A ball is dropped 5 meters from rest. Using conservation of energy, what is the final velocity of the ball? a.) 5.39 m/s b.) 4.45 m/s c.) 9.90 m/s d.) 1.23 m/s e.) None of the above

Answer:

Part B)

v = 4.98 m/s

Explanation:

Part a)

As the ball is rolling on the inclined the the friction force will be static friction and the contact point of the ball with the plane is at instantaneous rest

The point of contact is not slipping on the ground so we can say that the friction force work done would be zero.

So here in this case of pure rolling we can use the energy conservation

Part b)

By energy conservation principle we know that

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

so we will have

[tex]\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega'^2 + mgh[/tex]

here in pure rolling we know that

[tex]v = R\omega[/tex]

now from above equation we have

[tex]\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv'^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v'}{R})^2 + mgh[/tex]

now we have

[tex]\frac{1}{2}mv^2(1 + \frac{2}{5}) = \frac{1}{2}mv'^2(1 + \frac{2}{5}) + mgh[/tex]

now plug in all values in it

[tex]\frac{1}{2}m(6^2)(\frac{7}{5}) = \frac{1}{2}mv'^2(\frac{7}{5}) + m(9.81)(0.80)[/tex]

[tex]25.2 = 0.7v^2 + 7.848[/tex]

[tex]v = 4.98 m/s[/tex]

Final answer:

The conservation of energy can be applied to the rolling ball because rolling motion involves both kinetic energy and potential energy. By equating the initial total energy of the ball to the final kinetic energy of the ball, we can find its velocity after it has passed the hill.

Explanation:

(a) The conservation of energy can be applied to the rolling ball because rolling motion involves both kinetic energy and potential energy. Although rolling requires friction, the work done by friction is accounted for in the system's total energy. Therefore, the law of conservation of energy can still be applied.

(b) To find the velocity of the ball after it has passed the hill, we can apply the conservation of energy. At the top of the hill, the ball has both potential energy and kinetic energy. At the bottom of the hill, the ball only has kinetic energy. By equating the initial total energy of the ball to the final kinetic energy of the ball, we can solve for its velocity.

Answer:

168.32 mile

Explanation:

1 mile = 1.61 km

1.61 km = 1 mile

1 km = 1 / 1.61 mile

So, 271 km = 271 / 1.61 = 168.32 mile

Answer:

The answer is 1.1A

Explanation:

See the attached file

Answer:

14.82 Volts

Explanation:

[tex]V_{back}[/tex] = Back emf of the motor

[tex]V_{battery}[/tex] = battery voltage = 17.2 Volts

[tex]i_{locked}[/tex] = Current in locked condition = 12.3 A

R = resistance

In locked condition, using ohm's law

[tex]R = \frac{V_{battery}}{i_{locked}}[/tex]

[tex]R = \frac{17.2}{12.3}[/tex]

R = 1.4 Ω

[tex]i_{normal}[/tex] = Current in normal condition = 1.7 A

Back emf of the motor is given as

[tex]V_{back}[/tex] = [tex]V_{battery}[/tex] - [tex]i_{normal} R[/tex]

[tex]V_{back}[/tex] = 17.2 - (1.7 x 1.4)

[tex]V_{back}[/tex] = 14.82 Volts

Answer:

6.45 x 10^5 N/C

Explanation:

q = + e = 1.6 x 10^-19 C

m = 9.99 x 10^-27 kg

V = 13 kV = 13000 V

B = 1 T

Let v be the speed and E be the strength of electric field.

1/2 mv^2 = eV

v^2 = 2 e v / m

v^2 = (2 x 1.6 x 10^-19 x 13000) / (9.99 x 10^-27)

v = 6.45 x 10^5 m/s

As the charge particle is undeflected, the force due to magnetic field is counter balanced by the force due to electric field.

q E = q v B

E = v B = 6.45 x 10^5 x 1 = 6.45 x 10^5 N/C

The result will be the magnitude of the smallest electric field (E) required for the 6Li ions to pass through the magnetic field undeflected.

