under full braking.

How can we find Distance at velocity [tex]140m/s[/tex] ?

In case of [tex]70m/s[/tex] car stop under full braking after [tex]3m[/tex].

So we can find the acceleration by third equation of motion

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

Here [tex]v[/tex] is final velocity which is zero.

[tex]u[/tex] is initial velocity which is [tex]70m/s[/tex]

[tex]s[/tex] is displacement

So substitute the values in equation of motion

[tex]0=70^2+2*a*3\\-4900/6=a\\-816.6m/s^2=a[/tex]

In case of initial velocity [tex]140m/s[/tex] acceleration is same because same car and same road condition so acceleration is [tex]-816.6m/s^2[/tex]

We can find the distance by using the equation of motion

[tex]v^2=u^2+2as\\0=140^2+2*(-816.6)*s\\-19600/-1633.2=s\\12m=s[/tex]

Distance is [tex]12m[/tex]

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A visible white and red light indicate a second craft is coming from the starboard (right) side.

A navigation light often referred to as a running light or position light, is a source of illumination aboard a ship, plane, or spacecraft that provides information about the location, course, or condition of the vehicle. Red and green navigation lights help with traffic control by indicating the orientation of the craft.

All navigation light systems typically comprise one or more white lights as well as red and green sidelights that designate the boat's port and starboard sides.

Having a flashlight on board is also essential since you never know when a navigation light may go out.

The typical navigational lights:

Sidelights: Because they are visible to other vessels approaching from the side or head-on, these red and green lights are sometimes known as combo lights. The port (left) and starboard (right) sides of a ship are indicated by the red and green lights, respectively.

sternlight: Only from behind or almost behind the ship can one see the sternlight.

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3,89,988 cm/min is the linear velocity

Given,

Diameter of CD = 12 cm

So, Radius of CD = 6 cm

CD is spinning at 10350 rev/min

Firstly , convert rev/min into rad/min

1 rev = 2π radians

10350 rev/min = 10350 × 2π

                        = 64998 rad/min

Formula used,

[tex]v=rw[/tex] where,

[tex]v[/tex] is the Linear velocity

[tex]r[/tex] is the radius

[tex]w[/tex] is the angular velocity

[tex]v[/tex] = 6 cm × 64998rad/min

  = 3,89,988 cm/min

Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.

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Answer:

The spring constant is 647.05 N/m.

Explanation:

The weight of the mass is W=mg

                                            W=66*10

                                            W=660

The spring force is given by the formula F=kx

where k is the spring force constant.

The spring force is equal to the weight of the mass. Therefore equate W and F.

kx= 660

k=660/x

Difference in length x=2.1-0.08= 1.02 m

k=660/1.02

k=647.05 N/m

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a. Work done = 2. 0648 Joules

b. Electrical potential, V = 4. 03 × 10^10 volts

a. Work done = force × distance

Force = density × acceleration due to gravity

Force = 5. 80 × 10

Force = 58 Newton

We have that distance = 3. 56cm = 0. 0356 m

Work done = 58 × 0. 0356

Work done = 2. 0648 Joules

b. The formula for electric potential is given thus,

Electric potential, V = [tex]k\frac{q}{r}[/tex]

r = 0. 0356m

k = Coulomb's constant = [tex]8.98*10^9[/tex]kg⋅m3⋅s−4⋅A−2.

q= 1.60x10^-¹C

substitute into the formula

Electric potential, V = [tex]8. 98[/tex] × [tex]10^9[/tex] × [tex]\frac{1. 6 * 10^-^-1}{0. 0356}[/tex]

Electric potential, V = [tex]8. 98 * 10^9[/tex] × [tex]4. 494[/tex]

Electric potential, V = [tex]4. 03 * 10^1^0[/tex] volts

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The valid prediction is  f(x) ≥ 0 over the interval [–1, 1]

The question seems to be incomplete the complete question is given in the image!!!!

A function from a set X to a set Y is an assignment of an element of Y to each element of X. The set X is called the domain of the function and the set Y is called the codomain of the function.

