The Space Shuttle rocket ship starts vertically upward from the launch pad acceleration of 6.5 m/sec^2. (a) Find how much time it takes for the rocket ship to accelerate to 100 m/s. (b) At what height above the ground will the ship reach the velocity of 100 m/s?

Answer:

The initial temperature of the gas was of T1= 112ºC .

Explanation:

T1= ?

T2= 35 ºC = 308.15 K

V1= 0.225 L

V2= 0.18 L

T2* V1 / V2 = T1

T1= 385.18 K = 112ºC

Answer:

diameter is 13.46 mm

Explanation:

length of rod = 200 mm  = 0.2 m

compression force = 80,000 N

factor of safety = 2.5

Sy = 496 MPa

E = 71 GPa

to find out

diameter

solution

first we calculate the allowable stress i.e.  = Sy/factor of safety

allowable stress = 496/ 2.5= 198.4 MPa  198.4 × [tex]10^{6}[/tex] Pa

now we calculate the diameter d by the Euler's equation i.e.

critical load = [tex]\pi ^{2}[/tex] E × moment of inertia / ( K × length )²   ..........1

now we calculate the critical load i.e.  allowable stress × area

here area = [tex]\pi[/tex] /4 × d²  

so critical load = 198.4 ×  [tex]\pi[/tex] /4 × d²

and K = 1 for pin ends  

and moment of inertia is =   [tex]\pi[/tex] / 64 × [tex]d^{4}[/tex]

put all value in equation 1 and we get d

198.4 ×[tex]10^{6}[/tex] ×  [tex]\pi[/tex] /4 × d²  = [tex]\pi ^{2}[/tex] 71 × [tex]10^{9}[/tex]  × [tex]\pi[/tex] / 64 × [tex]d^{4}[/tex]  / ( 1 × 0.2 )²

155.8229× [tex]10^{6}[/tex]  × d²  =   700.741912× [tex]10^{9}[/tex]× 0.049087×  [tex]d^{4}[/tex] / 0.04

d=0.01346118 m

d = 13.46 mm

diameter is 13.46 mm

Answer:

h = 3.10 m

Explanation:

As we know that after each bounce it will lose its 11% of energy

So remaining energy after each bounce is 89%

so let say its initial energy is E

so after first bounce the energy is

[tex]E_1 = 0.89 E[/tex]

after 2nd bounce the energy is

[tex]E_2 = 0.89(0.89 E)[/tex]

After third bounce the energy is

[tex]E_3 = (0.89)(0.89)(0.89)E[/tex]

here initial energy is given as

[tex]E = mgH_o[/tex]

now let say final height is "h" so after third bounce the energy is given as

[tex]E_3 = mgh[/tex]

now from above equation we have

[tex]mgh = (0.89)(0.89)(0.89)(mgH)[/tex]

[tex]h = 0.705H[/tex]

[tex]h = 0.705(4.4 m)[/tex]

[tex]h = 3.10 m[/tex]

Answer:

Period of oscillation, T = 2 sec

Explanation:

It is given that,

Mass of the object, m = 5 kg

Spring constant of the spring, k = 50 N/m

This object is hanging from a coiled spring. We need to find the period of oscillation of the spring. The time period of oscillation of the spring is given by :

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

[tex]T=2\pi\sqrt{\dfrac{5\ kg}{50\ N/m}}[/tex]

T = 1.98 sec

or

T = 2 sec

So, the period of oscillation is about 2 seconds. Hence, this is the required solution.