Here's how to calculate the strength of the electric field required for the 6Li ions to pass through the magnetic field undeflected:

Force Balance:

For the ions to move undeflected, the magnetic force acting on them needs to be balanced by the electric force acting in the opposite direction.

Magnetic Force:

The magnetic force (F_magnetic) on a charged particle moving through a magnetic field is given by: F_magnetic = q * v * B

Where:

q is the charge of the particle (q = +e for 6Li ion)

v is the velocity of the particle

B is the magnetic field strength

Electric Force:

The electric force (F_electric) on the charged particle in an electric field (E) is: F_electric = q * E

Balancing Forces:

For undeflected motion: F_magnetic = F_electric

Substitute the expressions from steps 2 and 3: q * v * B = q * E

Solving for Electric Field (E):

Since the charge (q) of the ion appears on both sides, we can cancel it.

E = v * B

Finding Ion Velocity (v):

The ions are accelerated by a potential difference (V) of 13 kV (13000 V).

The potential difference is related to the ion's kinetic energy (KE) by: KE = q * V

Assuming the ion starts from rest, all the potential energy is converted to kinetic energy.

KE = 1/2 * m * v^2 (where m is the mass of the ion)

Solve for v:

Combine equations from steps 6: 1/2 * m * v^2 = q * V

Solve for v: v = sqrt( 2 * q * V / m )

Substitute v in the equation for E (from step 5):

E = sqrt( 2 * q * V / m ) * B

Plug in the values:

q = +e (elementary charge = 1.602 x 10^-19 C)

V = 13000 V

m = 9.99 x 10^-27 kg

B = 1.0 T

Calculate E using the above values and the constant for elementary charge.

Answer:

The tip of the needle to move from the highest point to the lowest point in 0.4 sec and the needle tip to travel a total distance in 0.9 sec.

Explanation:

Given that,

Frequency = 2.5 Hz

Amplitude = 1.27 cm

(a). We need to calculate the time

The frequency is the reciprocal of the time.

[tex]f=\dfrac{1}{T}[/tex]

[tex]T=\dfrac{1}{f}[/tex]

[tex]T=\dfrac{1}{2.5}[/tex]

[tex]T=0.4\ sec[/tex]

The time taken from highest point to lowest point

[tex]T'=\dfrac{T}{2}[/tex]

[tex]T'=\dfrac{0.4}{2}[/tex]

[tex]T'=0.2\ sec[/tex]

(b).  We need to calculate the time

The time taken in one cycle = 0.4 sec

The distance covered  in one sec= 4 times x amplitude

[tex]d=4\times1.27[/tex]

[tex]d=5.08\ m[/tex]

We need to calculate the speed

Using formula of speed

[tex]v=\dfrac{5.08}{0.4}[/tex]

[tex]v=12.7[/tex]

We need to calculate the time

[tex]t=\dfrac{11.43}{12.7}[/tex]

[tex]t= 0.9 sec[/tex]

Hence,  The tip of the needle to move from the highest point to the lowest point in 0.4 sec and the needle tip to travel a total distance in 0.9 sec.

The amount of time it took the tip of the needle to move from the highest point to the lowest point in its travel is equal to 0.2 seconds.

Given the following data:

Mathematically, the frequency of an object in simple harmonic motion is give by:

[tex]F=\frac{1}{T} \\\\T=\frac{1}{F}\\\\T=\frac{1}{2.5}[/tex]

T = 0.4 seconds.

Now, we can calculate the time it took the needle to move from the highest point to the lowest point:

[tex]t = \frac{T}{2} \\\\t = \frac{0.4}{2}[/tex]

Time, t = 0.2 seconds.

The distance traveled by the needle per seconds is given by:

[tex]d=4A\\\\d=4 \times 1.27\\\\[/tex]

Distance, d = 5.08 cm.

For the speed:

[tex]Speed = \frac{distance}{time} \\\\Speed =\frac{5.08}{0.4}[/tex]

Speed = 12.7 cm/s.

