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2 years ago

50 POINTS!

Physics

1 answer:

zysi [14]2 years ago

7 0

The object is moving with constant or uniform acceleration and in average speed

The object is de accelerating

The object deaccelerated and came to rest so fast.

The object moves slowly first then accelerated.

The object accelerated at first so fast then move with constant acceleration then again accelerated .

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A train travels 55 kilometers in 2 hours, and then 52 kilometers in 5 hours. What is its average speed?

bulgar [2K]

The formula for working out speed is distance ÷ time.

55 km ÷ 2 hours = 27.5 km/h (average speed for first part of journey)

52km ÷ 5 hours = 10.4 km/h (average speed for second part of journey)

(27.5 + 10.4) ÷ 2 = 18.95 km/h (average speed throughout the journey)

4 0

2 years ago

Read 2 more answers

How much work is accplished when a force of 250 n pushes a box accross the floor for a distance of 50 meters

7nadin3 [17]

You just multiply these two numbers, it's 1250J

4 0

2 years ago

A trained eye in the dark for an extended period of time may pick up a light stimulus from a light source, at the lowest radiate

frez [133]

Answer:

#_photons = 30 photons / s

Explanation:

Let's start by finding the energy of a photon of light, let's use the Planck relation

E = h f

the speed of light is related to wavelength and frequency

c = λ f

we substitute

E = h c /λ

E₀ = 6.63 10⁻³⁴ 3 10⁸/500 10⁻⁹

E₀ = 3.978 10⁻¹⁹ J

now let's use a direct proportion rule. If the energy of a photon is Eo, how many fornes has an energy E = 1.2 10⁻¹⁷ J in a second

#_photons = 1 photon (E / Eo)

#_photons = 1 1.2 10⁻¹⁷ /3.978 10⁻¹⁹

#_photons = 3.0 10¹

#_photons = 30 photons / s

3 0

2 years ago

12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

$v_{f} = v_{o}+a\cdot (t-t_{o})$ (1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v_{f}$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t_{o}$ - Initial time, measured in seconds.

$t$ - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust ( $v_{o} = 0\,\frac{m}{s}$ , $a = 30\,\frac{m}{s^{2}}$ , $t_{o} = 0\,s$ , $0\,s\le t< 30\,s$ )

$v = 30\cdot t$ (2)

Free fall ( $v_{o} = 900\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s}$ , $t_{o} = 30\,s$ , $30\,s \le t \le 120\,s$ )

$v = 900-9.81\cdot (t-30)$ (3)

Now we created the graph speed-time, which can be seen below.

5 0

2 years ago

The motion of a particle along a straight line is described by the equation x=6+4t2 -t 4 , where x is in meter and t is in secon

Aleks [24]

Answer:

The position of the particle is 6m

The velocity of the particle is 16 m/s in negative direction

The acceleration of the object is -40 m/s²

Explanation:

Given;

motion of the particle along a straight line as x = 6 + 4t² - t⁴

The position of the object when t = 2s

x = 6 + 4(2)² - (2)⁴

x = 6 + 16 - 16

x = 6m

The velocity of the object when t = 2s

Velocity = dx/dt

dx/dt = 8t - 4t³

when t = 2s

Velocity = 8(2) - 4(2)³

Velocity = 16 - 32

Velocity = -16m/s

Velocity = 16 m/s (in negative direction)

The acceleration of the object when t = 2s

Acceleration = d²x/dt² = 8 - 12t²

Acceleration = 8 - 12 (2)²

Acceleration = -40 m/s²

5 0

3 years ago