A rock is propelled off a pedestal that is 10 meters off the level ground. The rock leaves the pedestal with a speed of 18 meters per second at an angle above the horizontal of 20 degrees. How high does the rock get, and how far downrange from the pedestal does the rock land?

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:

The speed of the ball is 0.8 m/s.

(D) is correct option.

Explanation:

Given that,

mass of ball = 3.0 kg

length = 50 cm

We need to calculate the speed of the ball

Using relation between the centripetal force and tension

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

[tex]v^2=\dfrac{T\times r}{m}[/tex]

Where, m = mass

T = tension

r = radius

Put the value into the formula

[tex]v=\sqrt{\dfrac{3.8\times50\times10^{-2}}{3.0}}[/tex]

[tex]v=0.8\ m/s[/tex]

Hence, The speed of the ball is 0.8 m/s.

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:

7.66 (litres); 523.66 (N).

Explanation:

all the details are provided in the attached picture. Note, P_i means 'Weight of iron', ρ_i means "Density of iron', ρ_w means 'Density of water', F_f means 'Arhimede force'.

The answers are marked with red colour.

Answer:

2.6 m/s²

Explanation:

m = mass of the object = 7.8 kg

Consider the right direction as positive and left direction as negative

F = horizontal applied force towards right = 50 N

f = frictional force acting towards left = - 30 N

a = acceleration

Force equation for the motion of the object is given as

F + f = ma

50 - 30 = (7.8) a

a = 2.6 m/s²

Answer:

(a):The time what it takes his voice reach the earth via radio waves is t1= 1.28 seconds.

(b): The time what it takes his voice reach from Mars to the earth via radio waves is t2= 186.66 seconds.

Explanation:

d1= 3.85 *10⁸ m

d2= 5.6 *10¹⁰ m

C= 300,000,000 m/s

t1= d1/C

t1= 1.28 s

t2= d2/C

t2= 186.66 s

Answer:

3.014 x 10⁻⁸ N

Explanation:

q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C

v = speed of the jet = 685 m/s

B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T

θ = angle between the magnetic field and direction of motion = 90

magnitude of the magnetic force is given as

F = q v B Sinθ

F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90

F = 3.014 x 10⁻⁸ N

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.

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:

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:

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:

0.5 A

Explanation:

N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm

The induced emf is given by

e = - N dФ/dt

Where, dФ/dt is the rate of change of magnetic flux.

Ф = B A

dФ/dt = A dB/dt

so,

e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V

negative sign shows the direction of magnetic field.

induced current, i = induced emf / resistance = 0.2 / 0.4 = 0.5 A

Answer:

x = 0.95 m

y = 0.166 m

Explanation:

m = mass of the bullet = 24 g = 0.024 kg

v = speed of bullet before collision = 280 m/s

M = mass of the pendulum = 3.7 kg

L = length of the string = 2.8 m

h = height gained by the pendulum after collision

V = speed of the bullet and pendulum combination

Using conservation of momentum

m v = (m + M) V

(0.024) (280) = (0.024 + 3.7) V

V = 1.805 m/s

using conservation of energy

Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination

(m + M) g h = (0.5) (m + M) V²

(9.8) h = (0.5) (1.805)²

h = 0.166 m

y = vertical displacement = h = 0.166 m

x = horizontal displacement

horizontal displacement is given as

x = sqrt(L² - (L - h)²)

x =  sqrt(2.8² - (2.8 - 0.166)²)

x = 0.95 m

Answer:83.17 cm/s

Explanation:

Let positive x be negative direction and negative x be positive direction in this question

General equation of motion of SHM is

x=[tex]Asin\left ( \omega_{n}t\right )[/tex] --------1

where [tex]\omega [/tex]is natural frequency of motion given by

[tex]\omega_n[/tex]=[tex]\sqrt{\frac{k}{m}}[/tex]

Where K is spring constant

here A=23.7cm

And it is given it is again at 23.7 from equilibrium position having passed through the equilibrium once.

i.e. it covers this distance in [tex]\frac{T}{2}[/tex] sec

where T is the time period of oscillation i.e. returning to same place after T sec

therefore T=0.634 sec

differentiating equation 1 we get

v=[tex]A\omega_n[/tex]cos[tex]\left ( \omega_{n}t\right )[/tex]

and [tex]T\times \omega_n[/tex]=[tex]2\pi[/tex]

[tex]\omega_n[/tex]=9.911rad/s

[tex]v[/tex]=[tex]23.7\times 9.911cos\left (789.261\degree\right)[/tex]

v=83.17 cm/s

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:

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:

Power required, P = 0.49 kW

Explanation:

It is given that,

Mass of drum, m = 150 kg

Height, h = 20 m

Time, t = 1 min = 60 sec

Average power required is given by work done divided by total time taken. It is given by :

[tex]P=\dfrac{W}{t}[/tex]

W = F.d

Here, W = mgh

So, [tex]P=\dfrac{mgh}{t}[/tex]

[tex]P=\dfrac{150\ kg\times 9.8\ m/s^2\times 20\ m}{60\ s}[/tex]

P = 490 watts

or P = 0.49 kW

So, the average power required to raise the drum is 0.49 kilo-watts. Hence, this is the required solution.

