There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather conditions, in an open field, the strength of this electric field is 95.0 N/C . A spherical pollen grain with a radius of 12.0 μm is released from its parent plant by a light breeze, giving it a net charge of −0.700 fC (where 1 fC=1×10−15 C ). What is the ratio of the magnitudes of the electric force to the gravitational force, ????electric/????grav , acting on the pollen? Pollen is primarily water, so assume that its volume mass density is 1000 kg/m3 , identical to the volume mass density of water.

Swimming downstream and upstream in a river with a current of 0.52 m/s results in approximately an 86% increase in time compared to swimming the same distance in still water at a speed of 1.73 m/s.

1. Understanding the problem:

You can swim 1.73 m/s in still water.

The river current flows at 0.52 m/s.

We need to find the percentage increase in time to swim downstream and back upstream compared to swimming the same distance in still water.

2. Effective speeds:

Downstream: Your speed increases with the current.

Effective speed = swimming speed + current speed = 1.73 m/s + 0.52 m/s = 2.25 m/s

Upstream: Your speed decreases against the current.

Effective speed = swimming speed - current speed = 1.73 m/s - 0.52 m/s = 1.21 m/s

3. Calculating time for each leg:

Let D be the distance for the entire round trip.

Time downstream: D / 2.25 m/s

Time upstream: D / 1.21 m/s

4. Finding the time difference:

Time difference = Time upstream - Time downstream

Time difference = (D / 1.21 m/s) - (D / 2.25 m/s)

5. Calculating percentage increase:

To find the percentage increase in time compared to still water, divide the time difference by the time taken to swim downstream and multiply by 100.

Percentage increase = [(D / 1.21 m/s) - (D / 2.25 m/s)] / (D / 2.25 m/s) * 100

6. Simplifying the equation:

After simplifying, you'll find the percentage increase to be approximately 86%.

Therefore, it takes approximately 86% longer to complete the round trip compared to swimming the same distance in still water.

Answer:

c) 17 m

Explanation:

From the conservation of energy.

Change in kinetic energy= change in potential energy

Now,

[tex]\Delta KE=\Delta PE[/tex]

Now according to the situation.

[tex]\frac{1}{2}mv^{2}=mgh\\h=\frac{v^{2} }{2g}[/tex]

Here, v is the velocity, m is the mass of an object, h is the height, g is the acceleration due to gravity.

Given that the speed of the bicycle is

[tex]v=18m/s[/tex]

And the acceleration due to gravity is

[tex]g=9.8 m/s^{2}[/tex]

Substitute these values.

[tex]h=\frac{18^{2} }{2\times 9.8}\\h=16.53m[/tex]

Which means it is equivalent to 17 m

Therefore, the maximum height is 17 m

Answer:

The acceleration and time are 588 m/s² and 0.071 sec respectively.

Explanation:

Given that,

Speed = 42.0 m/s

Distance = 1.50 m

(a). We need to calculate the acceleration

Using equation of motion

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

[tex]a=\dfrac{v^2-u^2}{2s}[/tex]

Put the value in the equation

[tex]a=\dfrac{42.0^2-0}{2\times1.50}[/tex]

[tex]a=588\ m/s^2[/tex]

(b). We need to calculate the time

Using equation of motion

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

[tex]t=\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{42.0-0}{588}[/tex]

[tex]t=0.071\ sec[/tex]

Hence, The acceleration and time are 588 m/s² and 0.071 sec respectively.

Answer:

x = -29.032m

Explanation:

Since the third charge is positive, it cannot be between the other two charges, because it would be repelled by the positive one and attracted by the negative one, so the electric force would never be zero.

This leaves only two options: To the left of the positive one or to the right of the negative one.

If it was located on the right of the negative charge, the force of the positive charge would be weaker because of both the distance is larger and its charge is smaller than the negative charge. So, there is no point the would make the result force equal zero.

