Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 6.0 cm per year. (a) What distance does it move in 71 seconds at this speed?
m
(b) What is its speed in miles per million years?
mi/My

Answer:

10.29 cm/s

Explanation:

Discharge in to the bowl = 17.0 cm³/s

Diameter of the bowl, d₁ = 1.45 cm

Now,

Rate at which water level rise at its diameter = [tex]\frac{\textup{Discharge}}{\textup{Area of cross-section}}[/tex]

also,

Area of cross-section = [tex]\frac{\pi}{\textup{4}}\times1.45^2[/tex]

or

Area of cross-section = 1.651 cm²

Therefore,

Rate at which water level rise at its diameter = [tex]\frac{\textup{17}}{\textup{1.651}}[/tex]

or

Rate at which water level rise at its diameter = 10.29 cm/s

The rate at which the water level rises at a diameter of 1.45 cm is 8.44 cm/s.

To find the rate at which the water level rises at this diameter, we can use the concept of continuity equation. The continuity equation states that the flow rate of a fluid is constant at all points in a pipe or container.

Therefore, the rate at which the water level rises at a particular diameter can be found by dividing the flow rate by the cross-sectional area at that diameter.

Rate of water level rise = Flow rate / Cross-sectional area

Substituting the given values, we get:

Rate of water level rise = 17.0 cm³/s / (π * (d1/2)²)

Rate of water level rise = 17.0 cm³/s / (π * (1.45 cm/2)²)

Rate of water level rise = 8.44 cm/s

Answer:22.76 m/s

Explanation:

Given

Train length(L)=75 m

Front of train after travelling 125 m is 18 m/s

Time taken by the front of train to cover 125 m

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

[tex]18^2-0=2\times a\times 125[/tex]

[tex]a=1.296 m/s^2[/tex]

Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel  a distance of 125+75=200 m

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

[tex]v^2=2\times 1.296\times 200[/tex]

[tex]v=\sqrt{518.4}=22.76 m/s[/tex]

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

[tex]X=V_{o}*t + \frac{a*t^{2}}{2}[/tex]   Replacing the known values:

[tex]1500=6t+0.9*t^{2}[/tex]    Solving for t we get 2 possible answers:

t1 = -44.3s   and t2 = 67.7s    Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

Answer:

Mass of the airplane, m = 562.5 slugs

Explanation:

Given that,

Weight of the airplane, W = 18,000 lbs

The acceleration due to gravity is, [tex]g=32\ fp/s^2[/tex]

Let m is the mass of the airplane. The weight of an object is equal to the product of mass and acceleration due to gravity as :

W = m × g

[tex]m=\dfrac{W}{g}[/tex]

[tex]m=\dfrac{18000}{32}[/tex]

m = 562.5 slugs

So, the mass of the airplane is 562.5 slugs. Hence, this is the required solution.

Answer:

distance stop 1.52m,

velocity  4.0 m/s y^

Explanation:

The movement of the particle is two-dimensional since it has acceleration in the x and y axes, the way to solve it is by working each axis independently.

a) At the point where the particle begins to return its velocity must be zero (Vfx = 0)

     Vfₓ = V₀ₓ + aₓ t  

     t = -  V₀ₓ/aₓ

     t = - 2.4/(-1.9)

     t=  1.26 s

At this time the particle stops, let's find his position

     X1 = V₀ₓ t + ½ aₓ t²

     X1= 2.4 1.26 + ½ (-1.9) 1.26²

     X1= 1.52 m

At this point the particle begins its return

b) The velocity has component x and y

   As a section, the X axis x Vₓ = 0 m/s is stopped, but has a speed on the y axis

    Vfy= Voy + ay t

    Vfy= 0 + 3.2 1.26

    Vfy = 4.0 m/s

the velocity is  

    V = (0 x^ + 4.0 y^) m/s

c) In order to make the graph we create a table of the position x and y for each time, let's start by writing the equations

      X = V₀ₓ t+ ½  aₓ t²

      Y = Voy t + ½  ay t²

      X= 2.4 t + ½ (-1.9) t²

      Y= 0 + ½ 3.2 t²

      X= 2.4 t – 0.95 t²

      Y=   1.6 t²

With these equations we build the table to graph, for clarity we are going to make two distance graph with time, one for the x axis and another for the y axis

                       Chart to graph

              Time (s)     x(m)            y(m)

                 0                0               0

                 0.5             0.960       0.4

          1       1.45          1.6

                 1.50      1.46      3.6

                2.00      1.00      6.4

In the problem, we have a particle moving with constant acceleration. We can find the distance before it turns around using the equations of motion, and the velocity at that time is zero. To plot its position over time, again we have to use the equations of motion for each coordinate.

The problem given relates to motion in two dimensions specifically linear motion with constant acceleration. Let's address each part of the question.

https://brainly.com/question/29349889

#SPJ3

To find the charge that creates a 4.80 N/C electric field at a distance of 3.60 m, use the relationship from Coulomb's law, which after calculation gives a charge magnitude of 6.912 nC.

To find the magnitude of the charge that creates a 4.80 N/C electric field at a point 3.60 m away, we use Coulomb's law and the formula for the electric field E due to a point charge q. The relationship between the electric field E and the charge q is given by:

E = k * |q| / r²

where,
E is the electric field strength,
k is Coulomb's constant (8.9875 * 10⁹ N m²/C²),
q is the charge, and
r is the distance from the charge to the point where the electric field is measured.

Rearranging the formula to solve for the charge q, we get:

q = E * r² / k

Plugging in the given values:

q = (4.80 N/C) * (3.60 m)² / (8.9875  10⁹ N m²/C²)

After calculation, the magnitude of the charge that generates the electric field is:

q = 6.912 * 10⁻⁹ C or 6.912 nC

Answer:2.737 kN

Explanation:

Given

mass of log(m)=205 kg

ramp inclination[tex]=30^{\circ}[/tex]

coefficient of kinetic friction between log and ramp is ([tex]\mu _k[/tex])=0.9

log has an acceleration of 0.8 m/s^2

Let T be the tension in the rope

[tex]T-mgsin\theta -f_r=ma[/tex]

Where [tex]mgsin\theta [/tex]=Sin component of weight

[tex]f_r=friction\ Force=\mu _KN[/tex](Where N is Normal reaction)

[tex]T-mgsin\theta -\mu _k\left ( mgcos\theta \right )=ma[/tex]

[tex]T=m\left ( gsin\theta +\mu _kcos\theta +a\right )[/tex]

sin30=0.5

cos30=0.866

[tex]T=205\times \left ( 9.81\times 0.5+0.9\times 9.81\times 0.866+0.8\right )[/tex]

[tex]T=205\times 13.35=2.737 kN[/tex]

To find the tension in the rope, you need to first calculate the gravitational and frictional forces on the log on the inclined plane. Then using Newton's second law, set up an equation for net force as the sum of these forces and the tension. Solve for the tension to get the desired result.

The physics subject of this problem involves the understanding of Newton's second law, forces in equilibrium, the concept of tension, friction, and forces on inclined planes. Given are the mass of the log (m) = 205 kg, acceleration (a) = 0.8 m/s2, angle of inclination (θ) = 30 degrees, and coefficient of kinetic friction (μ) = 0.900.

Firstly, calculate the force due to gravity along the ramp (fgravity) = m*g*sin(θ), where g = 9.8 m/s2. Secondly, calculate the force due to friction (ffriction) = μ*m*g*cos(θ). The net force (Fnet) pulling the log up the ramp is the difference between the tension force (T), the force due to gravity along the ramp and the frictional force, i.e., Fnet = T - fgravity - ffriction.

Since we also know from Newton’s second law that Fnet = m*a, we can set these equal and solve for the tension: T = m*a + fgravity + ffriction. Plugging in the given values into this equation will provide the tension in the rope.

https://brainly.com/question/30470948

#SPJ3

Answer:

4.14 m

Explanation:

In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s .

Let in this last leg , u be the initial velocity.

s = ut + 1/2 g t²

2 = .2 u + .5 x 9.8 x .04

u = 9.02 m /s .

Let v be the final velocity in this leg

v² = u² + 2 g s

v² = (9.02)² + 2 x 9.8 x 2

= 81.36 +39.2

v = 10.97 m / s

Now consider the whole height from where the ball dropped . Let it be h.

Initial velocity u = 0

v² = u² +2gh

(10.97 )² = 2 x 9.8 h

h = 6.14 m

Height from window

= 6.14 - 2m

= 4.14 m

Answer:

Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

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

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr =[tex]\frac{38\times 1000}{3600}=10.56m/s[/tex]

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

[tex]0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2[/tex]

Now the force can be obtained using newton's second law as

[tex]Force=\frac{Weight}{g}\times a[/tex]

Applying values we get

[tex]Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons[/tex]

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

[tex]v=u+at[/tex] with symbols having the same meanings

Applying values we get

[tex]0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds[/tex]

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

Answer:

2.77 m/s^2

Explanation:

For simplicity, we will say that the x-axis is parallel to the inclined plane and the y-axis is perpendicular to the inclined plane. The force that the surface applies on the block N will be perpendicular to the plane. The force of friction can be expressed like this:

[tex]Fr = u*N[/tex]

Where u is the coefficient of kinetic friction.

The x-component and y-component of the weight force will be:

[tex]W_y = W*cos(37)\\W_x = W*sin(37)[/tex]

In the y-axis, the acceleration of the block is 0. Then:

[tex]N - W_y = 0\\N = W_y = W*cos(37) = m*g*cos(37)[/tex]

In the X-axis, the summation of forces would be:

[tex]W_x - Fr = m*a\\m*g*sin(37) - u*N = m*g*sin(37) - u*m*g*cos(37) = ma\\a = \frac{mg(sin(37)-u*cos(37))}{m} = 9.81m/s^2*(sin(37)-0.4*cos(37))=2.77m/s^2[/tex]

Explanation:

It is given that,

Height, h = 3.1 m

Initial speed of the rocket, u = 0

Final speed of the rocket, v = 28 m/s

(b) Let a is the acceleration of the rocket. Using the formula as :

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

[tex]a=\dfrac{(28)^2}{2\times 3.1}[/tex]

[tex]a=126.45\ m/s^2[/tex]

(a) Let t is the time taken to reach by the rocket to reach to a height of h. So,

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

[tex]t=\dfrac{28\ m/s}{126.45\ m/s^2}[/tex]

t = 0.22 seconds

(c) At t = 0.1 seconds, height of the rocket is given by :

[tex]h=ut+\dfrac{1}{2}at^2[/tex]

[tex]h=\dfrac{1}{2}\times 126.45\times (0.1)^2[/tex]

h = 0.63 meters

(d) Let v' is the speed of the rocket 0.10 s after launch.

So, [tex]v'=u+at[/tex]

[tex]v'=0+126.45\times 0.1[/tex]

v' = 12.64 m/s

Hence, this is the required solution.

Answer:

The current in the solenoid=0.463 Ampere

Explanation:

Given:

Number of turns , N=420

Length of the solenoid=0.575 m

Magnitude of Magnetic Field,[tex]B=4.22\times10^{-5}\ \rm T[/tex]

We know that the magnitude of the magnetic Field inside the solenoid is

[tex]B=\mu_0 ni[/tex]

Where

According to question

[tex]B=\mu_0 ni\\\\4.22\times10^{-5}=1.24664\times10^{-7}\times \dfrac{420}{0.575}\times i\\=0.463\ \rm Amp[/tex]

Hence the current is calculated.

Answer:

Yes! They can be correct

Explanation:

An isolated system is a system where its total energy and mass stay constant.

In this case, even though the volume changed, the mass remains constant (m = 1 kg), so there is no mass exchange, and we must think that the globe is completely closed.

Now, we don't know exactly what happens with energy, but I can give you an example where the total energy of the globe remains constant.

Imagine that the balloon of volume [tex]V_1[/tex] is at a certain height [tex]h_1[/tex], under some pressure, [tex]P_1[/tex]. If you lower the balloon to a height [tex]h_2[/tex], the pressure increases, and from the Boyle's law,

[tex]P_1 V_1 = P_2 V_2[/tex],

[tex]V_2 = V_1 \frac{P_1}{P_2}[/tex], where [tex]P_2 > P_1[/tex],

[tex]V_2 < V_1[/tex], just like the case you state, and there was no exchange of mass or energy related to the inner gas of the balloon, so yes, They can be correct.

Answer:

20,000 Ns

Explanation:

mass of exhaust gases, m = 50 kg

velocity of exhaust gases, v = 400 m/s

The momentum of a body is defined as the measurement of motion of body. mathematically, it is defined as the product of mass off the body an its velocity.

change in momentum of rocket = final momentum - initial momentum

                                                     =  m x v - 0

                                                     = 50 x 400

                                                     = 20,000 Ns

Final answer:

The change in momentum of the rocket (also known as impulse) during the blast is 20,000 kg·m/s. This is calculated by multiplying the mass of the exhaust gas (50 kg) by its average velocity (400 m/s).

Explanation:

The change in momentum of the rocket during the blast (also known as impulse) can be calculated using the conservation of momentum. The momentum of the exhaust gas expelled will be equal in magnitude and opposite in direction to the change in momentum of the rocket.

To calculate the change in momentum of the rocket, we use the formula:

Change in momentum = mass of exhaust × velocity of exhaust.

In this case, the mass of the exhaust gas is 50 kg, and the average velocity at which it is expelled is 400 m/s:

Change in momentum = 50 kg × 400 m/s = 20000 kg·m/s.

This is the momentum gained by the rocket in the opposite direction, due to Newton's third law of motion.

Answer:

(a) Acceleration of the shot put is 80[tex]m/s^{2}[/tex]

(b) Time taken for accelerating the shot is 0.2 s

(c) Horizontal force is 632 N

Solution:

As per the question:

Final speed of the shot put throw, v' = 16 m/s

Initial speed, v = 0 m/s

The distance covered by the shot put, d = 1.6 m

The shot put throw was at constant acceleration

Now,

(a) Acceleration of the shot put, [tex]a_{s}[/tex] is given by eqn 2 of motion:

[tex]v'^{2} = v^{2} + 2a_{s}d[/tex]

[tex]16^{2} = 0 + 2a_{s}\times 1.6[/tex]

[tex]a_{s} = 80 m/s^{2}[/tex]

(b) Time taken, t is given by eqn 1 of motion:

v' = v + [tex]a_{s}t[/tex]

16 = 0 + 80t

t = 0.2 s

(c) Horizontal component of force is given by:

[tex]F = ma_{s}[/tex]

where

m = mass = 7.9 kg

Now,

[tex]F = 7.9\times 80 = 632 N[/tex]

Answer:

Explanation:

The force acts on the throw during which hand moved by 1.6 m

So it is a case of uniformly accelerated motion

s = 1.6 m

u = 0

v = 16 m /s

v² = u² + 2 a s

16 x 16 = 0 +2 x a x 1.6

a = 80 m / s²

v = u + at

16 = 0 + at

16 = 80 t

t = 1 /5 = .2 s.

force = mass x acceleration

= 7.9 x 80

= 632 N.

Answer:

[tex]F=(3i+3.6j)\ N[/tex]

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, [tex]u=(1i+0j)\ m/s[/tex]

After 8 seconds, final velocity of the puck, [tex]v=(6i+6j)\ m/s[/tex]

Let the x and y component of force is given by [tex]F_x\ and\ F_y[/tex].

x component of force is given by :

[tex]F_x=m\times \dfrac{v-u}{t}[/tex]

[tex]F_x=4.8\times \dfrac{6-1}{8}[/tex]

[tex]F_x=3\ N[/tex]

y component of force is given by :

[tex]F_y=m\times \dfrac{v-u}{t}[/tex]

[tex]F_y=4.8\times \dfrac{6-0}{8}[/tex]

[tex]F_y=3.6\ N[/tex]

So, the component of the force is [tex]F=(3i+3.6j)\ N[/tex]. Hence, this is the required solution.

The components of the force exerted by the rocket engine on the puck are calculated using Newton's Second Law. The acceleration in each direction (X and Y) is determined by dividing the change in velocity by the change in time. The force is then found by multiplying the mass of the puck by the acceleration, resulting in a force of 3.00î N in the X direction and 3.60ĵ N in the Y direction.

The physics behind this problem involves the concept of Newton's Second Law of motion, which states that force equals mass times acceleration (F=ma). To solve this problem, we'll need to find the change in velocity and divide it by the change in time to determine the acceleration. Then, we'll use Newton's Second Law to find the force.

The initial velocity of the puck is 1.00î m/s and the final velocity is (6.00î + 6.0ĵ) m/s. Therefore, the change in velocity (final - initial) in the X direction is 5î m/s and in the Y direction is 6ĵ m/s. The time duration for this change is 8 seconds. So, the acceleration (change in velocity divided by time) is 5/8 = 0.625î m/s² in the X-direction and 6/8=0.75ĵ m/s² in the Y-direction.

Finally, using Newton's Second Law (F=ma), we can find the force in each direction by multiplying the mass of the puck (4.80 kg) by the acceleration. This results in a force of 0.625 * 4.80 = 3.00 N in the X direction (3.00î N) and 0.75 * 4.80 = 3.60 N in the Y direction (3.60ĵ N).

https://brainly.com/question/13447525

#SPJ3

Answer:

The hurricane will be 166.15km away from the island.

Explanation:

We can solve this problem with trigonometry and the formulas of constant velocity motion.

we need to find the displacement on the X and Y axis.

the speed on the X axis on the first trame of movemont is defined as:

[tex]Vx=V*cos(ang)[/tex]

we will take the reference to the left of the X axis as positive, so:

[tex]Vx=43*cos(60)\\Vx=21.5 km/h[/tex]

So the displacement on X is:

[tex]dx=Vx*t\\dx=21.5km/h*3h\\dx=64.5km[/tex]

we will do the same for the Y axis:

[tex]Vy=V*sin(ang)\\Vy=43sin(60)\\Vy=37.24km/h[/tex]

the Y displacement is:

[tex]dy=Vy*t\\dy=37.24km/h*3h\\dy=111.72km[/tex]

Then on the second trame of the movement we only have displacement on the Y axis.

[tex]dy2=Vy2*t\\dy2=23km/h*1.8h\\dy2=41.4km[/tex]

now we need the total displacement on the X and Y axis:

[tex]Y=dy+dy2\\Y=153.12km\\X=64.5km[/tex]

The magnitud of the total displacement will be:

[tex]D=\sqrt{(X)^2+(Y)^2}\\D=\sqrt{(64.5)^2+(153.12^2}\\ \\D=166.15km[/tex]

Final answer:

The distance of the hurricane from Grand Bahama 4.80 hours after passing over the island is 170.4 km.

Explanation:

To find the distance of the hurricane from Grand Bahama 4.80 hours after passing over the island, we need to calculate the distance it has traveled during that time.

First, let's calculate the distance traveled during the initial three hours when the hurricane was moving at a speed of 43.0 km/h. The distance traveled is given by:

Distance = Speed × Time = 43.0 km/h × 3 h = 129.0 km

Now, let's calculate the distance traveled during the remaining 1.80 hours when the hurricane was moving at a speed of 23.0 km/h. The distance traveled is given by:

Distance = Speed × Time = 23.0 km/h × 1.80 h = 41.4 km

Therefore, the total distance traveled by the hurricane is 129.0 km + 41.4 km = 170.4 km.

Answer:

Average speed of Olympic runner will be 23.08 km/hr

Explanation:

We have given distance d = 27.3 miles

We know that 1 mile = 1.6 km

So 27.3 mile = [tex]=27.3\times 1.6=43.68km[/tex]

Given time = 2 hour 45 minutes and 35 sec

We know that 1 hour = 60 minutes

So 45 minutes [tex]=\frac{45}{60}=0.75hour[/tex]

We also know that 1 hour = 3600 sec

So 1 sec [tex]=\frac{1}{3600}=2.777\times 10^{-4}hour[/tex]

So total time t =[tex]=2+0.75+0.000277=2.750277hour[/tex]

We know that speed [tex]=\frac{distance}{time }=\frac{63.48}{2.750277}=23.08km/hour[/tex]

To find the average speed of the runner in kilometers per hour, convert the time to hours and the distance to kilometers, then divide the distance by the time to get an average speed of approximately 15.932 km/h.

The calculation of a runner’s average speed in a long-distance event requires converting the given time and distance into a consistent set of units and then applying the formula for average speed. First, we convert the runner's time into hours and the distance into kilometers. The runner finished with a time of 2 hours, 45 minutes, and 35 seconds, which translates to 2.7597 hours when converted into hours entirely. The distance covered was 27.3 miles, which converts to approximately 43.941 kilometers (using the conversion factor where 1 mile = 1.60934 kilometers).

To find the average speed, we divide the total distance in kilometers by the total time in hours:

Average speed = Total distance / Total time

This results in an average speed of approximately 15.932 kilometers per hour (km/h).

Answer:

Speed in km/hr will be 109.412 km/hr

Explanation:

We have given speed of the car on a highway = 68 mi/hr

We have to find the speed in km/hr

For this first we have to change mi to km

We know that 1 mile = 1.609 km

Speed is the ratio of distance and time

So 68 mi/hr [tex]=68\times 1.609km/hr=109.412km/hr[/tex]

So the speed in km/hr will be 109.412 km /hr

Answer:

Explanation:

There is electric field between the plates whose value is given by the following expression

electric field E =  V /d where V is potential between the plates and d is distance between them

E = 300 / 5 x 10⁻³

=  60 x 10³ N/c

Force on electron = q E where q is charge on the electron

F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.

Acceleration a = force / mass

a = 96 x 10⁻¹⁶/ mass  = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹

= 10.55 x 10¹⁵ m / s²

For midway , distance travelled

s =  2.5 x 10⁻³ m

s      =  1\2 a t²

t = [tex]\sqrt{\frac{2s}{a\\ } }[/tex]

= [tex]\sqrt{\frac{2\times2.5\times10^{-3}}{ 10.55\times10^{15}}[/tex]

t = .474 x 10⁻¹⁸ s

For striking the plate time is calculated as follows

t = [tex][tex]\sqrt{\frac{2\times5\times10^{-3}}{ 10.55\times10^{15}}[/tex][/tex]

t = 0.67 x 10⁻¹⁸ s

The gravitational force that the moon produces on the Earth, when their centers are 3.88 x 10^8 m apart and the moon has a mass of 7.34 x 10^22 kg, can be calculated using Newton's Law of Universal Gravitation. The force comes out to be approximately 1.82 * 10^20 N.

To calculate the gravitational force exerted by the moon on the earth, we would use Newton's Law of Universal Gravitation. This law states that every point mass attracts every other point mass by a force pointing along the line intersecting both points. The force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between the point masses.

The law is mathematically expressed as: F = G * (m1 * m2)/d² , where F is the force of gravity between the two bodies, G is the gravitational constant (approximately 6.67 * 10^-11 N * m²/kg²), m1 and m2 are the masses of the two bodies, and d is the distance between the centers of the two bodies.

Thus, the gravitational force (F) between Earth and the Moon can be calculated as follows: F = (6.67 * 10^-11 N * m²/kg²) * (7.34 * 10^22 kg * 5.97 * 10^24 kg) / (3.88 * 10^8 m)^2. Calculating this gives a force of approximately 1.82 * 10^20 N. This is the gravitational force that the moon exerts on the Earth.

https://brainly.com/question/32609171

#SPJ3

We calculated that the gravitational force between the Moon and the Earth is approximately 1.95 ×[tex]10^{20}[/tex] N.

Gravitational Force Between the Moon and the Earth

To calculate the gravitational force between the Moon and the Earth, we use Newton's Universal Law of Gravitation. According to this law, the force of gravity ( F ) between two masses ( m1 and m2 ) is given by:

F = G x (m1 x m2) / r²

where:

G is the gravitational constant, approximately 6.67 × [tex]10^{-11}[/tex] N(m²/kg²)

m1 and m2 are the masses of the two objects

r is the distance between the centers of the two masses

Here, we have:

m1 (mass of the Earth) = 5.98 ×[tex]10^{24}[/tex]kg

m2 (mass of the Moon) = 7.34 × [tex]10^{22}[/tex] kg

r (distance between centers) = 3.88 × [tex]10^8[/tex] m

Now, we substitute these values into our formula:

F = (6.67 × [tex]10^{-11}[/tex] N(m²/kg²)) x (5.98 × [tex]10^{24}[/tex]kg x 7.34 × [tex]10^{22}[/tex] kg) / (3.88 × [tex]10^8[/tex] m)²

First, we calculate the numerator:

(6.67 × [tex]10^{-11}[/tex]) x (5.98 × [tex]10^{24}[/tex] x 7.34 × [tex]10^{22}[/tex]) = 2.92 × [tex]10^{37}[/tex] N m²/kg²

Next, we calculate the denominator:

(3.88 × [tex]10^8[/tex])² = 1.50 × [tex]10^{17}[/tex] m²

Finally, we divide the numerator by the denominator to get:

F = 2.92 × [tex]10^{37}[/tex] N m²/kg² / 1.50 × [tex]10^{17}[/tex] m² ≈ 1.95 × [tex]10^{20}[/tex] N

Therefore, the gravitational force that the Moon produces on the Earth is approximately 1.95 × [tex]10^{20}[/tex] N .

Answer:

32.176 ft/s²

9.807 m/s²

Explanation:

Data provided in the question:

Acceleration of the free falling object = 78979.4 mi/hr²

a) In ft/s

Now,

1 mi = 5280 foot

and,

1 hour = 3600 seconds

thus,

78979.4 mi/hr = [tex]\frac{\textup{78979.4}\times\textup{5280}}{\textup{1}\times\textup{3600}^2}[/tex]

or

78979.4 mi/hr = 32.176 ft/s²             ...............(1)

b) In m/s²

Now,

1 foot = 0.3048 m

thus,

we have from 1

32.17 ft/s² =  [tex]\frac{\textup{32.17}\times\textup{0.3048}}{\textup{1}\times\textup{1}^2}[/tex]

or

78979.4 mi/hr = 32.176 ft/s²  = 9.807 m/s²

Answer:

[tex]t=\sqrt{2h/g}-(1/g)*(\sqrt{v_{o}^2+2gh}-v_{o})[/tex]

Explanation:

First person:

[tex]y(t)=y_{o}-v_{o}t-1/2*g*t^{2}[/tex]

[tex]v_{o}=0[/tex]     the rock is dropped

[tex]y_{o}=h[/tex]    

[tex]y(t)=h-1/2*g*t^{2}[/tex]

after t1 seconds it hit the ground, y(t)=0

[tex]0=h-1/2*g*t_{1}^{2}[/tex]

[tex]t_{1}=\sqrt{2h/g}[/tex]

Second person:

[tex]y(t)=y_{o}-v_{o}t-1/2*g*t^{2}[/tex]

[tex]v_{o}[/tex]     the rock has a initial downward speed  

[tex]y_{o}=h[/tex]    

[tex]y(t)=h-v_{o}t-1/2*g*t^{2}[/tex]

after t2 seconds it hit the ground, y(t)=0

[tex]0=h-v_{o}t_{2}-1/2*g*t_{2}^{2}[/tex]

[tex]g*t_{2}^{2}+2v_{o}t_{2}-2h=0[/tex]

[tex]t_{2}=(1/2g)*(-2v_{o}+\sqrt{4v_{o}^2+8gh})[/tex]

the time t when the second person throws the rock after the first person release the rock is:

t=t1-t2

[tex]t=\sqrt{2h/g}-(1/g)*(\sqrt{v_{o}^2+2gh}-v_{o})[/tex]

Answer:

a) 2.7s

b) 29 m/s

Explanation:

The equation for the velocity  and position of a free fall are the following

[tex]v=v_{0}-gt[/tex] -(1)

[tex]x=x_{0}+v_{0}t-gt^{2}/2[/tex] - (2)

Since the hot-air ballon is descending at 2.1m/s and the camera is dropped at 42 m above the ground:

[tex]v_{0}=-2.1m/s[/tex]

[tex]x_{0}=42m[/tex]

To calculate the time which it takes to reach the ground we use eq(2) with x=0, and look for the positive solution of t:

[tex]t = \frac{1}{84}(2.1\pm\sqrt{2.1^{2} - 4\times42\times9.81/2} )[/tex]

        t = 2.71996

Rounding to two significant figures:

       t = 2.7 s

Now we calculate the velocity the camera had just before it lands using eq(1) with t=2.7s

[tex]v=-2.1-9.81*(2.71996)[/tex]

      v = -28.782 m/s

Rounding to two significant figures:

      v = -29 m/s

where the minus sign indicates the downwards direction

Answer: The thermal efficiency of the engine is 41.09 %.

Explanation:

Efficiency is the ratio of the useful work performed to the total energy expended or heat taken in.

Formula for thermal efficiency of engine is

[tex]\eta=1- \frac{Q_2}{Q_1}\times 100[/tex]

[tex]\eta[/tex] = efficiency

[tex]{Q_2}[/tex] = heat rejected = 266.7 kJ

[tex]{Q_1}[/tex] = heat extracted = 452.8 kJ

Putting in the values we get:

[tex]\eta=1- \frac{266.7 kJ}{452.8 kJ }\times 100[/tex]

[tex]\eta=0.41\times 100[/tex]

[tex]\eta =41.09\%[/tex]

The thermal efficiency of the engine is 41.09 %.

Answer:

a) At times t = 2 s and t = 6 s, the projectile will be at a height of 192 ft.

b) The projectile will return to the ground at t = 8 s.  

Explanation:

a) The height of the projectile is given by this equation:

s = -16·t² + 128 f/s·t  (see attached figure)

If the height is 192 ft, then:

192 ft = 16·t² + 128 ft/s· t

0 = -16·t² + 128 ft/s·t - 192 ft

Solving the quadratic equation:

t = 2 and t = 6

At times t = 2 s and t = 6 s, the projectile will be at a height of 192 ft.

b) When the projectile return to the ground, s = 0. Then:

0 = -16·t² + 128 ft/s·t

0 =t(-16·t + 128 ft/s)

t = 0 is the initial point, when the projectile is launched.

-16·t + 128 ft/s = 0

t = -128 ft/s / -16 ft/s² = 8 s

The projectile will return to the ground at t = 8 s.  

Answer: a) 121.54 nm; b) 434.08 nm ; c) 1281.9 nm

Explanation: In order to calculate these spectra lines we have to used the attached formula corresponding to these spectral series.

Final answer:

The transition wavelengths can be determined using the Rydberg formula, which relates the wavelength of a photon to the initial and final energy levels of the electron.

Explanation:

The transition wavelengths for the given series can be determined using the Rydberg formula, which relates the wavelength of a photon emitted or absorbed during a transition to the initial and final energy levels of the electron. The formula is:

1/λ = R(1/n₁² - 1/n₂²)

Where λ is the wavelength, R is the Rydberg constant, and n₁ and n₂ are the initial and final energy levels respectively.

For the given transitions:

a. the first member of the Lyman series: n₁ = 1, n₂ = 2

b. the third member of the Balmer series: n₁ = 3, n₂ = 2

c. the second member of the Paschen series: n₁ = 3, n₂ = 4

By substituting the values n₁ and n₂ into the formula and solving for λ, we can find the wavelength for each transition.

Answer:

The pressure at the bottom of the water layer is [tex]1.25\times10^{5}\ Pa[/tex]

Explanation:

Given that,

Oil layer = 120 cm

Water layer = 140 cm

We need to calculate the pressure at the bottom of the water layer

Using formula of pressure

[tex]P_{atm}=P_{atm}+P_{oil}+P_{w}[/tex]

[tex]P_{ab}=p_{atm}+\rho_{oil} gh+\rho_{w} gh[/tex]

Put the value into the formula

[tex]P_{ab}=1.013\times10^{5}+881\times9.8\times1.2+1000\times9.8\times1.4[/tex]

[tex]P_{ab}=125380.56\ Pa[/tex]

[tex]P_{ab}=1.25\times10^{5}\ Pa[/tex]

Hence, The pressure at the bottom of the water layer is [tex]1.25\times10^{5}\ Pa[/tex]

The pressure at the bottom of a 140 cm thick water layer can be calculated using the equation P = hρg, where h is the height of the water, ρ is the density of the water, and g is gravity. The result is a pressure of 13720 Pascal or Pa.

The pressure at the bottom of a fluid, such as water, is given by the equation P = hρg, where 'P' is the pressure, 'h' is the height (or depth) of the fluid, 'ρ' is the density of the fluid, and 'g' is the acceleration due to gravity. In the case of the water layer, the pressure is calculated by using the depth of the water, which is 140 cm or 1.4 m, the density of water, which is about 1000 kg/m³, and the acceleration due to gravity, which is approximately 9.8 m/s².

Therefore, the pressure at the bottom of the 140-cm-thick water layer would be P = (1.4 m) (1000 kg/m³) (9.8 m/s²), which equals 13720 Pascal or Pa. Note that we did not consider the 120 cm oil layer in this calculation since the question is asking specifically for the pressure at the bottom of the water layer. The oil layer contributes to the pressure at the bottom of the oil, not the water.

https://brainly.com/question/29039234

#SPJ3

Answer:

r = 5.335 meters

Explanation:

Given that,

Charge 1, [tex]q_1=-165\ \mu C[/tex]

Charge 2, [tex]q_2=115\ \mu C[/tex]

Force of attraction between two charges, F = 6 N

The force of attraction between two charges is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex], r is the separation between two charges

[tex]r=\sqrt{\dfrac{kq_1q_2}{F}}[/tex]

[tex]r=\sqrt{\dfrac{9\times 10^9\times 165\times 10^{-6}\times 115\times 10^{-6}}{6}}[/tex]

r = 5.335 m

So, the separation between two charges is 5.335 meters. Hence, this is the required solution.

Answer:

It will take 2.45 seconds.

Explanation:

A football field measured 120 yards, that is arround 0.068 miles, the car is moving with a constant speed so the formula we have to apply for this is:

[tex]t=\frac{x}{v}\\where\\x=distance\\v=speed\\[/tex]

[tex]t=\frac{0.068}{100}\\t=0.00068hrs[/tex]

in order to obtain the time in seconds:

[tex]tsec=0.00068hr*(\frac{3600sec}{1hr} )\\tsec=2.45 sec[/tex]