Final answer:

The average power required to lift a 150-kg drum to a height of 20 meters in 1 minute is calculated by first finding the work done and then dividing by time. The result is approximately 0.4905 kW.

Explanation:

Calculating Required Power to Lift a Drum

The problem involves finding the average power required to raise a certain weight to a specified height over a defined period of time. The weight in question is a 150-kg drum, the height is 20 meters, and the time is 1 minute (60 seconds).

To calculate power, you need to first calculate the work done, which is the product of force and displacement. Since we are lifting the object vertically, the force is the weight of the object, which is mass (m) times the acceleration due to gravity (g), and the displacement is the height (h). The formula for work (W) is:
W = m × g × h

The average power (P), then, is the work done over the time (t) it takes to do it:
P = W / t

Using the values given:

mass (m) = 150 kg

acceleration due to gravity (g) = 9.81 m/s²

height (h) = 20 m

time (t) = 60 s

First, calculate the work done:
W = 150 kg × 9.81 m/s² × 20 m = 29430 J

Now calculate the power required:
P = 29430 J / 60 s = 490.5 W

To convert this into kilowatts (kW), divide by 1000:
P = 0.4905 kW

Therefore, the average power required to lift the drum is approximately 0.4905 kW.

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:

286

Explanation:

p1v1/T1=p2v2/T2

then subtitude your values

T2=760*0.65*273/210*0.040

T2=135/8.4=16+273=289

The pressure-volume work done by the system during the expansion is 5.25 x 10³J.

The work that is done when a fluid is compressed or expanded is known as pressure-volume work. Pressure-volume work occurs whenever there is a change in volume but the outside pressure stays the same.

Here,

The pressure of the system of gas, P = 210 kPa

Initial volume of the system of gas, V₁ = 0.04 m³

Final volume of the system of gas, V₂ = 0.065 m³

The expression for the pressure-volume work done by a system of gas is given by,

Work done,

W = PΔV

W = P(V₂ - V₁)

Applying the values of P, V₁ and V₂,

W = 210 x 10³(0.065 - 0.04)

W = 210 x 10³x 0.025

W = 5.25 x 10³J

Hence,

The pressure-volume work done by the system during the expansion is 5.25 x 10³J.

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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:

2.17 x 10^8 m/s

Explanation:

Angle of incidence, i = 30 degree, refractive index of mineral oil, n = 1.38

Let r be the angle of refraction.

By use of Snell's law

n = Sin i / Sin r

Sin r = Sin i / n

Sin r = Sin 30 / 1.38

Sin r = 0.3623

r = 21.25 degree

Let the speed of light in oil be v.

By the definition of refractive index

n = c / v

Where c be the speed of light

v = c / n

v = ( 3 x 10^8) / 1.38

v = 2.17 x 10^8 m/s

Answer:

Moessbauer Effect = eggy eggs

Explanation:

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:

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:

0.324×10⁻³ m/s²

Explanation:

G=Gravitational constant=6.67408×10⁻⁸ m³/kg s

m₁=m₂=Mass of the ships=42000×10³ kg

r=Distance between the ships=93 m

The ships are considered particles

From Newtons Law of Universal Gravitation

[tex]F=G\frac {m_1m_2}{r^2}\\Here\ m_1=m_2\ hence\ m_1\times m_2=m_1^2\\\Rightarrow F=G\frac {m_1^2}{r^2}\\\Rightarrow F=\frac{6.67408\times 10^{-8}\times (42000\times 10^3)^2}{93^2}\\\Rightarrow F=13612.0674\ N\\[/tex]

[tex]F=ma\\\Rightarrow 13612.0674=42000\times 10^3a\\\Rightarrow a=0.324\times 10^{-3}\ m/s^2[/tex]

∴Acceleration of one of the liners toward the other due to their mutual gravitational attraction is 0.324×10⁻³

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:

Distance between two adjacent wave crests = 24m

Explanation:

Distance= speed × time

Distance traveled by waves in 60 seconds (15 crests)= 15 × distance

15 × distance = 6,0 (meters/second) × 60 seconds

distance = (360 meters) / 15 = 24 meters (between two adyacent waves)