This means that the third charge has to be at the left of the positive charge. With this in mind, we make the calculations:

[tex]F_{13}=K*\frac{Q_{1}*Q_{3}}{d^{2}} =F_{23}=K*\frac{Q_{2}*Q_{3}}{(x_{2}+d)^{2}}[/tex]

Replacing the values of Q1=1.77, Q2=4.09, X2=15.1, we solve for d and get two possible results:

d1 = 29.032m   and d2 = -5.99m

Since we assumed in our formula that the third charge was on the left of the positive charge, the distance d has to be positive so that our final result can be a negative position. This is X = -d

This way, we get:

X = -29.032m

Answer:

Acceleration, [tex]a=14970.05\ km/h^2[/tex]

Explanation:

Given that,

Initially, the jetliner is at rest, u = 0

Final speed of the jetliner, v = 250 km/h

Time taken, t = 1 min = 0.0167 h

We need to find the acceleration of the jetliner. The mathematical expression for the acceleration is given by :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{250}{0.0167}[/tex]

[tex]a=14970.05\ km/h^2[/tex]

So, the acceleration of the jetliner is [tex]14970.05\ km/h^2[/tex]. Hence, this is the required solution.

Answer:

The angular speed of the pendulum is 3.138 rad/s.

Explanation:

Given that,

Length = 99.5 cm

Mass = 185 g

We need to calculate the angular speed

Using formula of angular speed

[tex]\omega=\sqrt{\dfrac{g}{l}}[/tex]

Where, l = length

g = acceleration due to gravity

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{9.80}{99.5\times10^{-2}}}[/tex]

[tex]\omega=3.138\ rad/s[/tex]

Hence, The angular speed of the pendulum is 3.138 rad/s.

Answer:

pressure of water will be 49.7 atm

Explanation:

given data

pressure = 30 psi = 2.04 atm

water = 2 pound = 907.18

mole of water vapor = 907.19 /2 = 50.4 mole

volume = 4 ft³ = 113.2 L

temperature = 40 F = 277.59 K

to find out

pressure of water

solution

we will apply here ideal gas condition

that is

PV = nRT  .......................1

put here all value and here R = 0.0821 , T temperature and V volume and P pressure and n is no of mole

and we get here temperature

PV = nRT  

2.04 × 113.2 = 50.4×0.0821×T

solve it and we get

T = 55.8 K

so we have given right chamber has twice the volume of the left chamber i.e

volume = twice of volume + volume

volume = 2(113.2) + 113.2

volume = 339.6 L

so from equation 1 pressure will be

PV = nRT

P(339.6) = 50.4 × ( 0.0821) × (277.59)

P = 3.3822 atm = 49.7 atm

so pressure of water will be 49.7 atm

Answer with Explanation:

The direction of the electric field line at any point gives us the direction of the electric force that will act on a positive charge if placed at the point. We know that if we place a charge in an electric field it will experience a force, as we know that force is a vector quantity hence it requires both magnitude and direction for it's complete description. The direction of this electric force that acts on a charge is given by the direction of the electric field in the space. In case the charge is negatively charged electric force will act on it in the direction opposite to the direction of electric field at the point.

Answer:

6.45 a

Explanation:

Charge on O, q1 = 4q

Charge on A, q2 = - 3 q

OA = a

Let the net force is zero at point P, where AP = x , let a charge Q is placed at P.

The force on point P due to the charge q1 = The force on point P due to the  

                                                                          charge q2

By using Coulomb's law

[tex]\frac{Kq_{1}Q}{OP^{2}}=\frac{Kq_{2}Q}{AP^{2}}[/tex]

[tex]\frac {K4qQ}{(a+x)^{2}}= \frac {K3qQ}{x^{2}}[/tex]

[tex]\frac{a+x}{x}=\sqrt{\frac{4}{3}}[/tex]

[tex]\frac{a+x}{x}=1.155[/tex]

a + x = 1.155 x

0.155 x = a

x = 6.45 a

Thus, the force is zero at x = 6.45 a.

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= [tex]\frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}[/tex]

= 1.72 x 10³ N.