The given values for x = 3 , f(x) = -15 Here f (x) < 0

When x =0, f(x) = 5 Here f(x) > 0

So for f(x) ≤ 0 over the interval  (–[infinity], [infinity]) is the wrong option

When x = 2  f(x)=-5 here f(x) < 0

So , f(x) > 0 over the interval (–1, [infinity]) is also a wrong option

When x = 1 f(x)=0 here f(x) > 0

So , f(x) < 0 over the interval (0, 2) is also a wrong option

When x = -1 f(x) = 0

When x = 0 f(x) = 5

When x = 1 f(x) =0

So ,   f(x) ≥ 0 over the interval [–1, 1] is the valid prediction

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Answer:

8:P.E,9:gaining K.E,10:P.E,11:gaining K.E.

Explanation:

it gained potential energy at 8 because it was at its maximum height,and since potential energy is the energy that u posses due to ur height,thats the energy it possesed on the spot 8.as you can see on spot 9 its moving down so as its moving down,it gains kinetic energy,since its an energy possesed by a body due to its motion/movement,thats what it gained at spot 9, it gsained kineti energy because of its movement.and at 10 agian it would gain potential energy ecause of height and on point 11 as its heading upwards it gains kinetic energy because of movement.

The main force(s) acting on the puck after receiving the kick is a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motion.

Friction can be defined as that resistance that one surface or object encounters when moving over another.

So therefore, the main force(s) acting on the puck after receiving the kick is a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motion

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Answer:

The speed of sound depends on the medium thru which it travels in contrast to light (electromagnetic) which can travel thru a vacuum.

The speed of sound will depend on the density of the air thru which it travels - the density  of air is particularly affected by

humidity - water vapor in the air contributes to less density

temperature - warmer is less dense

motion of air - if the air is not motionless the speed will be affected

There are a few reasons why the value for the speed of sound may be a little different from the accepted value of 340 m/s such as the temperature and humidity of the air, the presence of other gases, and the shape of the object.

The temperature of the air: The speed of sound increases as the temperature of the air increases. So, if you measure the speed of sound in air that is warmer than 20 degrees Celsius, your value will be slightly higher than 340 m/s.

The humidity of the air: The speed of sound decreases as the humidity of the air increases. So, if you measure the speed of sound in air that is more humid than 0%, your value will be slightly lower than 340 m/s.

The presence of other gases: The speed of sound is slightly different in different gases. So, if you measure the speed of sound in a gas other than air, your value will be different from 340 m/s.

The shape of the object: The speed of sound is slightly different in different shapes of objects. So, if you measure the speed of sound in an object that is not a sphere, your value will be slightly different from 340 m/s.

In addition to these factors, there are also some experimental errors that can contribute to a difference between the actual value and the accepted value. These errors can be caused by things like inaccurate measurements of the distance or the time, or by small variations in the experimental setup.

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Answer:

Explanation:

The adaptation of a cheetah is shown in the image which shows it running is its long and strong legs to run fast which is denoted as option B.

These are the physical and behavioral features which ensures that they survive in their changing environment.

Cheetahs are known to have long and strong legs which is the reason why it runs at a very high speed.

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Answer: Option (B) is correct

Explanation: Hope it helps!

Good luck!

Answer:

The time taken is 555.56 s.

Explanation:

The average velocity is the ratio of the net velocity to the time required to achieve net velocity.

The formula of average velocity is,

v(av)=d/t

Given d=5.0 km which is 5*1000 m=5000 m and v(av)=9.0 m/s.

Therefore 9=5000/t

                 t=5000/9

                 t=555.56

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The normal in the water is 29.8°

Let

[tex]n_i=[/tex] the index of refraction for the material with incident light,

 [tex]n _r[/tex] = the index of refraction for the material with refracted light,

 [tex]\theta_i=[/tex] the angle of incidence

[tex]\theta_r[/tex] = the angle of refraction.

