3 Physics Questions Please Help


1. Which will hit the ground FASTER (greater ), a pencil that is dropped from the top of the table ( = 0) or a pencil that rolls of the edge of the table ( > 0)? (Assume the table is level and both pencils leave the table at the same instant.)

a. The pencil that rolls off the table will hit faster.

b. The pencil that is dropped will hit faster.

c. Both will hit at the same speed.

d. Can’t tell without knowing how fast the second pencil was rolling.


2. You are throwing a water balloon as hard as you can off the top of the bleachers. You want the balloon to hit the ground with as great a velocity as possible. Which of these actions will help the hit the ground fastest—that is, with greatest ?

a. Throwing the balloon UP at any angle of θ with an initial velocity vmax.

b. Throwing the balloon DOWN at the opposite angle (-θ) with the same initial velocity vmax.

c. Both of these options—choices (a) and (b)—will yield the same .

d. Cannot compare these options without knowing angle θ.


3. Batman, hanging from the top of a low building, sees a criminal a long-distance away pointing a gun over his head at him. (The gun has no sighting tool, and the criminal does not know physics.) Assume Batman reacts just as the criminal fires the gun. (Which of the following should Batman NOT do? Circle any/all BAD MOVES.

a. Stay where he is and hope the criminal will miss.

b. Retract his grapple and hope to dodge the bullet by quickly shooting upward.

c. Let go of the building and hope to dodge the bullet by falling downward.

d. Push off the building and hope to dodge the bullet by moving sideways.

Check: did you see the word NOT in the question?

Answer:

137.2J

Explanation:

Ep= mgh

Given,

and we know, g = 9.81 N/kg

Ep= 2 × 9.81 × 7

Ep= 137.2J

Answer:

5.5 is the correct answer

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Answer:

8

Explanation:

Finding time= t=change in velocity or distance/acceleration

so 32/9.5-5.5

32/4

8

Answer:

0.75 meters per second; north

Explanation:

speed (velocity)= distance/time

15 meters= distance

20 seconds= time

15/20= 0.75 meters per second; north

Hope this helps!

Answer:

D. an ellipse with the Sun at one focus

Explanation:

Based on Kepler's work, the orbit of a planet around the Sun is an ellipse with the Sun at one focus.

Hence according to Kepler's first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one focus.

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Answer:

  P_a = 3P₀ ,  P_b = 2 P₀ ,       P_c = 3 P₀

P_c> P_b> P_a

Explanation:

To determine which circuit consumes more energy than is given by the expression

           P = V I

           V = I R

           I = V / R

           P = V² / R

As all circuits have the same battery, the value of the resistance to which the battery is connected determines the consumption

Circuit A

In this circuit the three bulbs are in series so the total resistance is

         R_total = 3 R

the power dissipated is

         P_a = V² / 3R

if we call

         P₀ = V² / R

we substitute

         P_a = P₀/3

Circuit B

Two bulbs are connected in series

           R_total = 2 R

power is

        P_b = V2 / 2R

        P_b =  P₀/2

Circuit C

The 3 bulbs are connected, but in parallel, the resistance is

        1 / R_totak = 1 / R + 1 / R + 1 / R

        R_total = R / 3

        P _c = V2 3 / R

        P_c = 3 Po

By reviewing these results, we can sort the circuits

        P_c> P_b> P_a

Answer:

voltage

Explanation:

the current in a circuit, the voltage can be increased.

According to Ohm's law, when all other physical parameters, including temperature, are held constant, the voltage across a conductor is directly proportional to the current flowing through it.

From ohm's law, we can state that voltage is directly proportional to current.

Looking at the following formula:

I = V/R

Where V is voltage,

I is current, and

R is resistance

In the above equation, Current and Voltage are in a direct relationship such that if I is increased, V is also increased, and vice versa.

Therefore, To increase the current, the voltage should be increased.

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Answer:

the equipotential lines or surface are perpendicular to the electric field

Explanation:

In this exercise you are asked to explain the relationship between the direction of the equipotential lines and the electric field.