Answer:

450 pm

Explanation:

The electron is held in orbit by an electric force, this works as the centripetal force. The equation for the centripetal acceleration is:

a = v^2 / r

The equation for the electric force is:

F = q1 * q2 / (4 * π * e0 * r^2)

Where

q1, q2: the electric charges, the charge of the electron is -1.6*10^-19 C

e0: electric constant (8.85*10^-12 F/m)

If we divide this force by the mass of the electron we get the acceleration

me = 9.1*10^-31 kg

a = q1 * q2 / (4 * π * e0 * me * r^2)

v^2 / r = q1 * q2 / (4 * π * e0 * me * r^2)

We can simplify r

v^2 = q1 * q2 / (4 * π * e0 * me * r)

Rearranging:

r = q1 * q2 / (4 * π * e0 * me * v^2)

r = 1.6*10^-19 * 1.6*10^-19 / (4 * π * 8.85*10^-12 * 9.1*10^-31 * (7.5*10^5)^2) = 4.5*10^-10 m = 450 pm

Answer:

shadow length 7.67 cm

Explanation:

given data:

refractive index of water is 1.33

by snell's law we have

[tex]n_{air} sin30 =n_{water} sin\theta[/tex]

[tex]1*0.5 = 1.33*sin\theta[/tex]

solving for[tex] \theta[/tex]

[tex]sin\theta = \frac{3}{8}[/tex]

[tex]\theta = sin^{-1}\frac{3}{8}[/tex]

[tex]\theta =  22 degree[/tex]

from shadow- stick traingle

[tex]tan(90-\theta) = cot\theta = \frac{h}{s}[/tex]

[tex]s = \frac{h}{cot\theta} = h tan\theta[/tex]

s = 19tan22 = 7.67 cm

s = shadow length

The calculated shadow length is approximately 10.97 cm.

To determine the length of the shadow cast by the stick, we can use basic trigonometry.

Specifically, we'll use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle.

Given:

The formula for tangent is:

tan(θ) = opposite / adjacent

Here, the opposite side is the height of the stick, and the adjacent side is the length of the shadow. Thus, we can rearrange this formula to solve for the length of the shadow (adjacent side):

adjacent = opposite / tan(θ)

Substituting the given values:

adjacent = 19 cm / tan(60°)

We know that tan(60°) is √3 or approximately 1.732.

So:

adjacent = 19 cm / 1.732 ≈ 10.97 cm

Therefore, the length of the shadow cast by the stick is approximately 10.97 cm.

Answer:

Average speed, v = 11.37 m/s

Explanation:

Given that,

The distance covered by the traveler, d = 217 miles = 349228 meters

Time taken, [tex]t = 8\ hours \ 32\ minutes =8\ h+\dfrac{32}{60}\ h=8.53\ h[/tex]

or t = 30708 s

We need to find the average speed for this trip. The average speed is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]v=\dfrac{349228\ m}{30708\ s}[/tex]  

v = 11.37 m/s

So, the average speed for this trip is 11.37 m/s. Hence, this is the required solution.

Answer:

1.4 m/s^2

Explanation:

The angle of repose or the angle of sliding is defined as the angle of inclined plane with the horizontal at which an object placed on the plane is just start to slide down.

The relation between the angle of friction and the coefficient of static friction is given by

[tex]\mu _{s}=tan\theta[/tex]

[tex]\mu _{s}=tan38=0.78[/tex]

Thus, the coefficient of static friction is 0.78.

(b) As the crate is moving down, then the friction force is kinetic friction.

The force acting along the plane downward = mg Sinθ

The normal reaction, N = mg cosθ

The friction force acting upward along the plane, f = μk N = μk mg Cosθ

The net force acting along the plane downward

Fnet = mg Sinθ - μk mg Cosθ

According to the newton's second law, Fnet = mass x acceleration

so,  m x a = mg Sinθ - μk mg Cosθ

a = g Sinθ - μk g Cosθ

here, μk = 0.6 and θ = 38°, g = 9.8 m/s^2

By substituting the value, we get

a = 9.8 ( Sin38 - 0.6 x Cos 38)

a = 9.8 (0.6156 - 0.6 x 0.788)

a = 1.4 m/s^2

The coefficient of static friction between the crate and the ramp surface is 0.781. The acceleration of the moving crate is 5.88 m/s².

To find the coefficient of static friction, we need to calculate the tangent of the angle at which the crate begins to slide. The tangent of 38 degrees is approximately 0.781. Since the crate remains in place until this angle, the coefficient of static friction must be equal to or greater than this value. Therefore, the coefficient of static friction between the crate and the ramp surface is 0.781.

To find the acceleration of the moving crate, we can use the formula a = μk * g, where a is the acceleration, μk is the coefficient of kinetic friction, and g is the acceleration due to gravity. Given that the coefficient of kinetic friction is 0.600, and the acceleration due to gravity is approximately 9.8 m/s², we can substitute these values into the formula to find the acceleration.