Answer:

[tex]\frac{dr}{dt} = 0.14 in/s[/tex]

Explanation:

As the volume of the cylinder is constant here so we can say that its rate of change in volume must be zero

so here we can say

[tex]\frac{dV}{dt} = 0[/tex]

now we have

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

now find its rate of change in volume with respect to time

[tex]\frac{dV}{dt} = 2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}[/tex]

now we know that

[tex]\frac{dV}{dt} = 0 = \pi r(2h \frac{dr}{dt} + r\frac{dh}{dt})[/tex]

given that

h = 5 inch

r = 2 inch

[tex]\frac{dh}{dt} = - 0.7 in/s[/tex]

now we have

[tex]0 = 2(5) \frac{dr}{dt} + 2(-0.7)[/tex]

[tex]\frac{dr}{dt} = 0.14 in/s[/tex]

Final answer:

The change in radius δr/δt of a cylinder with a constant volume is found to be 0.28 inches per second when the height is decreasing at 0.7 inches per second and r = 2 inches, h = 5 inches.

Explanation:

The question involves using calculus to find the rate of change of the radius of a cylinder given a constant volume and a known rate of change in height. The rate of change of volume for a cylinder, which is 0 because the volume doesn't change, can be described as δV/δt = (πr²) (δh/δt) + (2πrh) (δr/δt) = 0. Given the height is decreasing at 0.7 in/sec, we can find the rate of change of the radius δr/δt. Using the known values of r = 2 inches and h = 5 inches, we can solve for δr/δt.

Starting with the equation for the volume of a cylinder V = πr²h, since the volume is constant, we take the derivative with respect to time to obtain 0 = π(2×r×δr/δt×h + r²×δh/δt). Substituting the known values gives 0 = π(2×2×δr/δt×5 + 2²×(-0.7)), which simplifies to 0 = 20πδr/δt - 5.6π. From this we can solve for δr/δt = 5.6π / 20π = 0.28 in/sec.

The rate of change of the radius is 0.28 inches per second when the height is decreasing at 0.7 inches per second, and the radius is 2 inches while the height is 5 inches.

In this situation, the neon gas should also have a temperature of 175 K, the same as the helium gas, given all the conditions and the principle of the ideal gas law.

The problem you're working on involves understanding the ideal gas law, which is PV = nRT. This equates the pressure (P), volume (V) of the gas, and temperature (T). There are two important points to consider. Firstly, the number of molecules or moles (n) and the ideal gas constant (R) aren't changing in this situation. Secondly, the volume and pressure are the same for both gases.

Given that the masses, volume, and pressure are the same for both helium and neon, we conclude that the number of moles (n) for helium and neon are equal (because the mass density is the same and for ideal gas mass density (ρ) = PM/RT, where M is molar mass). The temperature ratio should be the same as the ratio of Kelvin temperatures of two gases.

So, we can safely say that since the gases obey the ideal gas law, and all conditions are held constant aside from the identity of the gas and the temperature, the temperature of the neon gas must also be 175 Kelvin like the helium gas.

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Answer:

(i) false

(ii) true

(iii) true

(iv) false

Explanation:

(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.

So, this is false.

(ii) For monoatomic gas

Cp = 5R/2, Cv = 3R/2

So, thier ratio

Cp / Cv = 5 / 3 = 1.67

This statement is true.

(iii) for diatomic gases

Cp = 7R/2, Cv = 5R/2

Cp / Cv = 7 / 5 = 1.4

This statement is true.

(iv) It is false.

Answer:

a) [tex]a_c= 1.33 m/s^2 [/tex]

b) a= 1.79 m/s²

   θ = 41.98⁰

Explanation:

arc radius  = 12 m

constant speed = 4.00 m/s

(a) centripetal acceleration

     [tex]a_c=\frac{v^2}{R}[/tex]

     [tex]a_c=\frac{4^2}{12} [/tex]

                  = 1.33 m/s²

(b) now we have given

        [tex]a_t= \ 1.20 m/s^2 [/tex]

        now,

         [tex]a=\sqrt{a^2_c+ a^2_t}[/tex]

         [tex]a=\sqrt{1.33^2+ 1.20^2}[/tex]

            a= 1.79 m/s²

 direction

[tex]\theta = tan^{-1}(\frac{a_t}{a_r} )[/tex]

[tex]\theta = tan^{-1}(\frac{1.2}{1.33} )[/tex]

     θ = 41.98⁰

The centripetal acceleration of the hawk is 1.33 m/s².