Electric field lines go from a positive charge to a negative charge, the specific shape of these lines depends on the charge distribution, let's look at two specific cases:

* in the case of point charge the lines are radial

* In the case of electric plates, they are lines perpendicular to the plates.

Equipotential surfaces are related to the elective field by

            Ex = - dV / dx

            Ey = -dV / dy

            Ez = -dV / dz

             E = Exi ^ + Ey j ^ + Ezk ^

remember electric potential is a scalar and the electric field is a vector quantity, therefore the equipotential lines or surface are perpendicular to the electric field. Specifically for

* point charge the equipotential surfaces are concentric with the charge

* For electrical plates equipotential lines are parallel to the plates.

Answer:

2m/s

Explanation:

The formula for acceleration can be found by [ a = f/m ]

We are given the force [ 100N ] and the mass [ 50g ].

We can use these values to solve for the acceleration.

a = 100/50

a = 2m/s

Best of Luck!

Answer:

The total force on the cart is:  

F = F(applied) - F(friction) = (18 N) - (15 N) = 3.0 N  

The acceleration of the cart is:  

a = (change in velocity)/(time) = (1.0 m/s)/(5.0 s) = 0.20 m/s^2  

Using F = ma, the mass of the cart is:  

m = F/a = (3.0 N)/(0.20 m/s^2) = 15 kg

Answer:

Motion map #3

Explanation:

because with velocity negative the arrows have to be the same size in the reverse direction.

Answer:

The tiger will land at 3.71 meters from the base of the rock

Explanation:

Horizontal Motion

When an object is thrown horizontally with a speed vo from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:

[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]

The tiger leaps from a height of h=7.5 m with a speed of v=3 m/s. Substituting the values into the formula:

[tex]\displaystyle d=3\cdot\sqrt{\frac {2\cdot 7.5}{9.8}}=3.71\ m[/tex]

The tiger will land at 3.71 meters from the base of the rock

A book is sitting on a desk is due to the Gravitational force of Earth Gravity so D. the downward force the Earth pulls on the book.

The force of attraction between all masses that is objects in the universe, especially the attraction of the Earth's mass for bodies near its surface.

Einstein said that the shape of spacetime is responsible for the rise of force that we experience as gravity. A concentration of mass, such as the Earth or sun, bends space around it like a rock bends the flow of a river.

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Answer:

(a) E = 3.6 x 10³ N/C = 3.6 KN/C

(b) ΔV = 1044 Volts

(c) K.E = 1.67 x 10⁻¹⁶ J

(d) Vf = 4.47 x 10⁵ m/s

(e) a = 3.45 x 10¹¹ m/s²

(f) F = 5.76 x 10⁻¹⁶ N

(g) E = 3.6 x 10³ N/C = 3.6 KN/C  

(h)  Both values are same in part (h) and (a)

Explanation:

(a)

Electric field between oppositely charged plates is given as follows:

E = σ/ε₀

where,

E = Electric Field Intensity = ?

σ = surface charge density = 32 nC/m² = 3.2 x 10⁻⁸ C/m²

ε₀ = Permittivity of free space = 8.85 x 10⁻¹² C²/N.m²

Therefore,

E = (3.2 x 10⁻⁸ C/m²)/(8.85 x 10⁻¹² C²/N.m²)

E = 3.6 x 10³ N/C = 3.6 KN/C

(b)

E = ΔV/r

ΔV = Er

where,

r = distance between plates = 29 cm = 0.29 m

ΔV = Potential Difference = ?