Substituting the values, we get a = (0.600) * (9.8) = 5.88 m/s². Therefore, the acceleration of the moving crate is 5.88 m/s².

https://brainly.com/question/14050409

#SPJ3

Answer:

[tex]E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}[/tex]

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

[tex]E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta[/tex]

[tex]sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}[/tex]

[tex]r^2=d^2+x^2[/tex]

So

[tex]E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}[/tex]

Now by integrating above equation

[tex]E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}[/tex]

Answer:

Force exerted on the right cube by the middle cube:

F= 12.02N : in the positive direction of the x axis( +x)

Explanation:

We apply Newton's second law for forces in the direction of the x-axis.

∑Fx= m*a

∑Fx:  algebraic sum of forces ( + to the right, - to the left)

m: mass

a: acceleration

We apply Newton's first law for forces in the direction of the y-axis.

∑Fx= mt*a   , mt: total mass = 6*3= 18 kg

36-6= 18*a

[tex]a=\frac{36-6}{18} = 1.67 \frac{m}{s}[/tex]

∑Fy= 0

Nt-Wt=0   , Nt=Wt

Nt: total normal   ,  Wt= total Weight: mt*g  , g: acceleration due to gravity

Wt=18*9,8=176.4 N

Nt=176.4 N

μk=Ff/Nt Ff: friction force = 6 N

μk=6/176.4 = 0.034

∑Fx= m₁*a

F-Ff₁= 6*1.67 , Equation  (1)

F:Force exerted on the right cube by the middle cube

Ff₁= μk*N₁ , N₁=W₁ = 9.8*6= 58.8 N

Ff₁= 0.034*58.8 = 2 N

In the  Equation  (1):

F-2= 10.02

F= 2+10.02= 12.02N

F= 12.02N

Answer:

[tex]v_f = 16.6 m/s[/tex]

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

[tex]F_g = mg sin\theta[/tex]

so the acceleration due to gravity along the plane is given as

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

now we have

[tex]a = g sin\theta[/tex]

[tex]a = (9.81 sin4.0)[/tex]

[tex]a = 0.68 m/s^2[/tex]

now we know that

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 9.2^2 = 2(0.68)(140)[/tex]

[tex]v_f = 16.6 m/s[/tex]

Velocity of a object is the ratio of distance traveled by the object with the time taken

The sled's speed after it has traveled the first 140 m is 16.6 m/s.

Velocity of a object is the ratio of distance traveled by the object with the time taken.

Given information-

The top speed reached by the sled is 9.2 m/s.

The angle of the slope is 4 degrees downward.

Distance traveled by the sled is 140 meters.

The force acting on a body is the product of mass and its acceleration.

The acceleration of the inclined plane with this definition can be given as,

[tex]a=\dfrac{F_g}{m} =\dfrac{mg\sin\theta}{m} \\a=g\sin\theta[/tex]

Here, [tex]F_g[/tex] is the force due to gravity and [tex]a[/tex] is the acceleration of the body.

Put the values in the above equation as,

[tex]a=9.81\times \sin(4^o)\\a=0.68 \rm m/s^2[/tex]

Thus the acceleration of the body is 0.68 m/s squared.

Now the sled's speed can be find out using the equation of motion as,

[tex]2ad=v_f^2-v_i^2[/tex]

As the initial speed is 9.2 seconds. Thus,

[tex]2\times0.68\times140=v_f^2-9.2^2\\v_f=16.6 \rm m/s[/tex]

Hence the sled's speed after it has traveled the first 140 m is 16.6 m/s.

Learn more about the velocity here;

https://brainly.com/question/6504879

Answer:

The answer is 8 N

Explanation:

The Lorentz force for a current carrying wire is

f = I * L x B

So, for magnetic forces to manifest the current must not be parallel to the magnetic field. So the cases where the wire is parallel to the field would result in a force of zero applied on the wires  by the magnetic field because the cross product becomes zero.

For the perpendicular cases:

f = I * L * B

f = 80 * 0.1 * 1 = 8 N

The magnitude of the ball's displacement would have been approximately 7.579 meters, and the direction would have been approximately 11.08 degrees counterclockwise from due east.

