When an object moves with 2 dimensional motion, what happens to the vertical velocity?

Answer:

0.013 cm/minute

Explanation:

We are given;

Volume rate; dV/dt = 800 cm³/min

Now, volume of a sphere is given by the formula;

V = (4/3)πr³

We want to find the rate at which the radius of the balloon is changing. This is represented by dr/dt.

Now, dr/dt will be gotten from the relation;

dr/dt = dV/dt ÷ dV/dr

Now, dV/dr = 3 × (4/3)πr²

dV/dr = 4πr²

Thus;

dr/dt = 800/(4πr²)

At r = 70 cm

dr/dt = 800/(4π × 70²)

dr/dt = 0.013 cm/minute

Answer:

The average net force exerted on the car and riders by the magnets is 33751.8 newtons.

Explanation:

Let assume that car and its riders accelerate at constant rate, such that acceleration ([tex]a[/tex]), measured in meters per square second, can be found by using the following kinematic equation:

[tex]a = \frac{v_{f}-v_{o}}{t}[/tex] (Eq. 1)

Where:

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the car and its riders, measured in meters per second.

[tex]t[/tex] - Acceleration time, measured in seconds.

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 45\,\frac{m}{s}[/tex] and [tex]t = 6.8\,s[/tex], the average acceleration of the car is:

[tex]a = \frac{45\,\frac{m}{s}-0\,\frac{m}{s}}{6.8\,s}[/tex]

[tex]a = 6.618\,\frac{m}{s^{2}}[/tex]

By the Second Newton's Law, we find that average force exerted on the car and riders by the magnets ([tex]F[/tex]), measured in newtons, is:

[tex]F = m\cdot a[/tex] (Eq. 2)

Where [tex]m[/tex] is the mass of the car-riders system, measured in kilograms.

If we know that [tex]m = 5100\,kg[/tex] and [tex]a = 6.618\,\frac{m}{s^{2}}[/tex], the net force exerted on the car and riders is:

[tex]F = (5100\,kg)\cdot \left(6.618\,\frac{m}{s^{2}} \right)[/tex]

[tex]F = 33751.8\,N[/tex]

The average net force exerted on the car and riders by the magnets is 33751.8 newtons.

Answer: -0.10m/s

Explanation:

khan academy

Answer:

the law that the velocity of recession of distant galaxies from our own is proportional to their distance from us.

Explanation:

Hubble's law, also known as the Hubble–Lemaître law, is the observation in physical cosmology that galaxies are moving away from the Earth at speeds proportional to their distance. In other words, the farther they are the faster they are moving away from Earth.

Answer:

Explanation:

The chunk of ice will fall with acceleration of 9.8 m /s²

initial velocity u = 0 ,

v² = u² + 2 gH

H = 343 m

g = 9.8 m /s²

v is final velocity

v² = 0 + 2 x 9.8 x 343

v² = 6723

v = 82 m /s approx .

The velocity of the the chunk of ice as it hits the ground is 82m/s.

Given the data in the question;

Before the chunk of ice to fell off the roof, its was initially at rest, so

Final velocity as the ice hits the ground; [tex]v = \ ?[/tex]

To determine the final velocity, we use the third equation of motion:

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

Where v is final velocity, u is the initial velocity, s is distance or height of the building and a is acceleration due to gravity{ since the ice will be under gravity as it falls, ([tex]a = g = 9.8m/s^2[/tex]) }

We substitute our values into the equation

[tex]v^2 = [0^ 2] + [ 2\ *\ 9.8m/s^2\ *\ 343m]\\\\v^2 = 2\ *\ 9.8m/s^2\ *\ 343m\\\\v^2 = 6722.8m^2/s^2\\\\v = \sqrt{6722.8m^2/s^2}\\\\v = 81.99m/s\\\\v = 82m/s[/tex]

Therefore, the velocity of the the chunk of ice as it hits the ground is 82m/s.

Learn more: https://brainly.com/question/24679384

Explanation:

We are expected to solve for the total momentum before collision

the expression for momentum is given as

p = mv

p =momentum

m=mass

v=velocity

For small compact truck

m1=1000kg

v1=15m/s

For small truck

m1=2000kg

v1=10m/s

the total momentum can also be expressed as

Ptotal=m1v1+m2v2

Ptotal=1000*15+2000*10

Ptotal=15000+20000

Ptotal=35000kg•m/s

The direction is east because the momentum of the truck is more than the car

Answer:

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.

Explanation:

Electric current is the circulation of electric charges in an electric circuit, while electric current intensity (I) is the amount of electricity or electric charge (Q) that circulates through a circuit in unit time (t).

Then, the intensity of electric current is expressed as:

[tex]I=\frac{Q}{t}[/tex]

Where:

In this case:

Replacing:

[tex]7.4 A=\frac{Q}{222 s}[/tex]

Solving:

Q= 7.4 A* 222 s

Q= 1,642.8 C

A Coulomb represents about 6.24*10¹⁸ electrons, so you can apply the following rule of three: if 1 C represents 6.24*10¹⁸ electrons, 222 C how many electrons does it represent?

[tex]electrons=\frac{222 C*6.24*10^{18} }{1C}[/tex]

electrons= 1.39*10²¹

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.

Answer:

C. 25 N

Explanation:

w = Weight of box = 80 N

[tex]\mu_k[/tex] = Coefficient of kinetic friction = 0.25

[tex]\mu_s[/tex] = Coefficient of static friction = 0.5

F = Pulling force = 25 N

Static friction

[tex]f_s=\mu_s w\\\Rightarrow f_s=0.5\times 80=40\ \text{N}[/tex]

[tex]F<f_s[/tex]

The force of pulling is less than the static frictional force so this makes the friction force equal to the force of pulling.

So, the friction force on the box is 25 N.

Answer:

Since velocity is considered a vector quantity and vectors show you in what direction it goes the object wont be moving forward but backwards because its a negative and negatives go to the left and that would be considered backwards.

Explanation:

Hope this helps

Answer:

Anything that has the most kinetic has the least potential energy.

Explanation:

NEED BRAINLYEST PLZZ!!!!!!!!!!!!!!!

Answer:

object B has a greater volume and by2 g/cm3

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

From the question we have

force = 0.5 × 50

We have the final answer as

Hope this helps you

Answer:

The minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m

Explanation:

Given;

velocity of the proton, v = 2 x 10⁸ m/s

uncertainty in the velocity, = 1 % = 0.01 v = 0.01 x 2 x 10⁸ m/s = 2 x 10⁶ m/s

The momentum of the proton is given by;

ΔP = mv = (1.67 x 10⁻²⁷)(2 x 10⁶ ) = 3.34 x 10⁻²¹ kgm/s

The minimum uncertainty in the position of the proton is given by;

[tex]\delta x = \frac{h}{4\pi \delta P}\\\\ \delta x =\frac{6.626*10^{-34}}{4\pi *3.34*10^{-21}}\\\\ \delta x = 1.578*10^{-14} \ m[/tex]

Therefore, the minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m

We know, Current is given by :

[tex]I = \dfrac{EA}{\rho}[/tex]

Here, [tex]\rho[/tex] is resistivity and for iron.

[tex]\rho=9.71\times 10^{-8}\ \Omega\ m[/tex]

Putting [tex]\rho[/tex] in above equation, we get :

[tex]I = \dfrac{0.062\times \pi\times (2.3\times 10^{-3})^2}{4\times 9.71\times 10^{-8}}[/tex]

[tex]I=2.65\ A[/tex]

Therefore, the electron current is 2.65 A.

Hence, this is the required solution.

Answer:

v = 11.6 m/s

Explanation:

       [tex]v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{((7.5m/s)^{2} + (-8.9m/s)^{2}[/tex]

       ⇒ v = 11.6 m/s

Answer:

Explanation:

For the mass m1;

The sum of forces acting on the body is expressed according to Newton's second law as;

\sum Fx = m1ax

T - Ff = m1a .... 1

T is the tension

Ff is the frictional force acting on m1

For the mass m2:

\sumFy = m2a

W - T = m2a

m2g - T = m2a.... 2

W is the weight

g is the acceleration due to gravity

If the acceleration of the system is 0, the equation becomes;

From 2:

m2g - T = m2a.... 2

a = m2g-T/m2

From 1:

a = T-Ff/m1

Equate both accelerations

m2g-T/m2 = T-Ff/m1

Cross multiply

m1m2g - Tm1 = m2T-m2Ff

m1m2g+m2Ff = m2T+m1T

m1m2g+m2Ff = T(m2+m1)

T = m1m2g+m2Ff/m2+m1

Hence the tension in strings is expressed as

T = m1m2g+m2Ff/m2+m1

Answer:

120 north west

Explanation:

Answer:

Explanation:

We know that the relationship for the impulse and momentum is given as

Ft=mv

given data

mass m=0.04kg

t= 26.5ms= 26.5/1000= 0.0265seconds

velocity v= 41.3m/s

substituting in the expression we have

Ft=mv

make F subject of formula we have

F=mv/t

F=(0.04*41.3)/0.0265

F=1.652/0.0265

F=62.33N

Answer:

Explanation:

The question is incomplete. Here is the complete question.

At a construction site, a crane lifts a steel beam with a mass of 330 kg.

i. What is the gravitational potential energy added to the steel beam when the beam reaches a height of 18.4 m? (1 point)

Gravitational potential energy GPE = mgh

m is the mass of the steel beam = 330kg

g is the acceleration due to gravity = 9.81m/s²

h is the height the beam reaches = 18.4m

Substitute

GPE = 330 * 9.81 * 18.4

GPE = 3237.3 * 18.4

GPE = 59,566.32Joules

Hence the gravitational potential energy is 59,566.32Joules

Answer:

The increase in gravitational potential energy is approximately 60,000 J.

Explanation:

GPE=mgh

m=330

g=9.8^2

h=18.4

GPE=330*9.81*18.4

GPE=59,566.32 J

Hope I could help! :)

Answer:

Initial speed of train = 35.76 m/s (80 mph)

               I figured out the speed of the train (which equals 80mph) by viewing the movie. In one of the scenes, it is shown that the train reaches 80mph after Doc Ock desroys the break on the train.

                   1mph = 0.44704mps

                    80(0.44704) = 35.7632

                                      = 35.76m/s

   vf = 0m/s

   vi = 35.76m/s

Mass of train = approximately 137,100 kg (302,254 lbs)  (found in research)

Time taken to stop train = approximately 46 seconds

a = vf - vi

       t

  = 0 - 35.76

         46

  = -.7773913043

  = -.78 m/s^2

(The number is negative since the accelerarion of the train is slowing down)

According to Newton's Second Law, a = f/m. When rearranged, the formula reads f = m(a). As shown in the free body diagram, the weight of the train equals the force due to gravity.

(force = tension)

F =  m(a)

  = 137,100(-.78)

  = -106938 N

106938 N is the calculated tension of Spider-Man's webs

I also calculated the distance it took for Spider-Man to stop the train:

Approximate distance traveled by train:

d = 1/2(a)(t)^2

  = 1/2(-.78)(46)^2

  = 825.24 m

In order for Spider-Man to have been able to completely stop the train before falling off the edge of the tracks, I found that the tension of his webs must have equaled 106938N.

In order to determine the tension of Spider-Man's webs, I needed to know the mass of the train. Since the movie was unclear on what type of trian was used, I researched online to find the mass of the train. Knowing the mass of the train, i was then able to calculate the tension of Spider-Man's webs and the distance the train took to stop.

Explanation:

Answer:

Did he wash the shoes?-Were the shoes washed by him?

Answer:

B. fast-twitch fibers can deliver a quick burst of power. Slow-twitch fibers can maintain a contraction for a longer time.

Answer:

the person above is right, just came for answers lol :')

Explanation:

Answer:

His de Broglie wavelength is 8.35×10⁻³⁷ m

The uncertainty in his position is 1.15 × 10⁻³⁵ m

Explanation:

First, Convert 227-lb to kg and convert the unit of the speed from mi/h to m/s.

To convert 227-lb to kg,

1-lb = 0.453592 kg

∴ 227-lb = 227 ×  0.453592 kg

227-lb = 102.97 kg

To convert 17.25 ± 0.10 mi/h to m/s

1 mi = 1609.34 m

and 1 h = 3600 s

Therefore,

17.25 mi/h = (17.25 ×1609.34)/3600 m/s = 7.71 m/s

and 0.10 mi/h =  (0.10 ×1609.34)/3600 m/s = 0.044704 m/s

Hence, the speed 17.25 ± 0.10 mi/h = 7.71 ± 0.044704 m/s

Now

(a) To determine the de Broglie wavelength,

De Broglie wavelength is given by

λ = h/mv

Where λ is the de Broglie wavelength

h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)

m is the mass

and v is the speed (velocity)

From the question

m = 102.97 kg

v = 7.71 m/s

Therefore,

λ = 6.626×10⁻³⁴ / (102.97×7.71)

λ = 8.35×10⁻³⁷ m

Hence, his de Broglie wavelength is 8.35×10⁻³⁷ m

(b) To calculate the uncertainty in his position

From

Δx = h/(4πmΔv)

Where Δx is the uncertainty in the position

h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)

π is a constant ( π = 3.14)

m is the mass

Δv is the uncertainty in speed

From the question

m = 102.97 kg

Δv = 0.044704 m/s

Hence,

Δx = 6.626×10⁻³⁴  / (4×3.14×102.97×0.044704)

Δx = 1.15 × 10⁻³⁵ m

Hence, the uncertainty in his position is 1.15 × 10⁻³⁵ m.

Answer:

The angular velocity of the merry-go-round is [tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex].

Explanation:

After a careful reading of the statement, we notice that merry-go-round-child system is a system with no external force exerting on it, such that the Principle of Angular Momentum Conservation can applied to analyze the system:

[tex]m_{C}\cdot \left(\frac{v}{R} \right) = (I+m_{C}\cdot R^{2})\cdot \omega[/tex] (Eq. 1)

Where:

[tex]m_{C}[/tex] - Mass of the child, measured in kilograms.

[tex]v[/tex] - Initial speed of the child, measured in meters per second.

[tex]R[/tex] - Radius of the playground merry-go-round, measured in meters.

[tex]I[/tex] - Moment of inertia, measured in kilograms per square meter.

[tex]\omega[/tex] - Angular velocity of the merry-go-round-child system, measured in radians per second.

Then, we clear the final angular speed:

[tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex]

The angular velocity of the merry-go-round is [tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex].

Answer:

The ball would fall vertically for 112.45m

Explanation:

The vertical height of a horizontal throw is affected only by the speed of the throw and the acceleration due to gravity.

At this point, we may use this formula to determine the vertical height of the throw

vertical height = [tex]-v^2/2g[/tex]

The initial throwing speed has to be converted to m/s to ensure uniformity during the calculations. To do this we multiply by 1000 and divide by 3600

169.1km/hr = 46.972m/s

Maximum height = [tex]46.972^2 /2 \times 9.81=112.45[/tex] m

The pitch would make the ball fall vertically for 112.45m by the time it reached the home plate 18.2 m away

Answer:

less in magnitude and opposite in direction to their original velocities

Explanation:

The nature of collisions can be determined by the coefficient of restitution.

Coefficient of restitution is given by

[tex]e=\dfrac{\text{Kinetic energy after collision}}{\text{Kinetic energy before collision}}[/tex]

In the case described here the value of [tex]e[/tex] is in between 0 and 1.

This means that the kinetic energy after collision is less than the kinetic energy before collision.

If the kinetic energy after collision is less this means that the velocity is not balanced with the velocity before the collision however the direction of the masses are opposite Some amount of energy is dissipated as heat or through other forms of energy.

So, just after the collision, their velocities are less in magnitude and opposite in direction to their original velocities.