ΔV = (3.6 x 10³)(0.29)

V = 1044 Volts

(c)

Kinetic Energy of Proton = Work done on Proton

K.E = F r

but,  F = E q

K.E = E q r

where,

q = charge on proton = 1.6 x 10⁻¹⁹ C

Therefore,

K.E = (3600 N/C)(1.6 x 10⁻¹⁹ C)(0.29 m)

K.E = 1.67 x 10⁻¹⁶ J

(d)

K.E = (1/2)m(Vf² - Vi²)

where,

m = mass of proton = 1.67 x 10⁻²⁷ kg

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

1.67 x 10⁻¹⁶ J = (1/2)(1.67 x 10⁻²⁷ kg)(Vf² - (0 m/s)²]

Vf² = (1.67 x 10⁻¹⁶ J)(2)/(1.67 x 10⁻²⁷ kg)

Vf = √(20 x 10¹⁰ m²/s²)

Vf = 4.47 x 10⁵ m/s

(e)

2as = Vf² - Vi²

2(a)(0.29 m) = (4.47 x 10⁵ m/s)² - (0 m/s)²

a = (20 x 10¹⁰ m²/s²)/0.58 m

a = 3.45 x 10¹¹ m/s²

(f)

F = ma

F = (1.67 x 10⁻²⁷ kg)(3.45 x 10¹¹ m/s²)

F = 5.76 x 10⁻¹⁶ N

(g)

E = F/q

E = (5.76 x 10⁻¹⁶ N)/(1.6 x 10⁻¹⁹ C)

E = 3.6 x 10³ N/C = 3.6 KN/C

(h)

Both values are same in part (h) and (a)

Answer:

To the extent that Stellar spectra look like blackbodies, the temperature of a star can also be measured amazingly accurately by recording the brightness in two different filters. To get a stellar temperature: Measure the brightness of a star through two filters and compare the ratio of red to blue light.

Explanation:

Let [tex]u[/tex] be the initial velocity of the soccer ball at an angle of inclination of [tex]\theta_0[/tex] with the positive x-axis.

Given that:

[tex]\theta_0=45^{\circ}[/tex]

The horizontal distance covered by the projectile=20 m

Time of flight, [tex]t_f=2[/tex] seconds

Acceleration due to gravity, [tex]g= 10 m/s^2[/tex] downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, [tex]g= -10 m/s^2[/tex]

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, [tex]u_x=u\cos\theta_0[/tex].

The horizontal distance covered by the projectile [tex]= u_x\times t_f[/tex]

[tex]\Rightarrow u_x\times t_f=20[/tex]

[tex]\Rightarrow u_x\times 2=20[/tex]

[tex]\Rightarrow u_x=10[/tex] m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now,  [tex]u\cos(45^{\circ})=10[/tex] [as [tex]u_x=u\cos\theta_0[/tex]]

[tex]\Rightarrow u=10\sqrt{2}[/tex] m/s.

The vertical component of the initial velocity,

[tex]u_y= u\sin\theta_0[/tex]

[tex]\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})[/tex]

[tex]\Rightarrow u_y=10[/tex] m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

[tex]v=u+at[/tex]

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have [tex]u=u_y, a= -10 m/s^2[/tex].

So at any time instant, t.

[tex]v=u_y+(-10)t[/tex]

[tex]\Rightarrow v=10-10t[/tex]

The vertical component of the velocity, v, is the function of time and related as [tex]v=10-10t[/tex].

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

Complete Question

A number 14 copper wire has a diameter of 1.628 mm. Calculate the resistance of a 37.0 m long piece of such wire.

(Use 1.72×10-8 ?m for the resistivity of copper.)

For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor wiring, the limit is 15.0 A for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current?

What would be the voltage between the ends of the wire in the above problem?

What is the current density in the wire when it is carrying the maximum allowable current? (Current density is the current in the wire divided by the cross sectional area of the wire.)

What is the drift velocity of the electrons when the wire is carrying the maximum allowable current?

(The density of electrons in copper is 8.47×1028 m-3.)

Answer:

a

 [tex]R =0.3057 \ \Omega [/tex]

b

 [tex]P  = 68.78 \ W  [/tex]

c

 [tex]V  = 4.5855 \ V[/tex]  

d

 [tex]J  =   7.205 *10^{6}\ A/m^2 [/tex]

e

 [tex]v_d  =  0.0005316 \ m/s[/tex]

Explanation:

From the question we are told that

   The resistivity of copper is  [tex]\rho =  1.72* 10^{-8} \  \Omega \cdot m[/tex]

   The  current limit is  [tex]I  =  15 \ A[/tex]