The resultant acceleration  of the hawk at the given moment is 1.79 m/s².

The direction resultant acceleration of the hawk is 48⁰.

The given parameters;

The centripetal acceleration of the hawk is calculated as follows;

[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(4)^2}{12} \\\\a_c = 1.33 \ m/s^2[/tex]

The resultant acceleration is calculated as;

[tex]a = \sqrt{a_c^2 + a_t} \\\\a = \sqrt{(1.33)^2 + (1.2)^2} \\\\a = 1.79 \ m/s^2[/tex]

The direction of the acceleration is calculated as follows;

[tex]tan(\theta) = \frac{a_c}{a_t} \\\\\theta = tan^{-1} ( \frac{a_c}{a_t} )\\\\\theta = tan^{-1} ( \frac{1.33}{1.2} )\\\\\theta = 48^0[/tex]

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Answer:

a) Total displacement  = 3986.54 m

b) Average speeds

      Leg 1 ->  11.22 m/s

      Leg 2 ->  22.44 m/s

      Leg 3 ->  11.20 m/s

      Complete trip ->  21.63 m/s

Explanation:

a) Leg 1:

Initial velocity, u =  0 m/s

Acceleration , a = 2.04 m/s²

Time, t = 11 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    s = 0 x 11 + 0.5 x 2.04 x 11²

    s = 123.42 m

Leg 2:

We have equation of motion v = u + at

Initial velocity, u =  0 m/s

Acceleration , a = 2.04 m/s²

Time, t = 11 s

Substituting

   v = 0 + 2.04 x 11 = 22.44 m/s

We have equation of motion s= ut + 0.5 at²

Initial velocity, u =  22.44 m/s

Acceleration , a = 0 m/s²

Time, t = 2.85 min = 171 s

Substituting

   s= ut + 0.5 at²

    s = 22.44 x 171 + 0.5 x 0 x 171²

    s = 3837.24 m

a) Leg 3:

Initial velocity, u =  22.44 m/s

Acceleration , a = -9.73 m/s²

Time, t = 2.31 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    s = 22.44 x 2.31 + 0.5 x -9.73 x 2.31²

    s = 25.88 m

Total displacement = 123.42 + 3837.24 + 25.88 = 3986.54 m

Average speed is the ratio of distance to time.

b) Leg 1:

        [tex]v_{avg}=\frac{123.42}{11}=11.22m/s[/tex]

 Leg 2:

        [tex]v_{avg}=\frac{3837.24}{171}=22.44m/s[/tex]

Leg 3:

        [tex]v_{avg}=\frac{25.88}{2.31}=11.20m/s[/tex]

Complete trip:

        [tex]v_{avg}=\frac{3986.54}{11+171+2.31}=21.63m/s[/tex]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

Answer:

3kg

Explanation:

impulse = MV

then

m1v1=m2v2

when the values are subtitude

then

m2=1.2*25/10

m2=30kg//

Answer:

Electric charge, Q = 55.69 C

Explanation:

It is given that,

Electric current drawn by the compressor, I = 23.7 A

Time taken, t = 2.35 s

We need to find the electric charge passes through the circuit during this period. The definition of electric current is given by total charge divided by total time taken.

[tex]I=\dfrac{q}{t}[/tex]

Where,

q is the electric charge

[tex]q=I\times t[/tex]

[tex]q=23.7\ A\times 2.35\ s[/tex]

q = 55.69 C

So, the electric charge passes through the circuit during this period is 55.69 C. Hence, this is the required solution.

Answer:

The magnitude of the magnetic field inside the solenoid is [tex]7.3\times10^{-3}\ T[/tex].

