If an object is moving with a constant velocity of 700 cm/s and starts at an initial position of 20 cm at t=0, what will be the position at t= 0.9 s?

The concept of the centrifugal force is used for producing an effective imitation of gravity. It is an important concept in the mechanics.

when any body is executing circular motion about a fixed axis, an outward force act on that object to balance the body and to execute the circular motion. The outward force is known as the centrifugal force.

It is the pseudo force act in order to balance the centripetal force acting inward during the circular motion.

The centrifugal force is comes into the picture when any body executes circular motion about a fixed axis.

To simulate gravity for the inhabitants, the concept of the centrifugal force is used.

Hence, the concept of the centrifugal force is used for producing an effective imitation of gravity.

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Answer:

0.5

Explanation:

I think it is this beacause if u have the answer you have that

GPE = mgh

GPE = 57 x 9.8 x 381 = 2.12 x 10⁵ J

Consider,

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =m\langle v(t)\rangle=m\int_{-\infty}^{\infty}x\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial t}\Psi (x,t)+\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial t}\bigg\}dx}[/tex]

Multiply both sides by ih and simplification will yield

[tex]{:\implies \quad \displaystyle \sf ih\langle p\rangle =m\int_{-\infty}^{\infty}x\bigg[\Psi (x,t)\bigg\{\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi^{*}(x,t)}{\partial x^2}-V(x)\Psi^{*}(x,t)\bigg\}+\Psi^{*}(x,t)\bigg\{V(x)\Psi (x,t)-\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}\bigg\}\bigg]dx}[/tex]

Some simplification, Then Integrate by parts and then knowing the fact that the wave function vanishes for [tex]{\bf x\to \pm \infty}[/tex] will yield:

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =\dfrac{ih}{2}\int_{-\infty}^{\infty}\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial x}\Psi (x,t)-\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}\bigg\}dx}[/tex]

Integrating by parts and knowing the same fact by some simplification will yield:

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =-ih\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}dx}[/tex]

The momentum is thus contained within the wave function, so we can then deduce that:

[tex]{:\implies \quad \sf p\rightarrow -ih\dfrac{\partial}{\partial x}}[/tex]

[tex]{:\implies \quad \sf p^{n}\rightarrow \bigg(-ih\dfrac{\partial}{\partial x}\bigg)^{n}}[/tex]

[tex]{:\implies \therefore \quad \displaystyle \sf \langle p^{2}\rangle =-h^{2}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]

Now the kinetic energy

[tex]{:\implies \quad \displaystyle \sf \langle K\rangle =\dfrac{\langle p^{2}\rangle}{2m}=\dfrac{-h^2}{2m}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]

The classical formula for the total energy

[tex]{:\implies \quad \sf \dfrac{p^2}{2m}+V(x)=E}[/tex]

Multiplying this equation by [tex]{\sf \Psi (x,t)=\psi (x)exp\bigg(\dfrac{-iEt}{h}\bigg)}[/tex] and use the above equations and simplify it we will be having

[tex]{:\implies \quad \boxed{\bf{\dfrac{-h^2}{2m}\dfrac{d^{2}\psi (x)}{dx^{2}}+V(x)\psi (x)=E\psi (x)}}}[/tex]

This is the Famous Time-Independent Schrödinger wave equation

Note:- If I write all the explanation then the Answer box willn't allow me to submit the answer

A 95 kg fullback was running downfield at 8 m/s. A defensive player (110 kg) was running upfield at 6 m/s straight at the fullback. The momentum of each player before they impacted Fullback 760 kgm/s, Defender: 660 kgm/s.

In classical mechanics, the integral of a force, F, over the time period over which it acts, t, is referred to as the "impulse." Impulse is a vector quantity because force is one as well. When an item receives an impulse, it experiences an equivalent vector change in its linear momentum and also in the direction that results.

momentum = mass *velocity

momentum = 95*8 = 760 kg-m/sec Fullback

momentum = 110*6 = 660 kg-m/sec Defender

A 95 kg fullback was running downfield at 8 m/s. A defensive player (110 kg) was running upfield at 6 m/s straight at the fullback. The momentum of each player before they impacted Fullback 760 kgm/s, Defender: 660 kgm/s.

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Answer:

M g - T =  M a     acceleration of hanging block   (M = 2)

2 g - T = 2 a        where M = 2

T = 2 g - 2 a

T = 8 * a + 12 N       acceleration of blocks on table

2 g - 2 a = 8 a + 12

10 a = 2 g - 12  

a = .8 m/s^2   (assuming g = 10 m/s^2)

T = 2 g - 2 a = 2 * 10 - 2 * .8 = 20 - 1.6 = 18.4 N

Check: For blocks on table

a = (18.4 - 12) / 8 = 6.4 / 8 = .8 m/s^2

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Answer:

The specific heat capacity is defined as the quantity of heat (J) absorbed per unit mass (kg) of the material when its temperature increases 1 K (or 1 °C), and its units are J/(kg K) or J/(kg °C).

Answer:

true

Explanation:

Answer:

8 m/s

Explanation:

(16-0)/2 = 8 m/s

Answer:

Concave lens

.............

If the masses of the objects are both doubled to 600 kg each, the force between them will be 40N.

The Newton's law of universal gravitation states that, force of gravity between two objects is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

As such;

F = Km1m2/r^2

If the masses of the objects are doubled but the distance of separation remains the same, it follows that the force between the masses will also be doubled.  Hence, if the masses of the objects are both doubled to 600 kg each, the force between them will be 40N.