     The diameter is  [tex]d= 1.628 mm= 0.001628 \ m[/tex]

    The length of the wire is  [tex]l  =  37.0 m[/tex]

    The density of electron is  [tex]n  =   8.47*10^{28} m^{-3}[/tex]

Generally the radius is mathematically represented as

       [tex]r =  \frac{d}{2}[/tex]

=>     [tex]r =  \frac{0.001628}{2}[/tex]

=>     [tex]r =0.000814 \ m [/tex]

Generally the resistance is mathematically represented as

          [tex]R =  \frac{\rho  *  l }{A}[/tex]

Here A is the area which is mathematically represented as    

      [tex]A = 3.142 *   (0.00081)^2[/tex]

=>   [tex]A = 2.082 *10^{-6}\ m^2 [/tex]  

=>    [tex]R =  \frac{1.72* 10^{-8}  *  37.0  }{2.082 *10^{-6}}[/tex]

=>    [tex]R =0.3057 \ \Omega [/tex]

Generally the power dissipated is mathematically represented as

       [tex]P  =  I^2 R[/tex]

=>   [tex]P  =  (15)^2 *  0.3057 [/tex]

=>   [tex]P  = 68.78 \ W  [/tex]

Generally the voltage is mathematically represented as  

     [tex]V  =  I  * R[/tex]    

=>    [tex]V  =  15 * 0.3057[/tex]  

=>    [tex]V  = 4.5855 \ V[/tex]  

Generally the charge density is mathematically represented as

      [tex]J  =  \frac{I}{A}[/tex]

=>   [tex]J  =  \frac{15}{2.082 *10^{-6}}[/tex]

=>     [tex]J  =   7.205 *10^{6}\ A/m^2 [/tex]

Generally the drift velocity is mathematically represented as

     [tex]v_d  =  \frac{I}{ n  *  A *  e}[/tex]

Here e is the charge on an electron with value  [tex]e =  1.60 *10^{-19} \  C[/tex]

So

         [tex]v_d  =  \frac{15}{ 8.47*10^{28}  * 2.082 *10^{-6} *  1.60 *10^{-19} }[/tex]

 =>  [tex]v_d  =  \frac{15}{ 8.47*10^{28}  * 2.082 *10^{-6} *  1.60 *10^{-19} }[/tex]

 =>  [tex]v_d  =  0.0005316 \ m/s[/tex]

Answer:

Not D, every substance in existence has mass and is made from it in any given form.

B and C

Explanation:

The statements which are true about photons are:

Discussion:

The photon is a type of elementary particle. It is the quantum of the electromagnetic field including electromagnetic radiation such as light and radio waves, and the force carrier for the electromagnetic force.

Additionally, Photons are massless, so they always move at the speed of light in vacuum, 3 × 10⁸ m/s.

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Answer: B. 15.2s

Explanation: 600/39.5 = about 15.2

Answer:

Explanation:

The work done by an object can be found by using the formula

From the question

force = 25 N

distance = 3 m

We have

work done = 25 × 3

We have the final answer as

Hope this helps you

Answer:

Both light and sound can be described in terms of wave forms with physical characteristics like amplitude, wavelength, and timbre. Wavelength and frequency are inversely related so that longer waves have lower frequencies, and shorter waves have higher frequencies

The ability of both to cause interference is the resemblance between sound waves and light waves.

While sound waves are mechanical, light waves are electromagnetic. Unlike sound waves, which are longitudinal, light waves are transverse. Light waves can go across a vacuum. Since sound waves need a physical medium to go through, they cannot move in a vacuum.

Both sound and light waves adhere to the rule of reflection, which states that the angle of incidence and angle of reflection are equal. Therefore, choice 2 is the right one. While light waves are transverse, sound waves are longitudinal.

In that, they both have an amplitude, a wavelength, a period, and a frequency, light, and sound waves are comparable to one another.

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Answer: Force of magnitude 20.22lbf to the right at angle 8.5°

Explanation: The image named "Untitled" below shows the forces acting on the cart.