Explanation:

Given that,

Length = 81.0 cm

Radius = 1.70 cm

Number of turns = 1300

Current = 3.60 A

We need to calculate the magnetic field

Using formula of magnetic field inside the solenoid

[tex]B =\mu nI[/tex]

[tex]B =\mu\dfrac{N}{l}I[/tex]

Where, [tex]\dfrac{N}{l}[/tex]=Number of turns per unit length

I = current

B = magnetic field

Put the value into the formula

[tex]B =4\pi\times10^{-7}\times\dfrac{1300}{81.0\times10^{-2}}\times3.60[/tex]

[tex]B = 7.3\times10^{-3}\ T[/tex]

Hence, The magnitude of the magnetic field inside the solenoid is [tex]7.3\times10^{-3}\ T[/tex].

Answer:

Radius, r = 0.36 meters

Explanation:

It is given that,

Speed of cosmic ray electron, [tex]v=6.5\times 10^6\ m/s[/tex]

Magnetic field strength, [tex]B=10\times 10^{-5}\ T=10^{-4}\ T[/tex]

We need to find the radius of circular path the electron follows. It is given by :

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

[tex]r=\dfrac{mv}{qB}[/tex]

[tex]r=\dfrac{9.1\times 10^{-31}\ kg\times 6.5\times 10^6\ m/s}{1.6\times 10^{-19}\times 10^{-4}\ T}[/tex]

r = 0.36 meters

So, the radius of circular path is 0.36 meters. Hence, this is the required solution.

Answer:

Moessbauer Effect = eggy eggs

Explanation:

Answer:

T = 2.5 lb

N= -0.33 lb

Explanation:

given

r = 9 in

[tex]\dot{r} =-3.6 in/s and\ \ddot{r} = 0[/tex]

[tex]\dot{\theta} = 6.3\ rad/s and\ \ddot{\theta} = 2.1\ rad/s^2[/tex]

[tex]-T = m a_r = m(\ddot{r} -r{\dot{\theta}^2)[/tex]

[tex]N= m a_{\theta} = m(r\ddot{\theta}+2\dot{r}\dot{\theta}})[/tex]

[tex]T= mr{\dot{\theta}^2 = \frac{3}{386.4}(9)(6)^2 =2.5lb[/tex]

[tex]N= m(r\ddot{\theta}+2\dot{r}\dot{\theta}})=\frac{3}{386.4}[9(-2)+2(-2)(6)]=-0.326 lb[/tex]

Answer:

The electric field between the plates is 120 V/m.

(c) is correct option.

Explanation:

Given that,

Potential difference = 12 volt

Distance = 10 cm = 0.1 m

We need to calculate the electric field between the plates

Using formula of electric field

[tex]E = \dfrac{V}{d}[/tex]

Where, V = potential difference

d = distance between the plates

Put the formula

[tex]E =\dfrac{12}{0.1}[/tex]

[tex]E=120\ V/m[/tex]

Hence, The electric field between the plates is 120 V/m.

Answer:

Product side

Explanation:

When water vapor reacts reversibly with solid carbon to yield a mixture of hydrogen gas and carbon monoxide and we continually add more water vapor to the reaction the equilibrium of the reaction shifts to the product side.

Because gaseous water is reactant that appears in the reaction quotient expression.

[tex]H_{2}O+ C_{s}\leftrightharpoons H_{2}_{g}+ CO[/tex]

When we add more water vapor to the reaction the product formation is increased. The reaction goes in forward direction affecting the equilibrium.

Answer:

1.8 x 10⁻⁷ N/m

Explanation:

[tex]F_{max}[/tex] = maximum force with which the optical trap can pull = 11 x 10⁻¹⁵ N

x = stretch caused in DNA molecule due to the force = 0.06 x 10⁻⁶ m

k = spring constant of the spring

Maximum force is given as

[tex]F_{max}= k x[/tex]

[tex]11\times 10^{-15}= k (0.06\times 10^{-6})[/tex]

k = 1.8 x 10⁻⁷ N/m

The spring constant of the DNA molecule is calculated using Hooke's Law with the given force of 11 fN and stretch distance of 0.06 μm, resulting in a spring constant of approximately 183.33 N/m.