Applying Snell's law

[tex]n_isin\theta_i=n_rsin\theta_r[/tex]

Given

[tex]n_a=1, n_g=1.550, n_m=1.329n[/tex]

a)

From air to glass

[tex]n_a sin\theta_1[/tex][tex]=n_gsin\theta_2[/tex]

​[tex]1\\sin *41.3 = 1.550 *sin\theta_2[/tex]

A normal from glass to water

[tex]n_g sin\theta_2 =n_m sin \theta_3 \\1.550\cdot\sin25.2\degree=1.329\cdot\sin\theta_3\\1.550*sin25.2=1.329*sin\theta_3 \\\theta_3=\arcsin({1.550\cdot \sin25.2\degree \over 1.329})\\\approx29.8\theta_3 =arc sin( 1.3291*550*sin25) ≈ 29.8[/tex]

Hence the normal in the water is 29.8°

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The weight of the box is 300 newton whereas the acceleration is 3.3 m/s2.

The weight of the box is 300 newton because weight is a force which move towards the center of the earth due to gravity and we know that gravity is 10 m/s2.

The box is moving with 3.3 m/s2 if we put in the formula below.

acceleration= force/mass

A = 100/30

A = 3.3 m/s2

So we can conclude that the weight of the box is 300 newton whereas the acceleration is 3.3 m/s2.

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The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.

Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.

In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.

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Some health-related problems that are related to unhealthy body composition are:

This refers to the complications that a person has when his health system does not function optimally.

Hence, we can see that based on the fact that the body needs to have a healthy body composition such as a good Body Mass Index when this becomes unbalanced, it can bring the above health problems.

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0.22 m/s is the speed when the displacement is 1.75 cm.

Given:

A simple harmonic oscillator has;

Amplitude (A) =3.50 cm = 0.035 m

Maximum Speed (Vmax) = 26.0 cm/s = 0.26 m/s

Displacment (d) = 1.75cm =0.0175 m

The displacement d, whose maximum is the amplitude A , is expressed as:

∴ d = A Sin wt

   [tex]\frac{d}{A}[/tex] = Sin wt

   t = [tex]\frac{1}{W}[/tex] Sin⁻¹ ( [tex]\frac{d}{A}[/tex] )                  

  v = - Aw cos wt

  v = - Aw cos w [ [tex]\frac{1}{W}[/tex]sin⁻¹ ([tex]\frac{d}{A}[/tex]) ]

  v = - Aw cos [ sin⁻¹ ([tex]\frac{d}{A}[/tex]) ]

Speed, v = Vmax cos [ sin⁻¹ ([tex]\frac{d}{A}[/tex]) ]             ∵ Vmax = Aw

            v = 0.26 cos [ sin⁻¹ ( [tex]\frac{0.0175}{0.035}[/tex]) ]

            v = 0.22 m/s

Therefore, 0.22 m/s is the speed when the displacement is 1.75 cm.

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The speed of the second rock after the collision is 4.15m/s

Given the direction of rock in space is x , the direction with speed is 6m/s

After the collision the speed of the first rock is at 4 m/s

The angle given is 60 degrees with respect to x - axis

We need to calculate the speed of the second rock after the collision is

We know that

[tex]m_1=m_2[/tex]

[tex]v_1 = 6m/s\\v_2 = 4m/s[/tex]

θ = 60°

We need to find [tex]v_x[/tex] and [tex]v_y[/tex]

Therefore ,

[tex]m_1v_1= m_2v_2cos60+m_2v_2\\[/tex]

So, 6 = 4 cos 60° + [tex]v_2[/tex]

[tex]v_x = 6-4cos60\\v_x = 2m/s[/tex]

The same way we will find the momentum for [tex]v_y[/tex]

The total momentum is 0

0+0 = [tex]m_1(4)sin60 + m_2v_y[/tex]

[tex]v_y = -3.64 m/s[/tex]

Now ,

V (resultant) = [tex]\sqrt{(v_x)^{2} + (v_y)^2} \\v = \sqrt{(2)^2+(-3.64)^2} \\v = 4.15 m/s[/tex]