The second image, "Untitled2", demonstrate the vector diagram of the forces acting on the kart, in which:

f is force forward

f(left) is pull to the left

f(R) is the resulting force necessary for the kart to move

The drag force is not draw on the vector diagram because its value is already part of the force forward.

As we can see, the diagram forms a right triangle with f(R) as hypotenuse.

Using Pytagorean Theorem:

[tex][f(R)]^{2}=f^{2}+[f(left)]^{2}[/tex]

[tex][f(R)]^{2}=20^{2}+3^{2}[/tex]

[tex][f(R)]^{2}=409[/tex]

[tex]f(R)=\sqrt{409}[/tex]

f(R) = 20.22

Force forward equals 20.22lbf to the right.

The angle formed form the forward direction and the resulting force is

[tex]cos\theta=\frac{20}{20.22}[/tex]

cosθ = 0.989

θ = cos⁻¹(0.989)

θ = 8.5°

The forward force is to the right at angle 8.5° and magnitude 20.22lbf.

Answer:

74 N to the right

Explanation:

the forces are going in opposite horizontal directions, meaning that they are directly opposing each other. this means that you can subtract the force applied in the direction that is greater from the direction that is less to get the net force for the greater direction

this means 182 N - 108 N = 74 N to the right

In a game of tug of war the magnitude and direction of the net horizontal force on the rope is 74 N.

A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

In a game of tug of war, a rope is pulled by a force of 182 N to the right and by a force of 108 N to the left. The magnitude and direction of the net horizontal force on the rope is,

F = 182 - 108

F = 74 N

The magnitude and direction of the net horizontal force on the rope is  74 N.

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Answer:

Because water is polar and oil is nonpolar, their molecules are not attracted to each other. polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.

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Substances that can dissolve both polar and nonpolar substances are called amphiphilic or amphipathic. This unique property arises due to their molecular structure, which contains both polar and nonpolar regions.

Amphiphilic molecules typically have a hydrophilic (water-attracting) or polar end and a hydrophobic (water-repelling) or nonpolar end. This dual nature allows them to interact with and dissolve in both polar and nonpolar solvents. Let's explore the reasons behind this ability:

Polar interactions: The hydrophilic end of the amphiphilic molecule can form hydrogen bonds and electrostatic interactions with polar solvents (like water) due to the presence of charged or partially charged atoms (such as oxygen or nitrogen). This makes them soluble in polar substances.

Nonpolar interactions: The hydrophobic end of the amphiphilic molecule is typically a long hydrocarbon chain or a nonpolar group. These nonpolar regions can interact with other nonpolar substances (like oils, fats, or hydrophobic portions of biomolecules) through van der Waals forces and London dispersion forces.

Emulsification: The presence of both polar and nonpolar parts in an amphiphilic molecule allows it to act as an emulsifier. It can stabilize the interface between polar and nonpolar substances, preventing them from separating. For example, in the formation of an oil-in-water or water-in-oil emulsion, amphiphilic molecules form a barrier between the two immiscible substances, keeping them dispersed and forming a stable mixture.

Examples of amphiphilic molecules include surfactants (detergents), phospholipids (essential components of cell membranes), and lipoproteins (transporters of lipids in the bloodstream). These substances play vital roles in various biological processes and industrial applications due to their ability to interact with both polar and nonpolar substances.

Hence, Substances that can dissolve both polar and nonpolar substances are called amphiphilic or amphipathic. This unique property arises due to their molecular structure, which contains both polar and nonpolar regions.

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Answer: Yes

Explanation: Because Electromagnetic waves form a continuous spectrum and all types of electromagnetic wave travel at the same velocity through a vacuum (space) or air. The waves that form the electromagnetic spectrum are grouped in terms of their wavelength and their frequency.

Answer:

Well we all know that bread and cake taste different. We do this on purpose.

Explanation:

When we make cake we add plenty of sugar to make it the delectable dessert we all know and love while when making bread we focus more on the corn starch and yeast's.

Answer: 30.641

Explanation: If you get the ave which is 38.0 then the vi which is 7.359

38.0-7.359=30.641/0.0 which is 30.641