To calculate the spring constant (k) of a DNA molecule modeled as a spring, we can use Hooke's Law, which states that the force (F) applied to stretch or compress a spring is directly proportional to the displacement (x) it causes, as represented by the equation F = kx. The optical trap pulls with a maximum force of 11 fN and stretches the DNA molecule by 0.06 μm. Using the given values, the spring constant (k) can be calculated as:

k = F / x

Therefore, k = 11 fN / 0.06 μm, and to ensure the units are consistent, we convert 0.06 μm to meters (0.06 μm = 0.06 x 10-6 m).

k = 11 x 10-15 N / 0.06 x 10-6 m

k = (11 / 0.06) x 10-9 N/m

k ≈ 183.33 N/m

The spring constant of the DNA molecule is therefore approximately 183.33 N/m.

Answer:

A primitive solid is a 'building block' that you can use to work with in 3D. Rather than extruding or revolving an object, AutoCAD has some basic 3D shape commands at your disposal.

Explanation:

Answer:

The kinetic energy of an electron is [tex]1.54\times10^{-15}\ J[/tex]

Explanation:

Given that,

Distance = 0.1 nm

We need to calculate the momentum

Using uncertainty principle

[tex]\Delta x\Delta p\geq\dfrac{h}{4\pi}[/tex]

[tex]\Delta p\geq\dfrac{h}{\Delta x\times 4\pi}[/tex]

Where, [tex]\Delta p[/tex] = change in momentum

[tex]\Delta x[/tex] = change in position

Put the value into the formula

[tex]\Delta p=\dfrac{6.6\times10^{-34}}{4\pi\times10^{-10}}[/tex]

[tex]\Delta p=5.3\times10^{-23}[/tex]

We need to calculate the kinetic energy for an electron

[tex]K.E=\dfrac{p^2}{2m}[/tex]

Where, P = momentum

m = mass of electron

Put the value into the formula

[tex]K.E=\dfrac{(5.3\times10^{-23})^2}{2\times9.1\times10^{-31}}[/tex]

[tex]K.E=1.54\times10^{-15}\ J[/tex]

Hence, The kinetic energy of an electron is [tex]1.54\times10^{-15}\ J[/tex]

Answer:

Part a)

A = 66.2 m

Part b)

Angle = 38.35 degree

Explanation:

Part a)

Length of the vector is the magnitude of the vector

here we know that

[tex]A_x = 52.0 m[/tex]

[tex]A_y = 41.0 m[/tex]

now we have

[tex]A = \sqrt{A_x^2 + A_y^2}[/tex]

[tex]A = \sqrt{52^2 + 41^2}[/tex]

[tex]A = 66.2 m[/tex]

Part b)

Angle made by the vector is given as

[tex]tan\theta = \frac{A_y}{A_x}[/tex]

[tex]tan\theta = \frac{41}{52}[/tex]

[tex]\theta = 38.25 degree[/tex]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a linear function of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

[tex]m=\frac{Y-Y_{1}}{X-X_{1}}[/tex]  (1)

Where:

[tex]m=5.7[/tex] is the slope of the line

[tex]Y_{1}=2390.7pounds[/tex] is the airplane weight with  51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

[tex]X_{1}=51gallons[/tex] is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

[tex](X_{1},Y_{1})=(51,2390.7)[/tex]

Rewritting (1):

[tex]Y=m(X-X_{1})+Y_{1}[/tex]  (2)

As Y is a function of X:

[tex]Y=f_{(X)}=m(X-X_{1})+Y_{1}[/tex]  (3)

Substituting the known values:

[tex]f_{(X)}=5.7(X-51)+2390.7[/tex]  (4)

[tex]f_{(X)}=5.7X-290.7+2390.7[/tex]  (5)

[tex]f_{(X)}=5.7X+2100[/tex]  (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

[tex]f_{(81)}=5.7(81)+2100[/tex]  (7)

[tex]f_{(81)}=2561.7[/tex]  (8)   This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

Final answer:

To find the weight of the airplane with 81 gallons of fuel, calculate the additional fuel weight (30 gallons
* 5.7 pounds/gallon = 171 pounds) and add it to the initial weight (2390.7 pounds + 171 pounds = 2561.7 pounds).