For the time:

[tex]Time = \frac{distance}{speed} \\\\Time =\frac{11.43}{12.7}[/tex]

Time = 0.9 seconds.

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Answer:

346m/s

Explanation:

v1=335m/s

f1=200hz

f2=207hz

v2=?

v1/f1=v2/f2

335/200=v2/207

v2=335*207/200

v2=346m/s

The equivalent resistance of the combination is 15 ohm. Option C is correct.

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

Case 1: Two 20 ohm resistors are connected in parallel;

[tex]\rm \frac{1}{R_{eq}_1} =\frac{1}{R_1} +\frac{1}{R_2} \\\\ \rm \frac{1}{R_{eq}_1} =\frac{1}{20} +\frac{1}{20} \\\\ R_{eq}_1}= 10 \ ohm[/tex]

Case 2: Two 10-ohm resistors are connected in parallel.

[tex]\rm \frac{1}{R_{eq}_2} =\frac{1}{R_1} +\frac{1}{R_2} \\\\ \rm \frac{1}{R_{eq}_2} =\frac{1}{10} +\frac{1}{10} \\\\ R_{eq}_2}= 5 \ ohm[/tex]

Case 3; Two combinations are connected in series.

[tex]\rm R= R_{eq_1}+R_{eq_2}\\\\ \rm R= 10+ 5 \\\\ R=15 \ ohm[/tex]

The equivalent resistance of the combination is 15 ohm.

Hence, option C is correct.

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Final answer:

Two 20 ohm resistors in parallel have an equivalent resistance of 10 ohms, two 10 ohm resistors in parallel have an equivalent resistance of 5 ohms. When these two combinations are connected in series, the total equivalent resistance is the sum of both, which is c) 15 ohms.

Explanation:

To find the equivalent resistance of the given combinations of resistors, we first need to understand how resistors combine in parallel and in series.

When resistors are in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances:

Req = 1 / (1/R1 + 1/R2)

For two 20 ohm resistors in parallel:

Req = 1 / (1/20 + 1/20) = 1 / (2/20) = 10 ohms

For two 10 ohm resistors in parallel:

Req = 1 / (1/10 + 1/10) = 1 / (2/10) = 5 ohms

When we have resistors in series, their resistances simply add up:

Rtotal = Req1 + Req2

So, for the 10 ohm equivalent and the 5 ohm equivalent resistors in series:

Rtotal = 10 ohms + 5 ohms = 15 ohms

Therefore, the correct answer to the student's question is (c) 15 Ohm.

Answer:

[tex]F_2 = 1.10 \mu N[/tex]

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

now since it depends inverse on the square of the distance so we can say

[tex]\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}[/tex]

now we know that

[tex]r_2 = 18.2 mm[/tex]

[tex]r_1 = 12.2 mm[/tex]

also we know that

[tex]F_1 = 2.45 \mu N[/tex]

now from above equation we have

[tex]F_2 = \frac{r_1^2}{r_2^2} F_1[/tex]

[tex]F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)[/tex]

[tex]F_2 = 1.10 \mu N[/tex]

Answer:[tex]V_{air}=0.259m/s[/tex]

[tex]V_{water}=0.01293m/s[/tex]

Explanation:

Given data

Length of pipe[tex]\left ( L\right )[/tex]=13cm=0.13m

From tables at [tex]T=30^{\circ}[/tex]

Kinematic viscosity of air[tex]\left ( \mu\right )=1.6036\times 10{-5} m^{2}/s[/tex]

and reynolds number is given by

Re=[tex]\frac{V\times \characteristic\ length}{Kinematic visocity}[/tex]

Flow is laminar up to Re.no. 2100

Re=[tex]\frac{V\times L}{1.6036\times 10{-5}}[/tex]

2100=[tex]\frac{V\times 0.13}{1.6036\times 10{-5}}[/tex]

V=0.259 m/s

For water

Kinematic viscosity of water[tex]\left ( \mu\right )=0.801\times 10{-6}m^{2}/s[/tex]

2100=[tex]\frac{V\times 0.13}{0.801\times 10{-6}}[/tex]

V=0.01293 m/s

The angular velocity of the CD is 754.4 rad/s. The time it takes for the CD to rotate through 90 degrees is 0.0021 seconds. The average angular acceleration of the CD is 188.6 rad/s².

