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bixtya [17]
3 years ago
5

A ball is thrown horizontally from a 19 m -high building with a speed of 2.0 m/s . How far from the base of the building does th

e ball hit the ground?
Physics
1 answer:
bazaltina [42]3 years ago
3 0
Using the formula: ΔY = V₀y * t + (1/2) * ay * t²

Solve for time and get: 1.968s

Then use: v = d/t in the x-direction and get: d = 3.936
You might be interested in
P*V = n*R*T where P = pressure V = volume n = number of moles R = the universal gas constant T = temperature in degrees Kelvin T
seraphim [82]

The Gay-Lussac's law or Amonton's law states that the pressure of a given amount of a gas is directly propotional to its temperature if its volume is kept constant  .

            P∝T

and  

The Charles Law states that volume of given amount of gas at constant pressure  is directly propotional to temperature.

              V∝T

So, by Gay-Lussac's law if we increase the temperature the Pressure will increase and by Charles Law, if we increase the temperature the volume will increase.

Therefore, if the temperature of gas increases either the pressure of the gas, the volume of the gas, or both, will increase.

Hence,

Answer is option C

3 0
3 years ago
Read 2 more answers
What motivates the squire in lines 85- 90?
just olya [345]
<span>Answer: If you mean the Knight in the prologue, the man traveling with his son (the Squire) and a Yeoman, he is traveling to Canterbury to give thanks for his safe return from the wars in the Baltic. We're told that he has never been known to speak unkindly to anyone, a fact that sums up his chivalrous upbringing. Evidently he feels strongly motivated to live by a code of high standards and refined behavior.</span>
5 0
3 years ago
An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. At time t = 0 s, the object is at x0 = 3.9 m
solong [7]

Answer:

The object will move to Xfinal = 7.5m

Explanation:

By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:

Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s

With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:

Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =

= 3.9m + 3.6m = 7.5m

8 0
3 years ago
In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
4 years ago
The angle between incident ray and reflected ray is 130.what is the value of angle of incidence​
Artyom0805 [142]

Answer:

65

Explanation:

as i = r , so i + i = 130

so , i = 130/2 =65

7 0
3 years ago
Read 2 more answers
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