Explanation:

The question asks to calculate the weight of an airplane with a different amount of fuel in its tank, given the weight with a specific amount and the slope of weight increase per gallon of fuel added. To find the new weight, we first calculate the weight increase due to the additional fuel, then add this increase to the original weight of the airplane.

Initial weight with 51 gallons: 2390.7 pounds
Fuel increase: 81 gallons - 51 gallons = 30 gallons
Slope (rate of weight increase): 5.7 pounds per gallon
Additional weight from extra fuel: 30 gallons
* 5.7 pounds/gallon = 171 pounds
New weight with 81 gallons: 2390.7 pounds + 171 pounds = 2561.7 pounds

Answer:

1270.44 Hz

Explanation:

[tex]v_{L}[/tex]  = velocity of the our car = 35.0 m/s

[tex]v_{P}[/tex]  = velocity of the police car = ?

[tex]v_{S}[/tex]  = velocity of the sound = 343 m/s

[tex]f_{app}[/tex]  = frequency observed as police car approach = 1310 Hz

[tex]f_{rec}[/tex]  = frequency observed as police car go away = 1240 Hz

[tex]f[/tex]  = actual frequency of police siren

Frequency observed as police car approach is given as

[tex]f_{app}= \frac{(v_{s}-v_{L})f}{v_{s} -v_{P} }[/tex]

inserting the values

[tex]1310 = \frac{(343 - 35)f}{343 -v_{P} }[/tex]                          eq-1

Frequency observed as police car goes away is given as

[tex]f_{rec}= \frac{(v_{s} + v_{L})f}{v_{s} + v_{P} }[/tex]

inserting the values

[tex]1240 = \frac{(343 + 35)f}{343 + v_{P} }[/tex]                          eq-2

Dividing eq-1 by eq-2

[tex]\frac{1310}{1240} = \left ( \frac{343 - 35}{343 - v_{P} } \right )\frac{(343 + v_{P})}{343 + 35 }\\[/tex]

[tex]v_{P}[/tex]  = 44.3 m/s

Using eq-1

[tex]1310 = \frac{(343 - 35)f}{343 - 44.3 }[/tex]

f = 1270.44 Hz

Answer:

alternative E- none of these answers

Explanation:

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at a given point within the fluid, due to the force of gravity. Hydrostatic pressure increases in proportion to depth measured from the surface because of the increasing weight of fluid exerting downward force from above.

The formula is :

P= d x g x h

p: hydrostatic pressure (N/m²)

d: density (kg/m³) density of seawater is 1,030 kg/m³

g: gravity (m/s²) ≅ 9.8m/s²

h: height (m)

The hydrostatic pressure at 20,000 leagues under the sea is approximately 1,002,500,000,000 Pa.

The hydrostatic pressure at 20,000 leagues under the sea can be calculated using the equation for pressure in a fluid, which is given by P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth below the surface.

Since a league is the distance a person can walk in one hour, we need to convert it to meters. Assuming an average walking speed of 5 km/h, a league is equal to 5 km. Therefore, 20,000 leagues is equal to 100,000 km.

The pressure at this depth can be calculated using the known values: density of seawater is about 1025 kg/m³ and acceleration due to gravity is 9.8 m/s². Plugging in these values, we get P = (1025 kg/m³)(9.8 m/s²)(100,000,000 m) = 1,002,500,000,000 Pa.

Therefore, the correct answer is none of the provided options. The hydrostatic pressure at 20,000 leagues under the sea is approximately 1,002,500,000,000 Pa.