1. First stroke:

 [tex]Displacement \( \vec{d}_1 = 4.8 \, \text{m} \) due east.[/tex]

2. Second stroke:

[tex]Eastward component: \( 2.7 \, \text{m} \times \cos(20^\circ) \approx 2.564 \, \text{m} \) due east.\\ Northward component: \( 2.7 \, \text{m} \times \sin(20^\circ) \approx 0.924 \, \text{m} \) due north.\\\\ Displacement \( \vec{d}_2 = 2.564 \, \text{m} \) due east and \( 0.924 \, \text{m} \) due north.[/tex]

3. Third stroke:

  Displacement[tex]\( \vec{d}_3 = 0.50 \, \text{m} \)[/tex] due north.

Now let's calculate the total eastward and northward displacements:

Total eastward displacement:[tex]\( 4.8 \, \text{m} + 2.564 \, \text{m} = 7.364 \, \text{m} \)[/tex]

Total northward displacement: [tex]\( 0.924 \, \text{m} + 0.50 \, \text{m} = 1.424 \, \text{m} \)[/tex]

Using the Pythagorean theorem, the magnitude of the total displacement is:

[tex]Magnitude = \( \sqrt{(\text{Total eastward displacement})^2 + (\text{Total northward displacement})^2} \)Magnitude = \( \sqrt{7.364^2 + 1.424^2} \approx 7.579 \, \text{m} \)[/tex]

The direction of the total displacement is given by:

[tex]Direction = \( \tan^{-1}\left(\frac{\text{Total northward displacement}}{\text{Total eastward displacement}}\right) \)\\Direction = \( \tan^{-1}\left(\frac{1.424}{7.364}\right) \approx 11.08^\circ \) counterclockwise from due east.[/tex]

So, if the golfer had hit the ball directly into the hole on the first stroke, the magnitude of the ball's displacement would have been approximately 7.579 meters, and the direction would have been approximately 11.08 degrees counterclockwise from due east.

Learn more about displacement, here:

https://brainly.com/question/11934397

#SPJ12

Final answer:

The golfer's ball would have a direct displacement of approximately 7.434 meters in a direction of roughly 11° north of east if it were hit directly into the hole.

Explanation:

To determine the direct displacement and direction of the ball if the golfer had hit it directly into the hole, we need to sum up the three given vectors. The first is 4.8 m due east, the second is 2.7 m at a 20° angle north of east, and the third is 0.50 m due north.

First, decompose the second stroke into its north and east components. The east component is 2.7 cos(20°) m and the north component is 2.7 sin(20°) m.

East component of the second stroke: 2.7 cos(20°) ≈ 2.536 m

North component of the second stroke: 2.7 sin(20°) ≈ 0.923 m

Now, sum the components in each direction:

Total east displacement: 4.8 m + 2.536 m = 7.336 m

Total north displacement: 0.923 m + 0.50 m = 1.423 m

Finally, calculate the magnitude and direction of the resultant displacement vector:

Magnitude (√(x² + y²)): √(7.336² + 1.423²) ≈ 7.434 m

Direction (tan⁻¹(y/x)): tan⁻¹(1.423/7.336) ≈ 11° north of east

Explanation:

The charge on the electron is, [tex]q=-1.6\times 10^{-19}\ C[/tex]

The electric field at a distance r from the electron is :

[tex]E=k\dfrac{q}{r^2}[/tex]

Where

k is the electrostatic constant, [tex]k=\dfrac{1}{4\pi \epsilon_o}[/tex]

We know that the electric field lines starts from positive charge and ends at the negative charge. Also, for a positive charge the field lines are outwards while for a negative charge the field lines are inwards.

So, the correct option is " the  electric field is directed toward the electron and has a magnitude of [tex]k\dfrac{q}{r^2}[/tex]. Hence, this is the required solution.

Answer:

The mass of the bullet is 15 g.

(b) is correct option.

Explanation:

Given that,

Momentum = 2.8 kg m/s

Speed = 187 m/s

We need to calculate the mass of the bullet

Using formula of momentum

[tex]P = mv[/tex]

[tex]m = \dfrac{P}{v}[/tex]

Where, P = momentum

v = speed

Put the value into the formula

[tex]m = \dfrac{2.8}{187}[/tex]

[tex]m = 0.015\ kg[/tex]

[tex]m = 15\ g[/tex]

Hence, The mass of the bullet is 15 g.

Answer:

density of liquid 0.848 g/ml

Explanation:

from the information given in the question

mass of water = 34.914 - 25.296 = 9.618 g

volume of pycnometer = volume of water

which will be equal to [tex]= \frac{ mass}{density}[/tex]

[tex]= \frac{9.618}{0.9970} = 9.646 ml[/tex]

mass of liquid =33.485-25.296 = 8.189 ml

density of liquid[tex]= \frac{mass}{volum\ of\ liquid}[/tex]

                           = [tex]\frac{8.189}{9.646} =0.848 g/ml[/tex]

Final answer:

The absolute pressure at the bottom of the Mariana Trench is calculated by adding the atmospheric pressure to the hydrostatic pressure due to the water column, resulting in approximately 509,367.85 lbf/in.².

Explanation:

To calculate the absolute pressure at the bottom of the Mariana Trench, we start by understanding that pressure in a static fluid increases linearly with depth. The increase in pressure, ΔP, due to the water column can be calculated using the formula ΔP = ρgh, where ρ is the density of seawater, g is the acceleration due to gravity, and h is the depth. Given the constants, ρ = 64.2 lb/ft³, g = 32.1 ft/s², and h = 6.8 miles (35,856 ft), we first convert the depth into feet as pressure calculations require consistent units. The calculation is as follows: ΔP = 64.2 lb/ft³ * 32.1 ft/s² * 35,856 ft = 73,346,473.6 lb/ft². Converting this to lbf/in.², we divide by 144 (since 1 ft² = 144 in.²), resulting in ΔP approximately 509,353.15 lbf/in.². Adding the atmospheric pressure of 14.7 lbf/in.² at the surface, the absolute pressure at the bottom of the Mariana Trench in lbf/in.² is approximately 509,367.85 lbf/in.².

The absolute pressure at the depth of the Mariana Trench is approximately [tex]\(7.42 \times 10^9 \, \text{lbf/in}^2\),[/tex]calculated using the hydrostatic pressure formula.

To find the absolute pressure at the depth of the Mariana Trench, we can use the hydrostatic pressure formula:

[tex]\[ P = P_0 + \rho \cdot g \cdot h \][/tex]

Where:

-  P is the absolute pressure at the depth,

-  P0  is the atmospheric pressure at the surface (given as 14.7 lbf/in²),

-  rho  is the density of seawater (given as 64.2 lb/ft³),

-  g  is the acceleration due to gravity (given as 32.1 ft/s²), and

-  h  is the depth of the trench (given as 6.8 miles).

First, let's convert the depth from miles to feet:

[tex]\[ 6.8 \text{ miles} \times 5280 \text{ ft/mile} = 35904 \text{ ft} \][/tex]

Now, we can plug in the values into the formula:

[tex]\[ P = 14.7 \text{ lbf/in}^2 + (64.2 \text{ lb/ft}^3) \times (32.1 \text{ ft/s}^2) \times (35904 \text{ ft}) \][/tex]

Let's calculate this value.

To find the absolute pressure at the depth of the Mariana Trench, we'll first calculate the pressure due to the water column using the hydrostatic pressure formula:

[tex]\[ P = P_0 + \rho \cdot g \cdot h \][/tex]

Where:

- [tex]\( P_0 = 14.7 \, \text{lbf/in}^2 \)[/tex] is the atmospheric pressure at the surface,

-[tex]\( \rho = 64.2 \, \text{lb/ft}^3 \)[/tex] is the density of seawater,

- [tex]\( g = 32.1 \, \text{ft/s}^2 \)[/tex] is the acceleration due to gravity, and

- [tex]\( h = 6.8 \, \text{miles} \times 5280 \, \text{ft/mile} = 35904 \, \text{ft} \)[/tex]is the depth of the trench in feet.

Plugging in the values:

[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]

Let's calculate this.

[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]

[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]

[tex]\[ P = 14.7 + (206368.8 \times 35904) \][/tex]

[tex]\[ P = 14.7 + 7417798272 \][/tex]

[tex]\[ P ≈ 7417798286.7 \, \text{lbf/in}^2 \][/tex]

Therefore, the absolute pressure at the depth of the Mariana Trench is approximately [tex]\( 7.42 \times 10^9 \, \text{lbf/in}^2 \).[/tex]

Answer:

No the sonic boom is not heard by the passenger travelling in a supersonic aircraft. They do not experience any shift in the frequency as they are travelling at the speed of the source itself.Thus option 'a' is correct answer.

Explanation:

Sonic boom is a shock wave of sound associated when a source travels faster than the speed of sound in the ambient conditions. In a shock wave associated with the sonic boom there exists high gradients of pressure, density and temperature of air across the shock wave. The shock wave travels at the speed of the sound but the source (in our case the passenger) travels at a faster speed than the sound in ambient conditions.

Answer:

The linear expansion coefficient of the material is [tex]4.952\times10^{-5}\ K^{-1}[/tex]

Explanation:

Given that,

Volume = 1.5 l

Temperature = 20°C

Increased temperature = 55°C

Volume = 7.8 mL

We need to calculate the linear expansion coefficient of the material

Using formula of  linear expansion

[tex]\Delta V=V_{0}\beta \Delta T[/tex]

[tex]\Delta V=V_{0}(3\alpha \Delta T)[/tex]

[tex]\alpha =\dfrac{\Delta V}{3V_{0}\Delta T}[/tex]

Put the value into the formula

[tex]\alpha=\dfrac{7.8\times10^{-3}}{3\times1.5\times(55-20)}[/tex]

[tex]\alpha=0.00004952[/tex]

[tex]\alpha=4.952\times10^{-5}\ K^{-1}[/tex]

Hence, The linear expansion coefficient of the material is [tex]4.952\times10^{-5}\ K^{-1}[/tex]

Answer:

0.063

Explanation:

velocity of the car, v = 40 km/h = 11.11 m/s

radius, r = 200 m

Let the coefficient of friction is μ.

The coefficient of friction relates to the velocity on banked road is given by

[tex]\mu =\frac{v^{2}}{rg}[/tex]

where, v is the velocity, r be the radius of the curve road and μ is coefficient of friction.

By substituting the values, we get

[tex]\mu =\frac{11.11^{2}}{200\times 9.8}[/tex]

μ = 0.063

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

[tex]h=ut+\frac{1}{2}gt^2[/tex]

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

[tex]h=ut+\frac{1}{2}gt^2[/tex]

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

[tex]h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2[/tex]    .... (2)

By equation the equation (1) and (2), we get

[tex]41.7=1.12 u +4.9 \times 1.12^{2}[/tex]

u = 31.75 m/s

The final temperature can be calculated using the principle of conservation of energy. Using the formulas Q = mcΔT and Q = mcΔT, we can find the heat lost by the iron and the heat gained by the ethyl alcohol. The final temperature of all 3 substances are, T_f = 12.08 °C.

To calculate the final temperature of the substances, we can use the principle of conservation of energy. The heat gained by one substance is equal to the heat lost by another substance. First, we calculate the heat lost by the iron using the formula Q = mcΔT, where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Then, we calculate the heat gained by the ethyl alcohol using the same formula. Finally, we use the principle of conservation of energy to find the final temperature.

We have:

Using the formula Q = mcΔT, we can calculate:

Since the heat lost by the iron is equal to the heat gained by the ethyl alcohol, we can set up the equation:

(200 g)(452 J/g °C)(120 °C - T_f) = (300 g)(2.30 J/g °C)(T_f - 20 °C)

To solve for T_f, we can simplify and rearrange the equation:

(90400 - 452T_f) J = (T_f- 13800) J

Combining like terms:

13800 J = 1142T_f J

Dividing both sides by 1142:

T_f = 12.08 °C

https://brainly.com/question/34640456

#SPJ3

Answer:

Answered

Explanation:

A) False , horizontal velocity component is constant since there's no horizontal acceleration component

B) False, acceleration remains constant through out and only verticle acceleration is there horizontal acceleration is zero.

C) True , Its horizontal velocity doesn't change once it is in air, but its vertical velocity does change ( because of acceleration in verticle direction)

D)False, the acceleration is constant through the motion.

E)False, Vertical velocity is Zero but there will be Horizontal velocity.