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2 years ago

How long would it take for a person to sprint from the 0m line to the 100m line if they are traveling forward at 6m/s?

Physics

1 answer:

valina [46]2 years ago

8 0

Answer:

<em><u>This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.</u></em>

Explanation:

<h3>Derive the kinematic equations for constant acceleration using integral calculus.</h3><h3>Use the integral formulation of the kinematic equations in analyzing motion.</h3><h3>Find the functional form of velocity versus time given the acceleration function.</h3><h3>Find the functional form of position versus time given the velocity function.</h3>

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State how much energy is transferred in each of the following cases: 2 grams of steam at 100 degrees Celsius condenses to water

dusya [7]

Answer:

Explanation:

When 2 gms of steam condenses to water at 100 degree latent heat of vaporization is releases which is calculated as follows

Heat released = mass x latent heat of vaporization

= 2 x 2260 = 4520 J

When 2 gms of water at 100 degree is cooled to ice water at zero degree heat is releases which is calculated as follows

Heat released = mass x specific heat x( 100-0)

= 2 x 4.2 x 100 = 840 J

When 2 gms of water at zero degree condenses to ice at zero degree latent heat of fusion is releases which is calculated as follows

Heat released = mass x latent heat of fusion

= 2 x 334 = 668 J

When 2 grams of steam at 100 degrees Celsius turns to ice at 0 degrees Celsius heat released will be sum of all the heat released as mentioned above ie

4520 + 840 +668 = 6028 J

3 0

3 years ago

Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 5.5 m/s2

borishaifa [10]

Answer:

160.75 N

Explanation:

The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.

As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.

When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:

F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N

5 0

3 years ago

Newtons laws of motion

Viktor [21]

Answer:

Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.

Law 2. A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force.

Law 3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.

7 0

2 years ago

What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction

Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

$f'=f(\dfrac{v+v_{p}}{v-v_{s}})$

Where, f = frequency

v = speed of sound

$v_{p}$ = speed of passenger

$v_{s}$ = speed of source

Put the value into the formula

$f'=262\times(\dfrac{344+18}{344-30})$

$f'=302.05\ Hz$

Hence, The frequency is 302.05 Hz.

7 0

3 years ago

Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25

mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp = 1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0

2 years ago