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Answer:

a

0.57g

b

20.52

c

28.5

Explanation:

a

The bottle weighs 80g with tablets

If the bottle alone weighs 23g, the tablets weigh 57g

100 tablets weigh 57g, 1 tablet weighs 0.57g

b

0.57g is one tablet, so to achieve 36 tablets we must multiply by 36

0.57 multiplied by 36 is 20.52

c

To find 50 tablets we can use the same method we used before or a slightly faster method

100 tablets is 57g, so all we have to do is halve to find 50

57 divided by 2 is 28.5

Answer:

1. The faster the coil rotates the faster will it cut the magnetic field and the bigger will be the output voltage.

2. You could increase the output voltage of a generator by rotating it faster, increasing the number of turns on the coil or using stronger magnets,

Explanation:

Hopefully this helps if not im terribly sorry

1)The period of these ripples will be 0.2 sec.

2)The time period of sound wave will be 100 Hz.

3)The frequency of the sound wave will be 5×10⁶ Hz.

Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.

Frequency is given by the formula as,

Frequency of  ripples (f₁)=5 Hz

The period of these ripples will be;

T₁=1/f₁

T₁=1/5

T₁=0.2 sec

The time period of sound wave (T)=0.01 seconds

The time period of sound wave;

f=1/T

f=1/0.01

f=100 Hz

Time period of A radio wave = 0.0000002 seconds

The frequency of the sound wave is;

f=1/T

f=1/0.0000002

f=5×10⁶ Hz

Hence,the period of these ripples will be 0.2 sec.

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Answer:

Because it is the standard operational range for measuring the body temperature of a living individual.

Explanation:

When making a measuring equipment like a thermometer, you attempt to keep the reading range (Span) as small as possible so that the reading is as accurate as feasible. Because sensors are often rated in percent of span (in this example, span = 45 C–32 C = 13 degrees Celsius), One percent of 13 equals 0.13 degrees Celsius, thus if your sensor has a usual accuracy of 1%, your temperature reading is accurate to roughly 0.1 degrees Celsius.

Your precision would be 0.2 degrees Celsius if you doubled the span. The reading accuracy is now becoming unsatisfactory. Readability is another advantage of analog and mercury clinical thermometers. The markings for 0.1 deg C would be ten times closer together if a mercury thermometer had ten times the range (span). There is a limit to the device's readability.

Answer: reference point

Explanation: A reference point could be a put or question utilized for comparison to decide in case something is in movement.

The change in momentum of the object during the half period is -10 kgm/s.

The change in momentum of the object during the one period is 0.

The change in momentum of the object during the half period is calculated as follows;

ΔP(¹/₂) = Pf - Pi

where;

ΔP(¹/₂) = (-5 x ) - (5 x 1)  

ΔP(¹/₂) = -10 kgm/s

ΔP(1) = Pf - Pi

ΔP(1) = (5 x 1) - (5 x 1) = 0

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Answer:

12.65 Meters / Second (m/s)

Explanation:

the problem is really easy to solve, we begin by stating the kinetic energy formula which is:

KE = (1/2)(mass)(velocity)^2

We equate 1200 J to the formula, substitute the mass, and solve for velocity either using a CAS, app, or by hand

1200J = (0.5)(15kg)v^2

v = 12.65 m/s

8.3×10 to the power minus 7

Explanation:

lambda equals velocity over frequency.

Answer:

wait one second im about to go get some milk im thirsty

Explanation:

do you want any chicken on your salad

Answer:

Staying single makes it a whole lot easier to focus on yourself, gain self love, and work on yourself!

Explanation:

Given ,

Mass of the body (m) = 200g

Amplitude (A) = 20mm

Maximum force (F) = 0.6 N

Maximum velocity (vmax) =?

We know that

Fmax = mv2max/A

or , 0.6 = 200×10-3xv2max/20×10-3

or , 0.6×20/200= v2max

or , vmax = 0.24 m/s

∴ Maximum velocity = 0.24 m/s2

The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Mass of the body is,(m= 200g)

Amplitude is,(A  =20 mm)

Maximum force is,F= 0.6 N

To find;

Maximum velocity

Energy or work is equal to the product of force and displacement.

[tex]\rm KE = \frac{1}{2} mV_{max}^2 \\\\ \rm F_{max} \times A = \frac{1}{2} mV_{max}^2 \\\\ F_{max}= \frac{1}{2}\frac{ mV_{max}^2}{A} \\\\ 0.6= \frac{1}{2}\times \frac{ 0.2 \times V_{max}^2}{20 \times 10^{-3}} \\\\ V_{max}=0.3464 \ m/sec^2[/tex]

Hence,the value of maximum velocity will be 0.3464 m/s².

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Answer:

Mag isip ka po wag Kang tanong ng tanong kac kami po Yung nahihirapan mag hanap ng answer

Answer:

I think the letter B. But I'm not sure pero I try mo Lang po baka Yan po ang tamang sagot kac Yan po Yung nilagay na sagot saakin e Kaya try mo Lang po

A rifle recoils while firing a bullet. The speed of the rifle's recoil is small because the rifle has more mass than the bullet.

[tex]p_{1} =-p_2}[/tex]

[tex]m_{1} V_{1} =-M_{2} v_{2}[/tex]

Answer:

Explanation:

A rifle recoils while firing a bullet. The speed of the rifle's recoil is small  and typically fast because the rifle has more mass than the bullet.

Answer:

See below

Explanation:

The units of the specific heat hint at how to solve these types of problems

.88 J/(g-C)   * 10 g  *  (100- 50 C) = 440 J  

                                  ( see how the g and the C 'cancel out' ?)

Answer:

acts as a one-way switch for current