Hence the speed of the  second rock after the collision, in m/s is 4.15m/s

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a. The speed of the pendulum when it reaches the bottom is 0.9 m/s.

b. The height reached by the pendulum is 0.038 m.

c. When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

K.E = 100%P.E - 18%P.E

where;

K.E(bottom) = 0.82P.E

K.E(bottom) = 0.82(mgh)

K.E(bottom) = 0.82(1 x 9.8 x 0.05) = 0.402 J

K.E = ¹/₂mv²

2K.E = mv²

v² = (2K.E)/m

v² = (2 x 0.402)/1

v² = 0.804

v = √0.804

v = 0.9 m/s

P.E = 100%K.E - 7%K.E

P.E = 93%K.E

P.E = 0.93(0.402 J)

P.E = 0.374 J

P.E = mgh

h = P.E/mg

h = (0.374)/(1 x 9.8)

h = 0.038 m

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Thus, the speed of the pendulum when it reaches the bottom is 0.9 m/s.

The height reached by the pendulum is 0.038 m.

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

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In equation form, Or KE+PE=constant KEi+PEi=KEf+PEf where i and f denote initial and final values. This is known as the conservation of mechanical energy.

Κi + Ûg,i = kf + U,g,f

A closed system is a system where exists energy interactions with surroundings, but not mass interactions.

If we neglect any energy interactions from boundary work, heat, electricity, magnetism, and nuclear phenomena and assume that process occurs at a steady state and all effects from non-conservative forces can be neglected, then the equation of energy conservation is reduced to this form:

K + Ug = 0 (1)

Where:

K -Change in kinetic energy of the system, measured in joules.

Ug - Change in gravitational potential energy of the system, measured in joules.

If we know that K = Ki - Kf and Ug = Ug,i - Ug,f, then we get the following equation:

Ki + Ug,i = Kf +  Ug,f, (2)

Where i and f stand for initial and final states of each energy component.

Hence, the right answer is Ki + Ug,i = Kf +  Ug,f,.

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Answer:

A KEi + PEi = KEf + PEf

Explanation:

Answer: It's mostly found near Jamestown , New York

The purpose of the experiment is to describe how the plasma membrane contributes to how the cell functions.

The dependent variable is the variable bring investigated. In this case, the dependent variable is the egg circumference.

The independent variable is the variable controlled by the researcher.

The controlled variable has no influence on the results. An example is the concentration of vinegar.

A hypothesis is a proposition which can be tested by observations. In order to form a hypothesis, it's important to collect observations about the problem being examined.

The tools that can be used to collect data include observations and records taken.

The procedure to collect the data include:

The conclusions that can be drawn about how the sum of forces acting on an object affect its motion is that when the forces cancel out, then the resultant force will be zero, if not, the unbalanced force will result in the acceleration of the object.

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a.600 cm³

b. 2.5 g/cm³

c.  2.0775 kg

In Physics, this Archimedes' Eureka moment is called the Archimedes Principle, which states that when a body is immersed in a liquid, it experiences an upward buoyant force, which is equal to the weight of the liquid displaced by the body.

According to the question

Mass of Eureka is 60g

Cross sectional area  is 60 square centimeters

Mass of steel piece is 20g

Density = 8g /cubic centimeter

a)The total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can.

To find the mass of the water but we can easily find if we know the volume of the can.

In order to calculate the volume we would have to multiply the area of the cross section by the height.

Here,

60cm² x 10cm = 600 cm³

To find the mass that water has in this case we have to multiply the water's density by the volume,

so we get....

[tex]\frac{1g}{cm^3}[/tex] x 1000 = 1000g or 1kg

Now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can.

So we get....

1000g + 60g = 1060g or 1.06kg

b)The volume of the water that over flowed will be equal to the volume of the metal piece since, when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle".

Knowing this we just have to calculate the volume of the metal piece an that will be the answer.

So this time in order to find volume we will have to divide the total mass of the metal piece by its density.

So we get

20g ÷ [tex]\frac{8}{cm^3}[/tex] = 2.5 g/cm³

c)Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece.