To calculate the angular velocity of the CD, we can convert the given 7200 rev/min to radians per second. Since one revolution is equal to 2π radians, we can use the conversion factor to find the angular velocity. Thus, the angular velocity of the CD is 754.4 rad/s.

To calculate the time it takes for the CD to rotate through 90 degrees, we need to find the fraction of the total rotation that corresponds to 90 degrees. Since a full rotation is 360 degrees or 2π radians, 90 degrees is equal to π/2 radians. We can then use the formula Δθ = ωΔt to find the time it takes, where Δθ is the angle in radians, ω is the angular velocity, and Δt is the time. Rearranging the formula, we have Δt = Δθ/ω. Substituting the values, we get Δt = π/2 / 754.4 = 0.0021 seconds.

The average angular acceleration can be found using the formula α = (ωf - ωi) / Δt, where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and Δt is the time interval. The CD starts from rest and reaches full speed in 4 seconds, so the initial angular velocity is 0. Using the given full speed of 7200 rev/min, we can convert it to radians per second and use it as the final angular velocity. Thus, the average angular acceleration is α = (754.4 rad/s - 0 rad/s) / 4 s = 188.6 rad/s².

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Answer:

The conductivity of the wire is [tex]3.18\times10^{7}\ ohm-m[/tex].

Explanation:

Given that,

Diameter = 0.2 cm

Current = 20 A

Power = 4 W/m

We need to calculate the conductivity

We know that,

[tex]\sigma = \dfrac{1}{\rho}[/tex]

Using  formula of resistance

[tex]R = \dfrac{\rho l}{A}[/tex]....(I)

Where,

[tex]\rho[/tex] = resistivity

A = area

l = length

Using formula of power

[tex]P = i^2 R[/tex]

[tex]R = \dfrac{P}{i^2}[/tex]

Put the value of R in equation (I)

[tex]\dfrac{P}{i^2}=\dfrac{\rho l}{\pi r^2}[/tex]

[tex]\rho=\dfrac{P\pi r^2}{l\timesi^2}[/tex]

[tex]\sigma=\dfrac{l\times i^2}{P\pi r^2}[/tex]

Put the all values into the formula

[tex]\sigma=\dfrac{1\times(20)^2}{4\times3.14\times(0.1\times10^{-2})^2}[/tex]

[tex]\sigma=3.18\times10^{7}\ ohm-m[/tex]

Hence, The conductivity of the wire is [tex]3.18\times10^{7}\ ohm-m[/tex].

To solve for the minimum allowable conductivity of the wire, we calculate the resistance based on the maximum power dissipation and then rearrange the formula for resistance to solve for conductivity. The minimum allowable conductivity for the wire is 3.18
* 10^4 (
Ω
m)^-1.

Minimum Allowable Conductivity of the Wire

To find the minimum allowable conductivity of a 0.2 cm diameter wire carrying a 20-A current with a maximum power dissipation of 4W/m, we first need to calculate the resistance of a 1-meter length of the wire based on the power dissipation.

Since power dissipation (
P) along the wire is given by P = I^2
* R, where I is the current and R is the instantaneous resistance of the wire, we can rearrange to find R = P / I^2. Substituting the given values, we get R = 4W / (20A)^2 = 0.01
Ω/m.

The resistance R of a conductor is also given by R =
L / (
σ
* A), where L is the length, σ (sigma) is the conductivity, and A is the cross-sectional area. We can rearrange and solve for the conductivity σ = L / (R
* A).

For a wire with a diameter of 0.2 cm, the cross-sectional area A is
π
* (0.1 cm)^2. Converting to meters, A = 3.14 * (0.001 m)^2 = 3.14
* 10^-6 m^2.

Now, substitute L = 1 m, R = 0.01 Ω/m, and A = 3.14
* 10^-6 m^2 into the formula to find the conductivity σ. The minimum allowable conductivity can then be calculated as σ = 1m / (0.01 Ω/m
* 3.14
* 10^-6 m^2) which yields σ = 3.18
* 10^4 (
Ω
*m)^-1.

Answer:

1.34 x 10^3 Pa

Explanation:

density of oil = 0.85 x 10^3 kg/m^3

g = 9.81 m/s^2

height of oil column = 16.1 cm = 0.161 m

Pressure on the surface of water = height of oil column x density of oil x g

                                                      = 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa

Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.

Answer:

a) [tex]4.362580048\times 10^{-19}\ Joule[/tex]

b)[tex]0.27566898\times 10^{-19}\ Joule[/tex]

Explanation:

a) When, wavelength=λ=250 nm

[tex]\text{Work function of metallic caesium}=2.24\ eV\\=2.24\times 1.6021\times 10^{-19}\\=3.588704\times 10^{-19}\ Joule \ \text{(converting to SI units)}\\\lambda =\text {Wavelength of light}=250\ nm\\Energy\\E=\frac{hc}{\lambda}\\\text{Where h=Plancks constant}=6.62607004\times 10^{-34} m^2kg / s\\\text{c=speed of light}=3\times 10^8\ m/s\\\Rightarrow E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{250\times 10^{-9}}\\\Rightarrow E=0.07951284048\times 10^{-17} Joule\\[/tex]

[tex]\text{Kinetic energy}=\text{E - Work function}\\K.E.=(7.951284048\times 10^{-19})-(3.588704\times 10^{-19})\\K.E.=4.362580048\times 10^{-19}\ Joule\\[/tex]

b) When, λ=600 nm

[tex]E=\frac{hc}{\lambda}\\E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{600\times 10^{-9}}\\\Rightarrow E=3.313030502\times 10^{-19}\\\text{Kinetic energy}=\text{E - Work function}\\K.E.=(3.31303502\times 10^{-19})-(3.588704\times 10^{-19})\quad \text{Work function remains constant}\\K.E.=-0.27566898\times 10^{-19}\ Joule[/tex]

Answer:

q = 1.815 \times 10^{-8} C

Charge on one plate is positive in nature and on the other plate it is negative in nature.

Explanation:

E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm

According to the Gauss's theorem in electrostatics

The electric field between the two plates

[tex]E = \frac{\sigma }{\varepsilon _{0}}[/tex]

[tex]{\sigma }= E \times {\varepsilon _{0}}[/tex]

[tex]{\sigma }= 8.20 \times 10^{5} \times {8.854 \times 10^{-12}[/tex]

[tex]{\sigma }= 7.26 \times 10^{-6} C/m^{2}[/tex]

Charge, q = surface charge density x area

[tex]q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}[/tex]

q = 1.815 \times 10^{-8} C

The closest expression for [tex]\( F_2 \)[/tex] in terms of [tex]\( F_1 \)[/tex] is.

[tex]\[ \\F_2 = \frac{3}{4} F_1} \][/tex]

This means that after transferring 2 bricks from one bag to the other, the gravitational attraction [tex]\( F_2 \)[/tex] between the bags is [tex]\( \frac{3}{4} \)[/tex] of the initial attraction [tex]\( F_1 \).[/tex]

When initially each bag contains 4 bricks of mass [tex]\( M \)[/tex] each, the total mass in each bag is [tex]\( 4M \)[/tex]. The bags exert a gravitational attraction [tex]\( F_1 \)[/tex] on each other.

Let's denote:

-[tex]\( F_1 \):[/tex] Initial gravitational attraction between the bags when each bag has 4 bricks.

- [tex]\( F_2 \):[/tex] Gravitational attraction after transferring 2 bricks from one bag to the other.

Initial Situation Before Transfer

Each bag has 4 bricks so the mass of each bag is [tex]\( 4M \).[/tex]

Gravitational Attraction [tex]\( F_1 \)[/tex]

[tex]\[ F_1 = G \frac{(4M)(4M)}{d^2} \][/tex]

where [tex]\( G \)[/tex] is the gravitational constant and [tex]\( d \)[/tex] is the distance between the centers of the bags.

After Transferring 2 Bricks:

Now, one bag has 6 brick mass 6m and the other bag has 2 bricks mass2m

Gravitational Attraction[tex]\( F_2 \)[/tex]

[tex]\[ F_2 = G \frac{(6M)(2M)}{d^2} \]s[/tex]

Simplifying [tex]\( F_2 \)[/tex]

[tex]\[ F_2 = G \frac{12M^2}{d^2} \][/tex]

Relation between [tex]\( F_2 \)[/tex] and [tex]\( F_1 \)[/tex]

To find the relation between [tex]\( F_2 \)[/tex] and [tex]\( F_1 \)[/tex] we compare them

[tex]\[ \frac{F_2}{F_1} = \frac{G \frac{12M^2}{d^2}}{G \frac{16M^2}{d^2}} \][/tex]

[tex]\[ \frac{F_2}{F_1} = \frac{12M^2}{16M^2} \][/tex]

[tex]\[ \frac{F_2}{F_1} = \frac{3}{4} \][/tex]

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

initial gravitational potential energy = final total kinetic energy

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]

now we have

[tex]I = mk^2[/tex]

here k = radius of gyration of object

also for pure rolling we have

[tex]v = R\omega[/tex]

so now we will have

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})[/tex]

[tex]mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})[/tex]

[tex]v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}[/tex]

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

So it has less acceleration while moving on inclined plane for object which has more value of k

So it will take more time for the object to reach the bottom which will have more radius of gyration

Now we know that for hoop

[tex]mk^2 = mR^2[/tex]

k = R

For spherical shell

[tex]mk^2 = \frac{2}{3}mR^2[/tex]

[tex]k = \sqrt{\frac{2}{3}} R[/tex]

For solid sphere

[tex]mk^2 = \frac{2}{5}mR^2[/tex]

[tex]k = \sqrt{\frac{2}{5}} R[/tex]

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first

Answer:

3020.68 kg/m^3

Explanation:

r = 5.25 x 10^3 km = 5.25 x 10^6 m, d = 450 km = 450 x 10^3 m

T = 2.15 hours = 2.15 x 3600 second = 7740 second

Let the density of the planet is ρ and M be the mass of planet.

The formula for the orbital velocity is

[tex]v = \sqrt{\frac{GM}{r+d}}[/tex]

Time period is given by

[tex]T = {\frac{2\pi (r+d)}{v}}[/tex]

[tex]T = \frac{2\pi (r +d)^{1.5}}{\sqrt{GM}}[/tex]

[tex]7740= \frac{2\pi (5700\times 1000)^{1.5}}{\sqrt{6.67\times 10^{-11}M}}[/tex]

M = 1.83 x 10^24 kg

Density = mass / Volume

ρ = 1.83 x 10^24 / (4/3 x 3.14 x (5.25 x 10^6)^3)

ρ = 3020.68 kg/m^3

Answer:

They must have been traveling at 5333.33 km/h to cover that distance in 3 days.

That speed are 6,66 times higher than the speed of an aircraft jet.

Explanation:

d= 384000 km

t= 3 days = 3*24hr = 72hr

V= 384000km/72hr

V= 5333.33 km/h

comparison:

V1/V2= 5333.33/800

V1/V2= 6.66

Answer:

8.475 x 10^7 Pa

Explanation:

h = 8648 m, g = 9.8 m/s^2, density of water, d = 1000 kg/m^3

Pressure at a depth is defined as the product of depth of water , acceleration due to gravity and density of water.

P = h x d x g

P = 8648 x 1000 x 9.8 = 8.475 x 10^7 Pa

Answer:

Total pressure exerted at bottom =  119785.71 N/m^2

Explanation:

given data:

volume of water in bottle = 150 L = 0.35 m^3

Area of bottle = 2 ft^2

density of water = 1000 kg/m

Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water

Pressure due to water P = F/A

F, force exerted by water = mg

m, mass of water = density * volume

                             =  1000*0.350 = 350 kg

F  = 350*9.8 = 3430 N

A = 2 ft^2 = 0.1858 m^2  

so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2

Atmospheric pressure

At sea level atmospheric pressure is 101325 Pa

Total pressure exerted at bottom  = 18460.71 + 101325 = 119785.71 N/m^2

Total pressure exerted at bottom =  119785.71 N/m^2

Answer:

[tex]F_{applied} = 45.8 N[/tex]

Explanation:

When maximum force is applied on the crate along the plane so that it will not move then in that case friction force and component of the weight of the crate is along the plane opposite to the applied force

So here we will have

[tex]mgsin\theta + F_f = F_{applied}[/tex]

now we know that

[tex]F_f = \mu F_n[/tex]

also we know that

[tex]F_n = mg cos\theta[/tex]

so we will have

[tex]F_f = \mu ( mg cos\theta)[/tex]

now we will have

[tex]mg sin\theta + \mu (mg cos\theta) = F_{applied}[/tex]

[tex](5)(9.81)sin30 + (0.5)(5)(9.81)cos30 = F_{applied}[/tex]

so we will have

[tex]F_{applied} = 45.8 N[/tex]

The maximum force that can be applied parallel to the plane without moving the 5.0-kg crate on a 30° incline, considering a static friction coefficient of 0.5, is 45.7N.

To determine the maximum force that can be applied without moving the crate, we need to consider the effect of static friction, gravity, and the angle of the incline. The weight of the crate (W) is its mass (m) times acceleration due to gravity (g), which equals 5kg*9.8m/s² = 49N. However, this is the weight vertically down, so the force from gravity parallel to the incline is less, and we should multiply W by sin30⁰, getting 24.5N. The normal force (N) on the incline is W*cos30⁰, equal to 42.4N. Therefore, the maximum static friction force (fs) is coef. of static friction (μs) times N, which equals 0.5*42.4N = 21.2N. The max force applied to keep the crate from moving is the sum of the force of gravity and the static friction forces on the incline, which equals 24.5N + 21.2N = 45.7N.

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Answer:

[tex]KE_f = 372 J[/tex]

Explanation:

The forces on the box while it is sliding down are

1). Component of the weight along the inclined plane

2). Friction force opposite to the motion of the box

3). Applied force on the box

now we know that component of the weight along the inclined plane is given as

[tex]F_g = mgsin\theta[/tex]

[tex]F_g = (20.0)(9.8)sin30 = 98 N[/tex]

Now we know that other component of the weight of object is counterbalanced by the normal force due to inclined plane

[tex]F_n = mgcos\theta[/tex]

[tex]F_n = (20.0)(9.8)cos30 = 170 N[/tex]

now the kinetic friction force on the box is given as

[tex]F_k = \mu F_n[/tex]

[tex]F_k = 0.100(170) = 17 N[/tex]

now the Net force on the box which is sliding down is given as

[tex]F_{net} = F_g - F_k - F_{applied}[/tex]

[tex]F_{net} = 98 - 17 - 50 = 31 N[/tex]

now the work done by net force = change in kinetic energy of the box

[tex]F.d = KE_f - KE_i[/tex]

[tex]31(12) = KE_f - 0[/tex]

[tex]KE_f = 372 J[/tex]

Final answer:

The increase in the kinetic energy of the box is calculated by finding the net work done on the box as it slides down the incline, which is found to be 373.2 J. The closest answer from the options provided is 372 J.

Explanation:

We need to calculate the increase in the kinetic energy of the box as it slides down the incline. First, let's determine the forces acting on the box:

Gravitational force component along the incline: Fg = m*g*sin(θ) = 20.0 kg * 9.81 m/s2 * sin(30°) = 98.1 N

Kinetic friction force: [tex]F_{k}[/tex] = μk * N = μk * m*g*cos(θ) = 0.100 * 20.0 kg * 9.81 m/s2 * cos(30°) = 17.0 N

Applied force up the incline: Fa = 50.0 N

Next, calculate the net force on the box:

Fnet = Fg - [tex]F_{k}[/tex] - Fa = 98.1 N - 17.0 N - 50.0 N = 31.1 N

Now, calculate the work done by the net force, which equals the increase in kinetic energy:

Work = Fnet * d = 31.1 N * 12.0 m = 373.2 J

The closest answer to this calculated value is 372 J.

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

   Volume = 470 cm³

   Density = 3.55 g/cm³

   Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

   Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

   Volume = 470 cm³

   Density of liquid = 1 g/cm³

   [tex]\texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N[/tex]

c) Mass of the water displaced = Volume of body x Density of liquid

   Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

   Weight of rock under water = 16.37 - 4.61  =11.76 N

Answer:

Greatest amount of time to warm up: Aluminum

Then steel,

then copper

Explanation:

The higher the heat capacity, the longer it takes to warm the metal up.

Final answer:

To determine which metal heats up the fastest, you must consider their specific heat capacities. Copper, with the lowest specific heat capacity of 390 J/(kg°C), will heat up the fastest, followed by steel (450 J/(kg°C)) and aluminum (910 J/(kg°C)).

Explanation:

The concept of specific heat capacity is critical in understanding the rate at which different materials will heat up or cool down. Specific heat capacity, denoted by Cmetal, refers to the amount of heat needed to raise the temperature of one kilogram of a substance by one degree Celsius.

The metals in question are steel at 450 J/(kg°C), aluminum at 910 J/(kg°C) and copper at 390 J/(kg°C). The lower the specific heat capacity, the faster a material will reach a higher temperature when exposed to a constant heat source. Hence, according to their specific heat capacities, the metals rank in the following order in terms of heating up quickly: copper (390 J/(kg°C)), steel (450 J/(kg°C)), and aluminum (910 J/(kg°C)). This is because copper has the lowest specific heat capacity and therefore will require less energy to increase in temperature compared to steel and aluminum.

Answer:

Major term is 'things that provide intense gravity'

Minor term is 'extremely dense objects'

Middle term is 'neutron stars'

Explanation:

Answer:

130N

Explanation:

F=(M1+M2)V

F= (70+60)*1

F=130*1

F=130N//

Answer:

66.5 m/s

Explanation:

m = mass of the golf ball = 0.031 kg

F = magnitude of force applied to the ball = 6240 N

Acceleration experienced by the ball is given as

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

[tex]a = \frac{6240}{0.031}[/tex]

a = 201290.32 m/s²

d = distance for which the ball is in contact with the golf club = 0.011 m

v₀ = initial speed of the ball = 0 m/s

v = final speed of the ball = 0 m/s

Using the kinematics equation

v² = v₀² + 2 a d

v² = 0² + 2 (201290.32) (0.011)

v = 66.5 m/s

Answer:

542.06 K

Explanation:

v = rms speed of oxygen molecule = 650 m/s

M = molecular mass of the oxygen molecule = 32 g = 0.032 kg

R = universal gas constant = 8.314 J/(mol K)

T = temperature of the gas

Rms speed of oxygen molecule is given as

[tex]v = \sqrt{\frac{3RT}{M}}[/tex]

[tex]650 = \sqrt{\frac{3(8.314)T}{0.032}}[/tex]

T = 542.06 K

Explanation:

The resistance of a wire is given by :

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

Where

[tex]\rho[/tex] is the resistivity of the wire

l = initial length of the wire

A = initial area of cross section

If length and the area of cross section of the wire is doubled then new length is l' and A', l' = 2 l and A' = 2 A

So, new resistance of the wire is given by :

[tex]R'=\rho\dfrac{l'}{A'}[/tex]

[tex]R'=\rho\dfrac{l}{A}[/tex]

R' = R

So, the resistance of the wire remains the same on doubling the length and the area of wire.