Answer:

0.28 J

Explanation:

Since the air-hockey puck was initially at rest

KE₀ = initial kinetic energy of the air-hockey puck = 0 J

KE = final kinetic energy of the air-hockey puck

m = mass of air-hockey puck 0.5 kg

F = net force = 0.4 N

d = distance moved = 0.7 m

Using work-change in kinetic energy

F d = (KE - KE₀)

(0.4) (0.7) = KE - 0

KE = 0.28 J

Answer:

2 kg m^2

Explanation:

M = 1 kg, R = 1 m

The moment of inertia of the hoop about its axis perpendicular to its plane is

I = M R^2

The moment of inertia of the hoop about its edge perpendicular to it splane is given by the use of parallel axis theorem

I' = I + M x (distance between two axes)^2

I' = I + M R^2

I' = M R^2 + M R^2

I' = 2 M R^2

I' = 2 x 1 x 1 x 1 = 2 kg m^2

The moment of inertia of a hoop about an axis perpendicular to its plane at an edge is calculated using the Parallel-Axis Theorem and for a hoop with mass M = 1kg and radius R = 1m, it is 2 kg*m².

Calculation of the Moment of Inertia for a Hoop

To find the moment of inertia of a hoop with mass M = 1kg and radius R = 1m about an axis perpendicular to the hoop's plane at an edge, we use the Parallel-Axis Theorem. The moment of inertia of a hoop about its central axis (through its center, perpendicular to the plane) is MR². According to the Parallel-Axis Theorem, the moment of inertia about an axis parallel to this but passing through the edge of the hoop is given by I = MR₂ + MR₂ (because the distance from the central axis to the outer edge of the hoop is R). Thus, the moment of inertia for the hoop about the edge is 2MR₂ which simplifies to 2 * 1kg * (1m)² = 2 kg*m².

Answer:

Explanation:

As the dielectric is inserted between the plates of a capacitor, the capacitance becomes K times and the electric field between the plates becomes 1 / K times the original value. Where, K be the dielectric constant.

Answer:

C. 150 N to the left

Explanation:

If we take right to be positive and left to be negative, then:

∑F = -450 N + 300 N

∑F = -150 N

The net force is 150 N to the left.

Answer:

(C) 150 N to the left

Explanation:

It is given that,

Force acting in left side, F = 450 N

Force acting in right side, F' = 300 N

Let left side is taken to be negative while right side is taken to be positive. So,

F = -450 N

F' = +300 N

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

[tex]F_{net}=-450\ N+300\ N[/tex]

[tex]F_{net}=-150\ N[/tex]    

So, the net force on the table is 150 N and it is acting to the left side. Hence, the correct option is (c).

Answer:

a) [tex]4.1\times 10^{6} \frac{m}{s}[/tex]

b) [tex]9.2\times 10^{-8} s[/tex]

c) [tex]5.6\times 10^{-14} J[/tex]

d) 175000 volts

Explanation:

a)

[tex]q[/tex]  = magnitude of charge on the alpha particle = 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

[tex]m[/tex]  = mass of alpha particle = 4 x 1.67 x 10⁻²⁷ kg = 6.68 x 10⁻²⁷ kg

[tex]r[/tex]  = radius of circular path = 5.99 cm = 0.0599 m

[tex]B[/tex]  = magnitude of magnetic field = 1.43 T

[tex]v[/tex] = speed of the particle

Radius of circular path is given as

[tex]r = \frac{mv}{qB}[/tex]

[tex]0.0599 = \frac{(6.68\times 10^{-27})v}{(3.2\times 10^{-19})(1.43)}[/tex]

[tex]v = 4.1\times 10^{6} \frac{m}{s}[/tex]

b)

Time period is given as

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

[tex]T = \frac{2(3.14)(6.68\times 10^{-27})}{(3.2\times 10^{-19})(1.43)}[/tex]

[tex]T = 9.2\times 10^{-8} s[/tex]

c)

Kinetic energy is given as

[tex]K = (0.5)mv^{2}[/tex]

[tex]K = (0.5)(6.68\times 10^{-27})(4.1\times 10^{6})^{2}[/tex]

[tex]K = 5.6\times 10^{-14} J[/tex]

d)

ΔV = potential difference

Using conservation of energy

q ΔV = K

(3.2 x 10⁻¹⁹) ΔV = 5.6 x 10⁻¹⁴

ΔV = 175000 volts