We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 overflowed.

So now we the same process as in number a) just with a few adjustments.

[tex]\frac{1g}{cm^3}[/tex] x (1000 - 2.5) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can.

The mass of the water, and the mass of the metal piece.

Hence,

1060g + 997.5g + 20g = 2077.5 g or  2.0775 kg

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The heat needed is 132J

Given the mass of the gold m = 3 gram.

The initial temperature of the gold T = 21 degree celsius.

The final temperature of the gold  T' = 363 degree celsius.

Hence, the change in temperature  dT = T' - T

= 363 - 21 degree celsius

= 342 degree celsius.

The specific heat of the gold is given as c = 0.129 J/g°C

We are asked to calculate the heat required ( dQ ) to raise the temperature of gold.

The heat required for this is calculated as

dQ = mcDT

dQ = 3 × 342 × 0.129 J

dQ = 132.354 J ≈ 132 J

Hence, the Heat needed  is 132 J.

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Answer: B (132 joules)

Explanation:

4.719kJ is the work done by the gravitational force.

Given,

Mass of Spiderman, m = 80 kg

Acceleration due to gravity, g = 9.8[tex]m/s^{2}[/tex]

Formula used

F= mg

So, The force on Spiderman is= 80 × 9.8 = 784N

As Spiderman is dangling on the free end of a long rope, the other end of which is fixed to a tree limb above.

That's why only his vertical movement makes a difference in how much work gravity does.

So, we only consider the y coordinates here

As the rope is 12m long ,

y coordinate below the tree limb is –12m. (negative sign because the value is present below the origin.)

As it is given that the rope makes a 60.08° angle with the vertical.

So his final y coordinate is (–12.0m)cos60.08°

= -12×0.498 = -5.98m

Change in elevation (ΔS)=  –5.98m–(–12.0m)= 6.02m.

The work done by the gravity on Spiderman is:

W=FΔscosθ….. (1)

where, F = Force of gravity on Spiderman

ΔS = Change in the elevation

θ = angle between force and the displacement

Putting the values in equation (1)

W = 784 × 6.02 ×cos 180°

   = 784 × 6.02 × -1 = 4719.68 J = 4.719kJ

Hence, the work was done by the gravitational force on Spiderman in this maneuver is 4.719 kJ.

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If the current that is circuit is tripled,  the power is increased by a factor of 9. Option D

The term power refers to the rate of doing work. Now if we have to obtain the work done by the use of the formula; P = I^2R

Now if the current that is circuit is tripled, the power is now obtained by;

P = (3I)^2R = 9I^2R hence the power is increased by a factor of 9.

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The force on a charge in an electric field is  E = 1*10^-9  N/C.

         F = QE

and according to newton's  second law, the acceleration the charge will undergo is :

      a = F / m

Putting these two equations together we get

     E = m/q

For an electron with mass m= 9.11*10^-31 kg and charge  Q= 1.60*10^-19 C.

E = 9.11*10^-31 kg / 1.60*10^-19 C

E = 1*10^-9  N/C.

Electric discipline, electric-powered assets associated with each point in space when the price is present in any form. The magnitude and direction of the electrical subject are expressed with the aid of the value of E, called electric area power or electric powered discipline intensity or certainly the electrical area.

An electric-powered discipline is a place of the area around an electrically charged particle or objects wherein an electric charge might feel pressure. An electric-powered field is a vector amount and can be visualized as arrows going towards or far from expenses.

Examples of electric fields consist of the field produced in the dielectric of a parallel-plate capacitor (which creates an electrostatic discipline) and the electromagnetic wave produced by a radio broadcast monopole antenna (which creates a time-various area)

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Energy, heavier atoms atoms, neutron and a proton are the products of a fusion reaction.

Nuclear fusion is a type of reaction in which two or more atomic nuclei fuse or combine to form heavy nuclei and subatomic particles such as neutrons or protons and release of energy.

So we can conclude that Energy, heavier atoms, neutron and a proton are the products of a fusion reaction.

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Answer:

B C D

Explanation: