Một thang máy chuyển động thẳng đứng hướng xuống dưới chậm dần đều với gia tốc a= -4m/s2. Trần thang máy treo một vật nhỏ bằng một sợi dây mảnh, khoảng cách từ vật tới sàn thang máy h= 2m. Thang máy đang chuyển động thì dây đứt. Tính thời gian từ lúc dây đứt đến khi vật chạm sàn thang máy ? Lấy g= 10m/s2

Answer:

The answer is "The air will move faster on the narrow side".

Explanation:

The air on the top slows down in hypertensive. This is why light travels quicker on top. This results in air deflection downwards, required for its energy conservation to generate lift, and that is why air has to be moved quicker on the narrow side by the very same airflow per unit time as it departs.

Answer:

A) ΔV = 1.237 V

B) K.E = 1.237 eV

Explanation:

B)

The initial kinetic energy of the electron is given by the following formula:

[tex]K.E = \frac{1}{2}mv^2\\\\[/tex]

where,

K.E = Kinetic Energy of electron = ?

m = mass of elctron = 9.1 x 10⁻³¹ kg

v = speed of electron = 660000 m/s

Therefore,

[tex]K.E = \frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(660000\ m/s)^2[/tex]

K.E = 1.98 x 10⁻¹⁹ J

K.E = (1.98 x 10⁻¹⁹ J)([tex]\frac{1\ eV}{1.6\ x\ 10^{-19}\ J}[/tex])

K.E = 1.237 eV

A)

The energy applied by the potential difference must be equal to the kinetic energy of the electron, in order to stop it:

[tex]e\Delta V = K.E\\\\\Delta V = \frac{K.E}{e}[/tex]

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

Therefore,

[tex]\Delta V = \frac{1.98\ x\ 10^{-19}\ J}{1.6\ x\ 10^{-19}\ C}[/tex]

ΔV = 1.237 V

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Explanation:

The energy stored in a capacitor is given by

[tex]U = \dfrac{1}{2}QV = \dfrac{Q^2}{2C}[/tex]

In the case of a parallel plate capacitor, the capacitance C is given by

[tex]C = \dfrac{\epsilon_0A}{d}[/tex]

so we can rewrite the expression for the energy as

[tex]U = \dfrac{Q^2d}{2\epsilon_0 A}[/tex]

Increasing the charge from [tex]3\:\mu\text{C}\:\text{to}\:9\:\mu\text{C}[/tex] means that you're tripling the charge. The same thing is true when you increase the distance from 1.8 mm to 5.4 mm, i.e., you triple the separation distance. So the new energy U' is given by

[tex]U' = \dfrac{Q'^2d'}{2\epsilon_0 A}[/tex]

[tex]\:\:\:\:\:\:= \dfrac{(3Q)^2(3d)}{2\epsilon_0 A}[/tex]

[tex]\:\:\:\:\:\:= 27\left(\dfrac{Q^2d}{2\epsilon_0 A}\right)[/tex]

[tex]\:\:\:\:\:\:= 27U[/tex]

As we can see, tripling both the charge and the separation distance result in the 27-fold increase in its stored energy U.

Answer:

Total distance traveled by the sound wave = 2d . Thus we can conclude that the cliff should be at a distance of 877.15 metres from the source of sound to meet the required conditions

Explanation:

this is what i found on the web

Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]

Force:

[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]

Answer:

Waktu, t = 1.33 detik

Explanation:

Mengingat data berikut;

Kecepatan awal, Vo = 20 m/s

Kecepatan akhir, Vi = 80 m/s

Percepatan, a = 45 m/s²

Untuk menemukan interval perubahan kecepatan;

Dalam latihan ini, Anda diminta menghitung waktu untuk perubahan kecepatan.

Oleh karena itu, kita akan menggunakan persamaan gerak pertama;

Vi = Vo + at

Mengganti ke dalam rumus, kita memiliki;

80 = 20 + 45t

80 - 20 = 45t

45t = 60

Waktu, t = 60/45

Waktu, t = 1.33 detik

Answer:

a) Nonconservative Work

[tex]W_{disp} = 9\,J[/tex]

Final Gravitational Potential Energy

[tex]U_{f} = 0\,J[/tex]

Final Translational Energy

[tex]K_{f} = 35.131\,J[/tex]

b) Nonconservative Work

[tex]W_{disp} = 6.5\,J[/tex]

Final Gravitational Potential Energy

[tex]U_{f} = 12.259\,J[/tex]

Final Translational Energy

[tex]K_{f} = 25.373\,J[/tex]

c) Nonconservative Work

[tex]W_{disp} = 4\,J[/tex]

Final Gravitational Potential Energy

[tex]U_{f} = 24.518\,J[/tex]

Final Translational Energy

[tex]K_{f} = 15.614\,J[/tex]

Explanation:

The nonconservative work due to water resistance is defined by definition of work:

[tex]W_{disp} = F\cdot (y_{o}-y_{f})[/tex] (1)

Where:

[tex]W_{disp}[/tex] - Dissipate work, in joules.

[tex]F[/tex] - Resistance force, in newtons.

[tex]y_{o}[/tex] - Initial height, in meters.

[tex]y_{f}[/tex] - Final height, in meters.

The final gravitational potential energy ([tex]U_{f}[/tex]), in joules, is calculated by means of the definition of gravitational potential energy:

[tex]U_{f} = m\cdot g\cdot y_{f}[/tex] (2)

Where:

[tex]m[/tex] - Mass of the rock, in kilograms.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

The final translational kinetic energy ([tex]K_{f}[/tex]), in joules, is obtained by means of the Principle of Energy Conservation, Work-Energy Theorem and definitions of gravitational potential energy and translational kinetic energy:

[tex]m\cdot g\cdot y_{o} = U_{f} + K_{f} + W_{disp}[/tex] (3)

[tex]K_{f} = m\cdot g\cdot y_{o} - U_{f} - W_{disp}[/tex]

Lastly, the mechanical energy of the system ([tex]E[/tex]), in joules, is the sum of final gravitational potential energy, translational kinetic energy and dissipated work due to water resistance:

[tex]E = U_{f} + K_{f} + W_{disp}[/tex] (4)

Now we proceed to solve the exercise in each case:

a) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0\,m[/tex])

[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0\,m)[/tex]

[tex]W_{disp} = 9\,J[/tex]

Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0\,m[/tex])

[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0\,m)[/tex]

[tex]U_{f} = 0\,J[/tex]

Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 0\,J[/tex], [tex]W_{disp} = 9\,J[/tex])

[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -0\,J-9\,J[/tex]

[tex]K_{f} = 35.131\,J[/tex]

b) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0.50\,m[/tex])

[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0.5\,m)[/tex]

[tex]W_{disp} = 6.5\,J[/tex]

Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0.5\,m[/tex])

[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0.5\,m)[/tex]

[tex]U_{f} = 12.259\,J[/tex]

Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 12.259\,J[/tex], [tex]W_{disp} = 6.5\,J[/tex])

[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -12.259\,J-6.5\,J[/tex]

[tex]K_{f} = 25.373\,J[/tex]

c) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 1\,m[/tex])

[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 1\,m)[/tex]

[tex]W_{disp} = 4\,J[/tex]

Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 1\,m[/tex])

[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1\,m)[/tex]

[tex]U_{f} = 24.518\,J[/tex]

Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 24.518\,J[/tex], [tex]W_{disp} = 4\,J[/tex])

[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -24.518\,J-4\,J[/tex]

[tex]K_{f} = 15.614\,J[/tex]

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

Answer:

First remember that the distance between two points (a, b) and (c, d) is given by the equation:

[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]

Now let's define the position of the radar as:

(0mi, 0mi)

Then we can write the position of the plane as:

(480mi/h*t, 1mi)

where t is time in hours.

Then we can write the distance equation as:

[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]

Now we want to get:

the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.

So first we want to find the value of t such that:

d(3) = 3mi

We will look at the positive value of t, because at this point the plane is increasing its distance to the station.

[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]

The rate of change when the plane is 3 mi away from the station is:

d'(0.0059h)

remember that:

d'(t) = dd(t)/dt

We can write:

d(t) = h( g(t) )

such that:

h(x) = √x

g(t) = (480mi/h*t)^2 + (1mi)^2

then:

d'(t) = h'(g(t))*g'(t)

This is:

[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]

The rate of change at t = 0.0059h is then:

[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]

Answer:

True

Explanation:

This is because of the point where the forces are applied by our muscles and

the angle they have about the bones. Take for example the  diagram I uploaded.

If we do a free body diagram and a sum of torques, we would get that:

[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]

In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:

[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]

If we solve the equation for the force of the muscle we would get that:

[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]

Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.

Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:

[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]

which yields:

[tex]F_{muscle}=4.04 F_{hand}[/tex]

so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.

Answer:

Distance from sun, orbit and rotation. Presence of interior oceans, and elements that forms compressed water.

Explanation:

Answer:

Force = Mass × Acceleration

[tex]{ \tt{force = 2250 \times 4.3}} \\ = { \tt{9675 \: newtons}}[/tex]

Answer:

A current of 1ma flows through a copper wire, how many electron will pass a given point in one second? 1 Coulomb = 6.24 x 10^18 electrons (or protons)/1Sec which is also equal to 1 Amp/1 Sec. 1mA is 1/1000th of 1A so only 1/1000th of 6.24 x 10^18 electrons will pass a given point in 1 Sec.

Answer:

Resistance, R = 0.02 Ohms

Explanation:

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

Where;

V is the voltage or potential difference.

I is the current.

R is the resistance.

From the attachment, we would pick the following values on the graph of current against voltage;

Voltage, V = 0.5 V

Current = 25 A

To find resistance;

R = V/I

R = 0.5/2.5

Resistance, R = 0.02 Ohms

Note:

Resistance (R) is the inverse of slope i.e change in current with respect to change in voltage.

Answer:

b. approaches zero.

Explanation:

The phenomenon is known as length contraction.

Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.

let the length of the train = L

Let the length observed when the train is in motion = L₀

Apply Einstein's special theory of relativity;

[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]

from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)

As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L).  (L₀ < L)

Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero.  (L₀ = 0)

Thus, the length of this object, as measured by a stationary observer approaches zero

Answer:

?

Explanation:

i am sorry but there is no triangle

Explanation: edmentum sample answer

Answer:

Explanation:

F = ma so filling in:

60.0 = 4.50a and

a = 13.3 m/s/s

Answer:

α = 1930.2 rad/s²

Explanation:

The angular acceleration can be found by using the third equation of motion:

[tex]2\alpha \theta=\omega_f^2-\omega_i^2[/tex]

where,

α = angular acceleration = ?

θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s

Therefore,

[tex]2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}[/tex]

α = - 1930.2 rad/s²

negative sign shows deceleration

Explanation:

The sum of the voltages of the components connected in a series circuit is equal to the voltage across the battery.

[tex]V_T = V_1 + V_2 +V_3[/tex]

From Ohm's law ([tex]V=IR[/tex]) and in a series circuit, the amount of current flowing through the components is the same for all. So we can write [tex]V_T[/tex] as

[tex]V_T= 75\:\text{V} = I(170)+I(190) + 21\:\text{V}[/tex]

[tex]I(170+190)=54\:\text{V}[/tex]

[tex]I= \dfrac{54\:\text{V}}{360\:\text{ohms}}=0.15\:\text{A}[/tex]

Answer:

Explanation:

mass of the astronaut including the spacesuit, [tex]M=30[/tex]

distance of astronaut from the spaceship, d = 13 m

mass of the oxygen tank, m = 3 kg

Speed of tank with respect to spaceship, [tex]v=15~m/s[/tex]

a)

Using the conservation of linear momentum:

total momentum before collision = total momentum after collision

[tex]M.u=m.v+(M-m)v'[/tex]

[tex]0=3\times 15+(70-15)\times v'[/tex]

[tex]v'=0.82~m/s[/tex]

b)

She mush hold her breath until she reaches the spaceship, i.e.

[tex]t=d/v'[/tex]

[tex]t=13/0.82[/tex]

[tex]t=15.89~s[/tex]

Answer:

Ok, first, suppose that a given object is dropped or thrown down.

Then the acceleration of the object is the gravitational acceleration, given by:

a(t) = -9.8m/s^2

The velocity of the object can be integrated from that, it gives:

v(t) = (-9.8m/s^2)*t + V0

where V0 is the initial speed of the object, in the case that it is dropped,  V0 is equal to 0m/s

The position equation can be found integrating again:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t + p0

Where p0 is the initial position.

Now let's see our problem.

We know that in 8 frames, the flowerpot falls 0.84 of the height of the window, which is 1.27m

Then the distance that it falls is:

D = 0.84*1.27m = 1.07m

Now we also know that the camera captures at 30 fps

Then we have the relation:

30 frames = 1 second

1 = (1 second)/(30 frames)

With this, we can rewrite:

8 frames = 8 frames* (1 second)/(30 frames) = 0.267 seconds.

So now we know the distance that the pot fall, and the time in which it did fall.

Remember that the position equation for a falling object, is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t + p0

let's define p0 = 0m

Then the position equation of the pot is given by:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t

The distance that the pot would travel between t = 0s and t = t' is:

distance = p(t') - p(0s)

So, knowing that between t= 0s and t  = 0.267s, the pot did travel -1.07m (the negative sign is because it traveled downwards), we can find the initial speed of the pot.

-1.07m = p(0.267s) - p(0s)

-1.07m = (1/2)*(-9.8m/s^2)*(0.267s)^2 + V0*0.267s  

[-1.07m +  (1/2)*(9.8m/s^2)*(0.267s)^2]/0.267s = V0 = -2.7 m/s

This means that the pot was thrown downwards with an initial speed of 2.7 m/s

So we can conclude that the pot did not fall down on its own, someone threw it intentionally downwards.

This is the only thing that we can conclude with the given information.

Answer:

BFSkinner enfatizó la importancia de   creía en la importancia de desarrollar la psicología experimental y dejar atrás el psicoanálisis y las teorías acerca de la mente basadas en el simple sentido común.

Explanation:

The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.

An item or particle that is propelled in a gravitational influence, such as from the crust of the Ground, and moves along a curved route while solely being affected by gravity is said to be in projectile motion.

A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m.

The initial velocity is zero. Then the time taken to reach the ground is given as,

h = ut + 1/2at²

- 24 = 0×t + 1/2 (-9.8)t²

24 = 4.9t²

t² = 4.8979

t = 2.213 seconds

Then the horizontal speed is given as,

v = 18 / 2.213

v = 8.133 meters per second

The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.

Learn more about projectile motion:

https://brainly.com/question/29761109

#SPJ6

Answer:

The answer is "Option C".

Explanation:

It's evident from the figure below that after thirty minutes, not no more hydrogen can be created because all of the reactants have converted into products.

hydrogen gas created in cm cubes per period x = 20 seconds, y = 45 centimeters squared, and so on.

A reaction's terminus (the graph's flat line) indicates that no further products are being created during the reaction.

Answer:

initial velocity (u)=5m/s

final velocity (v)=?

acceleration (a)=3m/s^2

time (t)=4s

now,

acceleration (a)=v-u/t

3=v-5/4

3×4=v-5

12=v-5

12+5=v

17=v

v=17

Answer:

Option B

Explanation:

Generally

The forced response of a system is what the circuit does with the sources turned on, but with the initial conditions set to zero.

Therefore

Option B

Answer:

The cost is 297 cents.

Explanation:

Power of iron, P = 1140 W

Power of coffee pot, P' = 510 W

Voltage, V = 110 V

Time, t = 0.5 h each day

Cost = 12 cents per kWh

(a) Total energy

E = P x t + P' x t

E = 1140 x 0.5 x 60 x 60 + 510 x 0.5 x 60 x 60

E = 2052000 + 918000 = 2970000 J

1 kWh = 3.6 x 10^6 J

E = 0.825 kWh

For 30 days

E' = 0.825 x 30 = 24.75 kWh

So, the cost is

= 12 x 24.75 = 297 cents